problem solving on vectors

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Vector problems

Vector Problems

Here we will learn about more difficult vector problems, including vector routes involving midpoints, fractions and ratios of lengths. We will also look at parallel vectors.

There are also vector worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

What are vector problems?

Vector problems use vectors to solve a variety of different types of problems. Vectors have both a magnitude and direction and can be used to show a movement. A quantity which has just magnitude (size) is called a scalar.

For example,

We can write vectors in several ways,

• Using an arrow
• Using boldface  

Key facts for vector problems

In order to solve problems involving vectors it is helpful to use several key facts.

How to solve vector problems

In order to solve vector problems:

Write any information you know onto the diagram.

Decide the route.

Write the vector.

Simplify your answer.

Explain how to solve vector problems

Vector problems worksheet

Get your free vector problems worksheet of 20+ questions and answers. Includes reasoning and applied questions.

Related lessons on vectors

Vector problems is part of our series of lessons to support revision on vectors . You may find it helpful to start with the main vectors lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

• Magnitude of a vector  
• Column vector
• Vector notation
• Vector multiplication
• Vector addition
• Vector subtraction

Vector problem examples

Example 1: parallel lines.

The shape below is made from 8 equilateral triangles.

Find the vector \overrightarrow{QT} .

Parallel vectors of the same magnitude are the same. These triangles are all identical therefore we can label the corresponding vectors a and b .

2 Decide the route.

We need to find a route where we know the vectors.

3 Write the vector.

4 Simplify your answer.

Example 2: extended line

The line AB is extended to the point D so that the length AD is three times the length AB .

Find the vector \overrightarrow{AD} .

There is currently no further information we can add to the diagram.

We know AD is three times the length of AB . We need to find a route from A to B and then multiply it by three.

Example 3: midpoint

D is the midpoint of AC . Find the vector \overrightarrow{BD} .

When we are given information about midpoints, fractions of lines or ratios, it can be helpful to add this information onto the diagram. Here we are going to use the fraction  \frac{1}{2} to show that D is half way along the line.

We are trying to find a route from B to D . We know that \overrightarrow{BC}=\textbf{p} and we then need to get from C to D . We also know that CD=\frac{1}{2}CA .

We can find the vector \overrightarrow{CA} and then half it.

\begin{aligned} &\overrightarrow{CA}=-\textbf{p}+\textbf{q}\\\\ &\overrightarrow{CD}=\frac{1}{2}\overrightarrow{CA}=\frac{1}{2}(-\textbf{p}+\textbf{q}) \end{aligned}

Now we have a route from B to D .

Example 4: fraction of a line

Find the vector \overrightarrow{MN} .

When we are given information about midpoints, fractions of lines or ratios, it can be helpful to add this information onto the diagram. Here we know

\text{BM }=\frac{1}{4}\text{ BA}, \text{ AN} = \frac{1}{3} \text{ AC} .

We need to find a route from M to N . We can see that

\text{MA}=\frac{3}{4}\text{BA and AN}=\frac{1}{3}\text{AC} .

\overrightarrow{MA}=\frac{3}{4}\overrightarrow{BA}=\frac{3}{4}(-2\textbf{a})=-\frac{3}{2}\textbf{a}\\\overrightarrow{AN}=\frac{1}{3}\overrightarrow{AC}=\frac{1}{3}(6\textbf{b})=2\textbf{b}

We now have a route from M to N .

The answer here cannot be simplified.

Example 5: ratio

ED =2EH and the point J is such that GJ:JH = 2:1 .

Find the vector \overrightarrow{JD} .

When we are given information about midpoints, fractions of lines or ratios, it can be helpful to add this information onto the diagram. Here we know GJ:JH=2:1 . This means that  \text{GJ}=\frac{2}{3}\text{GH and JH}=\frac{1}{3}\text{GH} .

We also know that ED=2EH therefore HD is the same length as EH and in the same direction.

We know  \text{JH}=\frac{1}{3}\text{GH} .

We can find the vector  \overrightarrow{GH} .

\begin{aligned} &\overrightarrow{GH}=-\textbf{a}-\textbf{b}+2\textbf{a}=\textbf{a}-\textbf{b}\\\\ &\overrightarrow{JH}=\frac{1}{3}\overrightarrow{GH}=\frac{1}{3}(\textbf{a}-\textbf{b}) \end{aligned}

We now have a route from J to D .

Example 6: mix of information

E is the point on AB such that \text{EB }=\frac{1}{3}\text{ AB} .

D is the point on BC such that CD:DB=1:4 .

F is the midpoint of ED .

Find the vector \overrightarrow{EF} .

We want to find the the vector \overrightarrow{EF} .

We know that \text{EF}=\frac{1}{2}\text{ED} so we can start by finding the vector \overrightarrow{ED} .

\begin{aligned} &\overrightarrow{EB}=\frac{1}{3}\overrightarrow{AB}=\frac{1}{3}(6\textbf{a})=2\textbf{a}\\\\ &\overrightarrow{BD}=\frac{4}{5}\overrightarrow{BC}=\frac{4}{5}(10\textbf{b})=8\textbf{b} \end{aligned}

\overrightarrow{ED}=2\textbf{a}+8\textbf{b}

This answer cannot be simplified.

How to show two vectors are parallel

Two vectors are parallel if one is a multiple of the other. This is because if one vector is a multiple of another, it is a bigger or smaller version of the other.

In order to show two vectors are parallel:

Work out each vector.

Show that one is a multiple of the other by factorising.

Showing two vectors are parallel examples

Example 7: two lines are parallel.

Show that AB is parallel to CD .

We need to find the vector \overrightarrow{AB} .

\begin{aligned} &\overrightarrow{AB}=4\textbf{a}+\textbf{a}+2\textbf{b}-3\textbf{a}+2\textbf{b}\\\\ &\overrightarrow{AB}=2\textbf{a}+4\textbf{b} \end{aligned}

Since the vector  \overrightarrow{AB} is a multiple of the vector  \overrightarrow{CD} these vectors are parallel.

Example 8: two line segments form one straight line

ABCD is a parallelogram.

The line AB is extended to the point E such that \text{BE}=\frac{2}{3}\text{AB} .

The point F is on the line BC such that BF:FC=2:3 .

Show that DFE is a straight line.

We need to find the vectors \overrightarrow{DF} \text{ and } \overrightarrow{FE} .

\begin{aligned} &\overrightarrow{CF}=\frac{3}{5}\overrightarrow{CB}=3\textbf{b}\\\\ &\overrightarrow{FB}=\frac{2}{5}\overrightarrow{CB}=2\textbf{b} \end{aligned}

\begin{aligned} &\overrightarrow{DF}=6\textbf{a}+3\textbf{b}\\\\ &\overrightarrow{FE}=4\textbf{a}+2\textbf{b} \end{aligned}

Since \overrightarrow{DF} is a multiple of \overrightarrow{FE} the vectors are parallel. They also share the point F meaning they form a straight line.

Common misconceptions

• Using the wrong sign

Remember to make the vector negative when going backwards along it.

• Mistakes with ratios

If two parts of a line are in the ratio 1:4, this means one part is ⅕ of a line and the other part ⅘ .

Practice vector problem questions

1. ABCD is a parallelogram. The line AD is extended to the point E so that AE=3AD .

Find the vector  \overrightarrow{CE} .

2. \overrightarrow{PR}=4\textbf{a}, \overrightarrow{PQ}=3\textbf{b}

M is the midpoint of QR .

Find the vector \overrightarrow{RM} .

3. \overrightarrow{AB}=3\textbf{b}, \overrightarrow{BC}=2\textbf{a}

AD=2BC and the point E is on the line CD such that \text{ED}=\frac{1}{3}\text{CD} .

Find the vector \overrightarrow{AE} .

We know \text{DC}=\frac{1}{3}\text{DC} so we need to find the vector \overrightarrow{DC} .

4. \overrightarrow{AC}=9\textbf{a}, \overrightarrow{AB}=8\textbf{b}

The point N is such that \text{AN}=\frac{1}{3}\text{AC} , and the point P is such that AP:PB=1:3 . M is the midpoint of BN . Find the vector \overrightarrow{PM} .

We know that BM=\frac{1}{2}BN so we need to find the vector \overrightarrow{BN} .

5. \overrightarrow{AB}=2\textbf{a}+6\textbf{b}, \overrightarrow{AC}=3\textbf{a}+2\textbf{b}

D is the midpoint of BC and E is the point on AD such that AE:ED=1:3 .

We need to begin by finding the vector \overrightarrow{AD} .

6. \overrightarrow{AB}=4\textbf{a}, \overrightarrow{AE}=8\textbf{b} D is the midpoint of AE .

4\textbf{b}-4\textbf{a}=4(\textbf{b}-\textbf{a}) so \overrightarrow{CE} is parallel to \textbf{b}-\textbf{a} .

Vector problems GCSE questions

1. \overrightarrow{AB}=6\textbf{b},\overrightarrow{BC}=9\textbf{a},\overrightarrow{DF}=-3\textbf{a}+3\textbf{b}

\overrightarrow{BC} =3\overrightarrow{AD} – they are multiples of each other therefore parallel.

2. \overrightarrow{AD}=12\textbf{a}, \overrightarrow{CD}=8\textbf{b}

E is the midpoint of AB .

3. \overrightarrow{AB}=4\textbf{a}, \overrightarrow{BC}=2\textbf{b}

D is the midpoint of AB.

Point E is such that AE:EC=3:1 .

Determine whether DEF is a straight line.

DEF is not a straight line since \overrightarrow{DE} and \overrightarrow{EF} and are not multiples of each other and therefore are not parallel.

Learning checklist

You have now learned how to:

• Solve complex problems involving vectors

The next lessons are

• Loci and construction
• Transformations
• Circle theorems

Still stuck?

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Privacy Overview

This is a vector:

A vector has magnitude (size) and direction :

The length of the line shows its magnitude and the arrowhead points in the direction.

Play with one here:

We can add two vectors by joining them head-to-tail:

And it doesn't matter which order we add them, we get the same result:

Example: A plane is flying along, pointing North, but there is a wind coming from the North-West.

The two vectors (the velocity caused by the propeller, and the velocity of the wind) result in a slightly slower ground speed heading a little East of North.

If you watched the plane from the ground it would seem to be slipping sideways a little.

Have you ever seen that happen? Maybe you have seen birds struggling against a strong wind that seem to fly sideways. Vectors help explain that.

Velocity , acceleration , force and many other things are vectors.

Subtracting

We can also subtract one vector from another:

• first we reverse the direction of the vector we want to subtract,
• then add them as usual:

A vector is often written in bold , like a or b .

Calculations

Now ... how do we do the calculations?

The most common way is to first break up vectors into x and y parts, like this:

The vector a is broken up into the two vectors a x and a y

(We see later how to do this.)

Adding Vectors

We can then add vectors by adding the x parts and adding the y parts :

The vector (8, 13) and the vector (26, 7) add up to the vector (34, 20)

Example: add the vectors a = (8, 13) and b = (26, 7)

c = (8, 13) + (26, 7) = (8+26, 13+7) = (34, 20)

When we break up a vector like that, each part is called a component :

Subtracting Vectors

To subtract, first reverse the vector we want to subtract, then add.

Example: subtract k = (4, 5) from v = (12, 2)

a = v + − k

a = (12, 2) + −(4, 5) = (12, 2) + (−4, −5) = (12−4, 2−5) = (8, −3)

Magnitude of a Vector

The magnitude of a vector is shown by two vertical bars on either side of the vector:

OR it can be written with double vertical bars (so as not to confuse it with absolute value):

We use Pythagoras' theorem to calculate it:

| a | = √( x 2 + y 2 )

Example: what is the magnitude of the vector b = (6, 8) ?

| b | = √( 6 2 + 8 2 ) = √(36+64) = √100 = 10

A vector with magnitude 1 is called a Unit Vector .

Vector vs Scalar

A scalar has magnitude (size) only .

Scalar: just a number (like 7 or −0.32) ... definitely not a vector.

A vector has magnitude and direction , and is often written in bold , so we know it is not a scalar:

• so c is a vector, it has magnitude and direction
• but c is just a value, like 3 or 12.4

Example: k b is actually the scalar k times the vector b .

Multiplying a vector by a scalar.

When we multiply a vector by a scalar it is called "scaling" a vector, because we change how big or small the vector is.

Example: multiply the vector m = (7, 3) by the scalar 3

It still points in the same direction, but is 3 times longer

(And now you know why numbers are called "scalars", because they "scale" the vector up or down.)

Multiplying a Vector by a Vector (Dot Product and Cross Product)

How do we multiply two vectors together? There is more than one way!

• The scalar or Dot Product (the result is a scalar).
• The vector or Cross Product (the result is a vector).

(Read those pages for more details.)

More Than 2 Dimensions

Vectors also work perfectly well in 3 or more dimensions:

Example: add the vectors a = (3, 7, 4) and b = (2, 9, 11)

c = (3, 7, 4) + (2, 9, 11) = (3+2, 7+9, 4+11) = (5, 16, 15)

Example: what is the magnitude of the vector w = (1, −2, 3) ?

| w | = √( 1 2 + (−2) 2 + 3 2 ) = √(1+4+9) = √14

Here is an example with 4 dimensions (but it is hard to draw!):

Example: subtract (1, 2, 3, 4) from (3, 3, 3, 3)

(3, 3, 3, 3) + −(1, 2, 3, 4) = (3, 3, 3, 3) + (−1,−2,−3,−4) = (3−1, 3−2, 3−3, 3−4) = (2, 1, 0, −1)

Magnitude and Direction

We may know a vector's magnitude and direction, but want its x and y lengths (or vice versa):

You can read how to convert them at Polar and Cartesian Coordinates , but here is a quick summary:

Sam and Alex are pulling a box.

• Sam pulls with 200 Newtons of force at 60°
• Alex pulls with 120 Newtons of force at 45° as shown

What is the combined force , and its direction?

Let us add the two vectors head to tail:

First convert from polar to Cartesian (to 2 decimals):

Sam's Vector:

• x = r × cos( θ ) = 200 × cos(60°) = 200 × 0.5 = 100
• y = r × sin( θ ) = 200 × sin(60°) = 200 × 0.8660 = 173.21

Alex's Vector:

• x = r × cos( θ ) = 120 × cos(−45°) = 120 × 0.7071 = 84.85
• y = r × sin( θ ) = 120 × sin(−45°) = 120 × -0.7071 = −84.85

Now we have:

(100,173.21) + (84.85, −84.85) = (184.85, 88.36)

That answer is valid, but let's convert back to polar as the question was in polar:

• r = √ ( x 2 + y 2 ) = √ ( 184.85 2 + 88.36 2 ) = 204.88
• θ = tan -1 ( y / x ) = tan -1 ( 88.36 / 184.85 ) = 25.5°

They might get a better result if they were shoulder-to-shoulder!

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Solving Problems with Vectors

We can use vectors to solve many problems involving physical quantities such as velocity, speed, weight, work and so on.

The velocity of moving object is modeled by a vector whose direction is the direction of motion and whose magnitude is the speed.

A ball is thrown with an initial velocity of 70 feet per second., at an angle of 35 ° with the horizontal. Find the vertical and horizontal components of the velocity.

Let v represent the velocity and use the given information to write v in unit vector form:

v   = 70 ( cos ( 35 ° ) ) i + 70 ( sin ( 35 ° ) ) j

Simplify the scalars, we get:

v   ≈ 57.34 i + 40.15 j

Since the scalars are the horizontal and vertical components of v ,

Therefore, the horizontal component is 57.34 feet per second and the vertical component is 40.15 feet per second.

Force is also represented by vector. If several forces are acting on an object, the resultant force experienced by the object is the vector sum of these forces.

Two forces F 1 and F 2 with magnitudes 20 and 30   lb , respectively, act on an object at a point P as shown. Find the resultant forces acting at P .

First we write F 1 and F 2 in component form:

v ≈ 57.34 i + 40.15 j

F 1 = ( 20 cos ( 45 ° ) ) i + ( 20 sin ( 45 ° ) ) j = 20 ( 2 2 ) i + 20 ( 2 2 ) j = 10 2 i + 10 2 j F 2 = ( 30 cos ( 150 ° ) ) i + ( 30 sin ( 150 ° ) ) j = 30 ( − 3 2 ) i + 30 ( 1 2 ) j = − 15 3 i + 15 j

So, the resultant force F is

F = F 1 + F 2 = ( 10 2   i + 10 2 j ) + ( − 15 3   i + 15 j ) = ( 10 2 − 15 3 ) i + ( 10 2 + 15 ) j ≈ − 12 i + 29 j

A force is given by the vector F = ⟨ 2 , 3 ⟩ and moves an object from the point ( 1 , 3 ) to the point ( 5 , 9 ) . Find the work done.

First we find the Displacement.

The displacement vector is

D = ⟨ 5 − 1 , 9 − 3 ⟩ = ⟨ 4 , 6 ⟩ .

By using the formula, the work done is

W = F ⋅ D = ⟨ 2 , 3 ⟩ ⋅ ⟨ 4 , 6 ⟩ = 26

If the unit of force is pounds and the distance is measured in feet, then the work done is 26 ft-lb.

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10.2: An Introduction to Vectors

• Last updated
• Save as PDF
• Page ID 4214

• Gregory Hartman et al.
• Virginia Military Institute

Many quantities we think about daily can be described by a single number: temperature, speed, cost, weight and height. There are also many other concepts we encounter daily that cannot be described with just one number. For instance, a weather forecaster often describes wind with its speed and its direction ("\(\ldots\) with winds from the southeast gusting up to 30 mph \(\ldots\)"). When applying a force, we are concerned with both the magnitude and direction of that force. In both of these examples, direction is important. Because of this, we study vectors , mathematical objects that convey both magnitude and direction information.

One "bare--bones'' definition of a vector is based on what we wrote above: "a vector is a mathematical object with magnitude and direction parameters.'' This definition leaves much to be desired, as it gives no indication as to how such an object is to be used. Several other definitions exist; we choose here a definition rooted in a geometric visualization of vectors. It is very simplistic but readily permits further investigation.

Definition 51 Vector

A vector is a directed line segment.

Given points \(P\) and \(Q\) (either in the plane or in space), we denote with \(\vec{PQ}\) the vector from \(P\) to \(Q\). The point \(P\) is said to be the initial point of the vector, and the point \(Q\) is the terminal point .

The magnitude , length or norm of \(\vec{PQ}\) is the length of the line segment \(\overline{PQ}\):

\[\norm{\vec{PQ}} = \norm{\overline{PQ}}.\]

Two vectors are equal if they have the same magnitude and direction.

Figure 10.18 shows multiple instances of the same vector. Each directed line segment has the same direction and length (magnitude), hence each is the same vector.

We use \(\mathbb{R}^2\) (pronounced "r two'') to represent all the vectors in the plane, and use \(\mathbb{R}^3\) (pronounced "r three'') to represent all the vectors in space.

Consider the vectors \(\vec{PQ}\) and \(\vec{RS}\) as shown in Figure 10.19. The vectors look to be equal; that is, they seem to have the same length and direction. Indeed, they are. Both vectors move 2 units to the right and 1 unit up from the initial point to reach the terminal point. One can analyze this movement to measure the magnitude of the vector, and the movement itself gives direction information (one could also measure the slope of the line passing through \(P\) and \(Q\) or \(R\) and \(S\)). Since they have the same length and direction, these two vectors are equal.

This demonstrates that inherently all we care about is displacement ; that is, how far in the \(x\), \(y\) and possibly \(z\) directions the terminal point is from the initial point. Both the vectors \(\vec{PQ}\) and \(\vec{RS}\) in Figure 10.19 have an \(x\)-displacement of 2 and a \(y\)-displacement of 1. This suggests a standard way of describing vectors in the plane. A vector whose \(x\)-displacement is \(a\) and whose \(y\)-displacement is \(b\) will have terminal point \((a,b)\) when the initial point is the origin, \((0,0)\). This leads us to a definition of a standard and concise way of referring to vectors.

Definition 52 Component Form of a Vector

• The component form of a vector \(\vec{v}\) in \(\mathbb{R}^2\), whose terminal point is \((a,\,b)\) when its initial point is \((0,\,0)\), is \(\langle a,b\rangle.\)
• The component form of a vector \(\vec{v}\) in \(\mathbb{R}^3\), whose terminal point is \((a,\,b,\,c)\) when its initial point is \((0,\,0,\,0)\), is \(\langle a,b,c\rangle.\)

The numbers \(a\), \(b\) (and \(c\), respectively) are the components of \(\vec v\).

It follows from the definition that the component form of the vector \(\vec{PQ}\), where \(P=(x_1,y_1)\) and \(Q=(x_2,y_2)\) is

\[\vec{PQ} = \langle x_2-x_1, y_2-y_1\rangle;\]

in space, where \(P=(x_1,y_1,z_1)\) and \(Q=(x_2,y_2,z_2)\), the component form of \(\vec{PQ}\) is

\[\vec{PQ} = \langle x_2-x_1, y_2-y_1,z_2-z_1\rangle.\]

We practice using this notation in the following example.

Example \(\PageIndex{1}\): Using component form notation for vectors

• Sketch the vector \(\vec v=\langle 2,-1\rangle\) starting at \(P=(3,2)\) and find its magnitude.
• Find the component form of the vector \(\vec w\) whose initial point is \(R=(-3,-2)\) and whose terminal point is \(S=(-1,2)\).
• Sketch the vector \(\vec u = \langle 2,-1,3\rangle\) starting at the point \(Q = (1,1,1)\) and find its magnitude.
• Using \(P\) as the initial point, we move 2 units in the positive \(x\)-direction and \(-1\) units in the positive \(y\)-direction to arrive at the terminal point \(P\,'=(5,1)\), as drawn in Figure 10.20(a). The magnitude of \(\vec v\) is determined directly from the component form: \[\norm{\vec v} =\sqrt{2^2+(-1)^2} = \sqrt{5}. \]
• Using the note following Definition 52, we have \[\vec{RS} = \langle -1-(-3), 2-(-2)\rangle = \langle 2,4\rangle.\] One can readily see from Figure 10.20(a) that the \(x\)- and \(y\)-displacement of \(\vec{RS}\) is 2 and 4, respectively, as the component form suggests.
• Using \(Q\) as the initial point, we move 2 units in the positive \(x\)-direction, \(-1\) unit in the positive \(y\)-direction, and 3 units in the positive \(z\)-direction to arrive at the terminal point \(Q' = (3,0,4)\), illustrated in Figure 10.20(b). The magnitude of \(\vec u\) is: \[\norm{\vec u} = \sqrt{2^2+(-1)^2+3^2} = \sqrt{14}.\]

Now that we have defined vectors, and have created a nice notation by which to describe them, we start considering how vectors interact with each other. That is, we define an algebra on vectors.

Definition 53 VECTOR ALGEBRA

• Let \(\vec u = \langle u_1,u_2\rangle\) and \(\vec v = \langle v_1,v_2\rangle\) be vectors in \(\mathbb{R}^2\), and let \(c\) be a scalar. (a) The addition, or sum, of the vectors \(\vec u\) and \(\vec v\) is the vector \[\vec u+\vec v = \langle u_1+v_1, u_2+v_2\rangle.\] (b) The scalar product of \(c\) and \(\vec v\) is the vector \[c\vec v = c\langle v_1,v_2\rangle = \langle cv_1,cv_2\rangle.\]
• Let \(\vec u = \langle u_1,u_2,u_3\rangle\) and \(\vec v = \langle v_1,v_2,v_3\rangle\) be vectors in \(\mathbb{R}^3\), and let \(c\) be a scalar. (a) The addition, or sum, of the vectors \(\vec u\) and \(\vec v\) is the vector \[\vec u+\vec v = \langle u_1+v_1, u_2+v_2, u_3+v_3\rangle.\] (b) The scalar product of \(c\) and \(\vec v\) is the vector \[c\vec v = c\langle v_1,v_2,v_3\rangle = \langle cv_1,cv_2,cv_3\rangle.\]

In short, we say addition and scalar multiplication are computed "component--wise.''

Example \(\PageIndex{2}\): Adding vectors

Sketch the vectors \(\vec u = \langle1,3\rangle\), \(\vec v = \langle 2,1\rangle\) and \(\vec u+\vec v\) all with initial point at the origin. Solution

We first compute \(\vec u +\vec v\).

\[\begin{align*} \vec u+\vec v &= \langle 1,3\rangle + \langle 2,1\rangle\\ &= \langle 3,4\rangle. \end{align*}\] These are all sketched in Figure 10.21.

As vectors convey magnitude and direction information, the sum of vectors also convey length and magnitude information. Adding \(\vec u+\vec v\) suggests the following idea:

\[\text{"Starting at an initial point, go out \(\vec u\), then go out \(\vec v\)."}\]

This idea is sketched in Figure 10.22, where the initial point of \(\vec v\) is the terminal point of \(\vec u\). This is known as the "Head to Tail Rule'' of adding vectors. Vector addition is very important. For instance, if the vectors \(\vec u\) and \(\vec v\) represent forces acting on a body, the sum \(\vec u+\vec v\) gives the resulting force. Because of various physical applications of vector addition, the sum \(\vec u+\vec v\) is often referred to as the resultant vector , or just the "resultant.''

Analytically, it is easy to see that \(\vec u+\vec v = \vec v+\vec u\). Figure 10.22 also gives a graphical representation of this, using gray vectors. Note that the vectors \(\vec u\) and \(\vec v\), when arranged as in the figure, form a parallelogram. Because of this, the Head to Tail Rule is also known as the Parallelogram Law: the vector \(\vec u+\vec v\) is defined by forming the parallelogram defined by the vectors \(\vec u\) and \(\vec v\); the initial point of \(\vec u+\vec v\) is the common initial point of parallelogram, and the terminal point of the sum is the common terminal point of the parallelogram.

While not illustrated here, the Head to Tail Rule and Parallelogram Law hold for vectors in \(\mathbb{R}^3\) as well.

It follows from the properties of the real numbers and Definition 53 that \[\vec u-\vec v = \vec u + (-1)\vec v.\] The Parallelogram Law gives us a good way to visualize this subtraction. We demonstrate this in the following example.

Example \(\PageIndex{3}\): Vector Subtraction

Let \(\vec u = \langle 3,1\rangle\) and \(\vec v=\langle 1,2\rangle.\) Compute and sketch \(\vec u-\vec v\).

The computation of \(\vec u-\vec v\) is straightforward, and we show all steps below. Usually the formal step of multiplying by \((-1)\) is omitted and we "just subtract.''

\[\begin{align*} \vec u-\vec v &= \vec u + (-1)\vec v \\ &= \langle 3,1\rangle + \langle -1,-2\rangle\\ &= \langle 2,-1\rangle. \end{align*}\]

Figure 10.23 illustrates, using the Head to Tail Rule, how the subtraction can be viewed as the sum \(\vec u + (-\vec v)\). The figure also illustrates how \(\vec u-\vec v\) can be obtained by looking only at the terminal points of \(\vec u\) and \(\vec v\) (when their initial points are the same).

Example \(\PageIndex{4}\): Scaling vectors

• Sketch the vectors \(\vec v = \langle 2,1\rangle\) and \(2\vec v\) with initial point at the origin.
• Compute the magnitudes of \(\vec v\) and \(2\vec v\).
• We compute \(2\vec v\): \[\begin{align*}2\vec v &= 2\langle 2,1\rangle\\&= \langle 4,2\rangle.\end{align*}\] Both \(\vec v\) and \(2\vec v\) are sketched in Figure10.24. Make note that \(2\vec v\) does not start at the terminal point of \(\vec v\); rather, its initial point is also the origin.
• The figure suggests that \(2\vec v\) is twice as long as \(\vec v\). We compute their magnitudes to confirm this. \[\begin{align*}\norm{\vec v} &= \sqrt{2^2+1^2}\\&= \sqrt{5}.\\\norm{2\vec v}&=\sqrt{4^2+2^2} \\&= \sqrt{20}\\&= \sqrt{4\cdot 5} = 2\sqrt{5}.\end{align*}\] As we suspected, \(2\vec v\) is twice as long as \(\vec v\).

The zero vector is the vector whose initial point is also its terminal point. It is denoted by \(\vec 0\). Its component form, in \(\mathbb{R}^2\), is \(\langle 0,0\rangle\); in \(\mathbb{R}^3\), it is \(\langle 0,0,0\rangle\). Usually the context makes is clear whether \(\vec 0\) is referring to a vector in the plane or in space.

Our examples have illustrated key principles in vector algebra: how to add and subtract vectors and how to multiply vectors by a scalar. The following theorem states formally the properties of these operations.

THEOREM 84 PROPERTIES OF VECTOR OPERATIONS

The following are true for all scalars \(c\) and \(d\), and for all vectors \(\vec u\), \(\vec v\) and \(\vec w\), where \(\vec u\), \(\vec v\) and \(\vec w\) are all in \(\mathbb{R}^2\) or where \(\vec u\), \(\vec v\) and \(\vec w\) are all in \(\mathbb{R}^3\):

• \(\underbrace{\vec u+\vec v = \vec v+\vec u}_{Commutative Property}\)
• \(\underbrace{\vec u+\vec v)+\vec w = \vec u+(\vec v+\vec w)}_{Associative Property}\)
• \(\underbrace{\vec v+\vec 0 = \vec v}_{Additive Identity}\)
• \((cd)\vec v= c(d\vec v)\)
• \(\underbrace{c(\vec u+\vec v) = c\vec u+c\vec v}_{Distributive Property}\)
• \(\underbrace{(c+d)\vec v = c\vec v+d\vec v}_{Distributive Property}\)
• \(0\vec v = \vec 0\)
• \(\norm{c\vec v} = |c|\cdot\norm{\vec v}\)
• \(\norm u = 0\) if, and only if, \(\vec u = \vec 0\).

As stated before, each vector \(\vec v\) conveys magnitude and direction information. We have a method of extracting the magnitude, which we write as \(\norm{\vec v}\). Unit vectors are a way of extracting just the direction information from a vector.

Definition 54 Unit Vector

A unit vector is a vector \(\vec v\) with a magnitude of 1; that is,

\[\norm{\vec v}=1.\]

Consider this scenario: you are given a vector \(\vec v\) and are told to create a vector of length 10 in the direction of \(\vec v\). How does one do that? If we knew that \(\vec u\) was the unit vector in the direction of \(\vec v\), the answer would be easy: \(10\vec u\). So how do we find \(\vec u\)?

Property 8 of Theorem 84 holds the key. If we divide \(\vec v\) by its magnitude, it becomes a vector of length 1. Consider:

\[\begin{align*} \Big\| \frac{1}{\norm{\vec v}}\vec v\Big\| &= \frac{1}{\norm{\vec v}}\norm{\vec v} & \text{ (we can pull out \( \frac{1}{\norm{\vec v}}\) as it is a scalar)}\\ &= 1. \end{align*} \]

So the vector of length 10 in the direction of \(\vec v\) is \( 10\frac{1}{\norm{\vec v}}\vec v.\) An example will make this more clear.

Example \(\PageIndex{5}\): Using Unit Vectors

Let \(\vec v= \langle 3,1\rangle\) and let \(\vec w = \langle 1,2,2\rangle\).

• Find the unit vector in the direction of \(\vec v\).
• Find the unit vector in the direction of \(\vec w\).
• Find the vector in the direction of \(\vec v\) with magnitude 5.
• We find \(\norm{\vec v} = \sqrt{10}\). So the unit vector \(\vec u\) in the direction of \(\vec v\) is \[\vec u = \frac{1}{\sqrt{10}}\vec v = \langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\rangle.\]
• We find \(\norm{\vec w} = 3\), so the unit vector \(\vec z\) in the direction of \(\vec w\) is \[\vec u = \frac13\vec w = \langle \frac13,\frac23,\frac23\rangle.\]
• To create a vector with magnitude 5 in the direction of \(\vec v\), we multiply the unit vector \(\vec u\) by 5. Thus \(5\vec u = \langle 15/\sqrt{10},5/\sqrt{10}\rangle\) is the vector we seek. This is sketched in Figure 10.25.

The basic formation of the unit vector \(\vec u\) in the direction of a vector \(\vec v\) leads to a interesting equation. It is: \[\vec v = \norm{\vec v}\frac{1}{\norm{\vec v}}\vec v.\] We rewrite the equation with parentheses to make a point:

This equation illustrates the fact that a vector has both magnitude and direction, where we view a unit vector as supplying only direction information. Identifying unit vectors with direction allows us to define parallel vectors .

Definition 55 Parallel Vectors

• Unit vectors \(\vec u_1\) and \(\vec u_2\) are parallel if \(\vec u_1 = \pm \vec u_2\).
• Nonzero vectors \(\vec v_1\) and \(\vec v_2\) are parallel if their respective unit vectors are parallel.

It is equivalent to say that vectors \(\vec v_1\) and \(\vec v_2\) are parallel if there is a scalar \(c\neq 0\) such that \(\vec v_1 = c\vec v_2\) (see marginal note).

Note : \(\vec 0\) is directionless; because \(\norm{0}=0\), there is no unit vector in the "direction'' of \(\vec 0\).

Some texts define two vectors as being parallel if one is a scalar multiple of the other. By this definition, \(\vec 0\) is parallel to all vectors as \(\vec 0 = 0\vec v\) for all \(\vec v\).

We prefer the given definition of parallel as it is grounded in the fact that unit vectors provide direction information. One may adopt the convention that \(\vec 0\) is parallel to all vectors if they desire.

If one graphed all unit vectors in \(\mathbb{R}^2\) with the initial point at the origin, then the terminal points would all lie on the unit circle. Based on what we know from trigonometry, we can then say that the component form of all unit vectors in \(\mathbb{R}^2\) is \(\langle \cos\theta,\sin\theta\rangle\) for some angle \(\theta\).

A similar construction in \(\mathbb{R}^3\) shows that the terminal points all lie on the unit sphere. These vectors also have a particular component form, but its derivation is not as straightforward as the one for unit vectors in \(\mathbb{R}^2\). Important concepts about unit vectors are given in the following Key Idea.

KEY IDEA 48 UNIT VECTORS

• The unit vector in the direction of \(\vec v\) is \[ \vec u = \frac1{\norm{\vec v}} \vec v.\]
• A vector \(\vec u\) in \(\mathbb{R}^2\) is a unit vector if, and only if, its component form is \(\langle \cos\theta,\sin\theta\rangle\) for some angle \(\theta\).
• A vector \(\vec u\) in \(\mathbb{R}^3\) is a unit vector if, and only if, its component form is \(\langle \sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta\rangle\) for some angles \(\theta\) and \(\varphi\).

These formulas can come in handy in a variety of situations, especially the formula for unit vectors in the plane.

Example \(\PageIndex{6}\): Finding Component Forces

Consider a weight of 50lb hanging from two chains, as shown in Figure 10.26. One chain makes an angle of \(30^\circ\) with the vertical, and the other an angle of \(45^\circ\). Find the force applied to each chain.

Solution Knowing that gravity is pulling the 50lb weight straight down, we can create a vector \(\vec F\) to represent this force. \[\vec F = 50\langle 0,-1\rangle = \langle 0,-50\rangle.\]

We can view each chain as "pulling'' the weight up, preventing it from falling. We can represent the force from each chain with a vector. Let \(\vec F_1\) represent the force from the chain making an angle of \(30^\circ\) with the vertical, and let \(\vec F_2\) represent the force form the other chain. Convert all angles to be measured from the horizontal (as shown in Figure 10.27), and apply Key Idea 48. As we do not yet know the magnitudes of these vectors, (that is the problem at hand), we use \(m_1\) and \(m_2\) to represent them.

\[\vec F_1 = m_1\langle \cos 120^\circ,\sin120^\circ\rangle\]

\[\vec F_2 = m_2\langle \cos 45^\circ,\sin45^\circ\rangle\]

As the weight is not moving, we know the sum of the forces is \(\vec 0\). This gives:

\[\begin{align*} \vec F + \vec F_1 + \vec F_2 & = \vec 0\\ \langle 0,-50\rangle + m_1\langle \cos 120^\circ,\sin120^\circ\rangle + m_2\langle \cos 45^\circ,\sin45^\circ\rangle &=\vec 0 \end{align*}\]

The sum of the entries in the first component is 0, and the sum of the entries in the second component is also 0. This leads us to the following two equations: \[\begin{align*} m_1\cos120^\circ + m_2\cos45^\circ &=0 \\ m_1\sin120^\circ + m_2\sin45^\circ &=50 \end{align*}\] This is a simple 2-equation, 2-unkown system of linear equations. We leave it to the reader to verify that the solution is \[m_1=50(\sqrt{3}-1) \approx 36.6;\qquad m_2=\frac{50\sqrt{2}}{1+\sqrt{3}} \approx 25.88.\]

It might seem odd that the sum of the forces applied to the chains is more than 50lb. We leave it to a physics class to discuss the full details, but offer this short explanation. Our equations were established so that the vertical components of each force sums to 50lb, thus supporting the weight. Since the chains are at an angle, they also pull against each other, creating an "additional'' horizontal force while holding the weight in place.

Unit vectors were very important in the previous calculation; they allowed us to define a vector in the proper direction but with an unknown magnitude. Our computations were then computed component--wise. Because such calculations are often necessary, the standard unit vectors can be useful.

Definition 56 STANDARD UNIT VECTORS

• In \(\mathbb{R}^2\), the standard unit vectors are \[\vec i = \langle 1,0\rangle \quad \text{and}\quad \vec j = \langle 0,1\rangle.\]
• In \(\mathbb{R}^3\), the standard unit vectors are \[\vec i = \langle 1,0,0\rangle \quad \text{and}\quad \vec j = \langle 0,1,0\rangle \quad \text{and}\quad \vec k = \langle 0,0,1\rangle.\]

Example \(\PageIndex{7}\): Using standard unit vectors

• Rewrite \(\vec v = \langle 2,-3\rangle\) using the standard unit vectors.
• Rewrite \(\vec w = 4\vec i - 5\vec j +2\vec k\) in component form.
• \[\begin{align}\vec v &= \langle 2,-3\rangle \\&= \langle 2,0\rangle + \langle 0,-3\rangle \\&= 2\langle 1,0\rangle -3\langle 0,1\rangle\\&= 2\vec i - 3\vec j\end{align}\]
• \[\begin{align}\vec w &= 4\vec i - 5\vec j +2\vec k\\&= \langle 4,0,0\rangle +\langle 0,-5,0\rangle + \langle 0,0,2\rangle \\&= \langle 4,-5,2\rangle\end{align}\]

These two examples demonstrate that converting between component form and the standard unit vectors is rather straightforward. Many mathematicians prefer component form, and it is the preferred notation in this text. Many engineers prefer using the standard unit vectors, and many engineering text use that notation.

Example \(\PageIndex{8}\): Finding Component Force

A weight of 25lb is suspended from a chain of length 2ft while a wind pushes the weight to the right with constant force of 5lb as shown in Figure 10.28. What angle will the chain make with the vertical as a result of the wind's pushing? How much higher will the weight be?

Solution The force of the wind is represented by the vector \(\vec F_w = 5\vec i\). The force of gravity on the weight is represented by \(\vec F_g = -25\vec j\). The direction and magnitude of the vector representing the force on the chain are both unknown. We represent this force with \[\vec F_c = m\langle \cos\varphi,\sin\varphi\rangle = m\cos\varphi\, \vec i + m\sin\varphi\,\vec j\] for some magnitude \(m\) and some angle with the horizontal \(\varphi\). (Note: \(\theta\) is the angle the chain makes with the vertical ; \(\varphi\) is the angle with the horizontal .)

As the weight is at equilibrium, the sum of the forces is \(\vec0\): \[\begin{align*} \vec F_c + \vec F_w + \vec F_g &= \vec 0\\ m\cos\varphi\, \vec i + m\sin\varphi\,\vec j + 5\vec i - 25\vec j &=\vec 0 \end{align*}\]

Thus the sum of the \(\vec i\) and \(\vec j\) components are 0, leading us to the following system of equations: \[5+m\cos\varphi = 0\] \[-25+m\sin\varphi = 0\label{eq:vect8}\]

This is enough to determine \(\vec F_c\) already, as we know \(m\cos \varphi = -5\) and \(m\sin\varphi =25\). Thus \(F_c = \langle -5,25\rangle.\) We can use this to find the magnitude \(m\): \[m = \sqrt{(-5)^2+25^2} = 5\sqrt{26}\approx 25.5\text{lb}.\] We can then use either equality from Equation \ref{eq:vect8} to solve for \(\varphi\). We choose the first equality as using arccosine will return an angle in the \(2^\text{nd}\) quadrant: \[5 + 5\sqrt{26}\cos \varphi = 0 \quad \Rightarrow \quad \varphi = \cos^{-1}\left(\frac{-5}{5\sqrt{26}}\right) \approx 1.7682\approx 101.31^\circ.\]

Subtracting \(90^\circ\) from this angle gives us an angle of \(11.31^\circ\) with the vertical.

We can now use trigonometry to find out how high the weight is lifted. The diagram shows that a right triangle is formed with the 2ft chain as the hypotenuse with an interior angle of \(11.31^\circ\). The length of the adjacent side (in the diagram, the dashed vertical line) is \(2\cos 11.31^\circ \approx 1.96\)ft. Thus the weight is lifted by about \(0.04\)ft, almost 1/2in.

The algebra we have applied to vectors is already demonstrating itself to be very useful. There are two more fundamental operations we can perform with vectors, the dot product and the cross product. The next two sections explore each in turn.

Contributors and Attributions

Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License.  http://www.apexcalculus.com/

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Mechanics: Vectors and Projectiles

Calculator pad, version 2, vectors and projectiles: problem set.

Coach Sweeney walks 26 yards to the north along the sideline, pauses, and walks 12 yards back to the south.

a. Determine the distance which Coach moved. b. Determine Coach's resultant displacement.

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a. Rosa Boat is paddling upstream at 1.25 m/s relative to the water in a river which is flowing at 0.50 m/s relative to the banks of the river. What is the resultant velocity of Rosa's boat (relative to the banks)? b. Rosa Boat is paddling downstream at 1.25 m/s relative to the water in a river which is flowing at 0.50 m/s relative to the banks of the river. What is the resultant velocity of Rosa's boat (relative to the banks)?

The takeoff speed of a military aircraft from an aircraft carrier is approximately 170 mi/hr relative to the air. They acquire this speed through a combination of a catapult system present on the aircraft carrier and the aircraft's jet propulsion system. A common strategy is to head the carrier and the plane into the wind. If a plane is taking off from an aircraft carrier which is moving at 40 mi/hr into a 20 mi/hr headwind, then what speed relative to the deck of the aircraft carrier must it obtain to takeoff?

Claire de Iles is shopping. She walks 16 m to the end of an aisle. She then makes a right hand turn and walks 21 m down the end aisle. Determine the magnitude of Claire's resultant displacement.

Jim Nazium is walking from lunch to his PE class. He exits the lunchroom and walks 43 m west. He then turns and walks 72 m north down the hallway leading to the locker room. Determine the magnitude and direction of Jim's resultant displacement.

On her trip from home to school, Karla drives along three streets after exiting the driveway. She drives 1.85 miles south, 2.43 miles east and 0.35 miles north. Determine the magnitude of Karla's resultant displacement.

Sheila is captain of the Varsity cross country team. During the after-school practice on Tuesday, she led the team on the following run from school to a nearby park where they met the coach for a meeting: 0.68 miles, north; 1.09 miles east; 1.56 miles north; 0.32 miles, west. Determine the magnitude and direction of the team's resultant displacement.

During the Vector Addition lab, Mac and Tosh start at the classroom door and walk 40.0 m, north, 32.5 m east, 15.5 m south, 68.5 m west, and 2.5 m, north. Determine the magnitude and direction of the resultant displacement of Mac and Tosh.

Avery, the quarterback of South's Varsity football team, made the most amazing pass in the Homecoming game against cross town rival North. He threw the pass from the exact center of the field to the corner of the end zone, where Jamaal caught it for the game winning score. If the football field is 160 feet wide (sideline to sideline) and it is 60 yards from midfield to the back of the end zone, then how far did the ball travel from Avery's hands to Jamaal's hands.

Problem 10:

Consider the map of the United States below. Given the scale that 1 cm = 340 km, a protractor and a ruler can be used to determine the magnitude and direction for the following trips. All directions are expressed using the counter-clockwise from east convention. For each trip, use the sine, cosine and tangent functions to determine the horizontal and vertical components of the displacement. Be sure to indicate E, W, N, or S as the direction for each component.

Problem 11:

The pilot of a plane flying due north is notified by the flight controller that there is a second plane flying south at about the same altitude and located in the same general area. The pilot is told that the southward bound plane is currently located at a position which is 13.5 km, 102° from her own plane.

a. How many kilometers to the north is the second plane? b. How many kilometers to the west is the second plane? c. If the two planes both have an airspeed of 290. km/hr, then how much time will elapse before the planes are side by side ?

Problem 12:

A spelunker (person who explores caves) determines that the cave entrance is located 349 m, 253° from her current position. How far south and how far west from her current position is the cave entrance?

Problem 13:

Avery, South's quarterback, throws a pass 36.5 yards at 21° W of S before it is caught by Mitchell with a diving catch. Assuming that the field runs north and south, and that Avery threw the pass from 7.2 yards behind the line of scrimmage, how many yards were gained on the play?

Problem 14:

Mia Ander exits the front door of her home and walks along the path shown in the diagram at the right (not to scale). The walk consists of four legs with the following magnitudes:

A = 88 m B = 272 m C = 136 m D = 183 m

Determine the magnitude and direction of Mia's resultant displacement.

Problem 15:

Dora is exploring a cave. She starts at the entrance and makes the following straight line movements:

68 m, south 112 m, 25° north of west (155° CCW) 34 m, south 182 m, 17° south of east (343° CCW)

Determine Dora's position relative to the entrance of the cave. That is, how far and in what direction is Dora from the cave entrance?

Problem 16:

Taylor and Drew finish their last class on the day before Spring break and decide to take a spontaneous road trip. Their trip involves the following movements:

42 miles, 67° north of west (113° CCW) 61 miles, west 23 miles, 17° west of south (253° CCW)

Taylor's car breaks down after the last leg of the trip. How far and in what direction are Taylor and Drew from campus?

Problem 17:

A weather report shows that a tornado was sighted 12 km south and 23 km west of your town. The storm is reported to be moving directly towards your town at a speed of 82 km/hr.

a. What distance from your town was the tornado sighted? b. Approximately how much time (in minutes and hours) will elapse before the violent storm arrives at your town?

Problem 18:

An airplane begins its journey into Canada from a destination located 285 mi south of the border. The plane flies along a straight-line path at 189 mi/h in a direction of 20.5 degrees west of north. Determine the number of minutes before the plane crosses the border. Assume that the border is aligned directly east and west in the region where the flight takes place.

Problem 19:

Glenda and Harold are attempting to cross a river in a kayak. The river flows due east at 1.9 m/s. Glenda and Harold head the kayak due north and row at 2.4 m/s (relative to the water). The river is 38 m wide at this location.

a. Determine the resultant velocity of the boat - both magnitude and direction. b. Determine the time for Glenda and Harold to cross the river. c. How far downstream will the boat be when Glenda and Harold reach the opposite shore?

Problem 20:

Ty Ridlegs boards a paddle boat and heads the boat westward directly across a river. The river flows south at 48 cm/s. Ty paddles the boat with a speed of 98 cm/s.

a. Determine the resultant velocity of the boat - both magnitude and direction. b. If the river is 22 m wide at this location, then how much time does it take Ty to cross the river? Assume that Ty keeps his paddle boat headed west. c. How far downstream will Ty be when he reaches the other side of the river?

Problem 21:

Dylan and Sophia are walking along Bluebird Lake on a perfectly calm day. Dylan, determined to impress Sophia by his ability to skip rocks, picks up the flattest rock he can find and gives it a sidearm launch from the edge of the water. The rock acquires a completely horizontal velocity of 26 m/s from a height of 0.45 m above the water surface.

a. How much time does it take the rock to fall to the water surface?

Problem 22:

In an effort to create a cannonball-style splash, eight-year old Matthew runs off the edge of the board of the high dive at 4.6 m/s and falls 2.3 m to the water below.

a. Determine the time for Matthew to fall the 2.3 m to the water. b. What horizontal distance from the edge of the board will Matthew plunge into the water? c. With what speed does Matthew enter the water?

Problem 23:

Ima Peode wishes to throw a 2.8-kg pumpkin horizontally off the top of the school roof in order to hit Mr. H's car. The car is parked a distance of 13.4 m away from the base of the building below the point where Ima is standing. The building's roof is 10.4 m high. Assuming no air resistance, with what horizontal speed must Ima toss the pumpkin in order to hit Mr. H's car.

Problem 24:

The La Quebrada Cliff Divers provide daily entertainment for the crowds at Acapulco, Mexico. As a group of professional high divers, they dive off the cliff of La Quebrada and fall 45.1 m (148 feet) to the water below. More than an act of bravery, the cliff divers must time their dive so that they hit the water when the crest of an incoming wave has arrived. Determine the speed with which Pedro must run off the cliff in order to land in the water a horizontal distance of 17.8 m from the edge of the cliff.

Problem 25:

An emergency relief plane is dropping a care package from a plane to a group of medical personnel working for a relief agency in an African village. The package is designed to land in a small lake, inflate an attached raft upon impact, and finally resurface with the raft side down. The plane will be moving horizontally with a ground speed of 59.1 m/s. The package will be dropped a horizontal distance of 521 m from the intended target location. At what altitude above the pond must the plane be flying in order to successfully accomplish this feat?

Problem 26:

The Choo Choo Restaurant in DesPlaines, IL is a 50’s style diner which is notorious for the delivery of food from the kitchen to the dining room by an O-scale model train. Dinner baskets filled with hot dogs, hamburgers, French fries and the like are mounted to the tops of flatbed train cars and transported to table tops. On Matthew’s fifth birthday, a French fry rolled off the top of the pile on a tight turn moving at a speed of 1.25 m/s and fell to the floor.

a. Determine the time for the French fry to fall 113 cm from the top of the pile to the floor. b. Determine the horizontal displacement of the fry from the edge of the track. c. Determine the speed of the French fry upon striking the floor.

Problem 27:

Aaron Agin and Bud Derfenger are lab partners who last year earned a reputation for breaking beakers, spilling acid, mixing the wrong chemicals, breaking thermometers and accidentally lighting Sophia’s hair on fire with a Bunsen burner. And now to the delight of the physics class, Mr. H has made the mistake of allowing them to partner again. In a recent lab which utilized expensive tracks and carts, Aaron and Bud lived up to their reputation. Despite strong warnings from Mr. H, they allowed a cart to roll off the track and then off the table with a speed of 208 cm/s. The crash of the cart to the floor a horizontal distance of 96.3 cm from the table’s edge turned the entire classroom silent. Use this information to determine the height of the lab tables in Mr. H’s lab.

Problem 28:

Sharon Steady and Al Wayskachon won South’s recent egg toss contest held during Homecoming week. In their winning toss, Sharon gave the egg an underhand toss, releasing it with a velocity of 8.06 m/s at an angle of 30° to the horizontal. To the pleasure of the crowd, Al caught the egg at the same height as the toss without even a fracture to its shell.

a. Calculate the horizontal and vertical components of the initial velocity. b. Calculate the time for the egg to reach the midpoint of the trajectory. c. Calculate the total time the egg is in the air. d. Calculate the horizontal distance which the egg traveled from Sharon to Al. e. Calculate the height of the egg (relative to the releast point) when it was at the peak of its trajectory.

Problem 29:

Li Ping Phar, the famous Chinese ski jumper, leaves the ramp with an initial velocity of 34.9 m/s at an angle of 35°.

a. Determine the total time of flight. b. Determine the horizontal displacement. c. Determine the peak height (relative to the starting height). Assume that Li lands at the same height as the top of the ramp and that Li is a projectile.

Problem 30:

A tennis player stretches out to reach a ball that is just barely above the ground and successfully 'lobs' it over her opponent's head. The ball is hit with a speed of 18.7 m/s at an angle of 65.1 degrees.

a. Determine the time that the ball is in the air. b. Determine the maximum height which the ball reaches. c. Determine the distance the ball travels horizontally before landing.

Problem 31:

On New Year’s eve of 2007, Robbie Maddison set the world record for the longest motorcycle jump, traveling 98.3 m through the air from ramp to ramp. (The record has since been broken several times by Maddision himself.) Assuming a launch angle of 45°, insignificant air resistance and a landing location at the same height as the launch height, determine the speed with which Maddison left the ramp.

Problem 32:

Mr. Udadi takes his three children to the park for some summertime recreation. Olive Udadi is enjoying swinging and jumping. On one jump, Olive leaves the swing at a 30° angle to the horizontal with a speed of 2.2 m/s. She lands on the ground a horizontal distance of 1.09 m from the launch location.

a. Determine the horizontal and the vertical components of the initial velocity. b. Determine the time which Olive is in the air. c. Determine the vertical height (relative to the landing location) from which Olive jumps from the swing.

Problem 33:

In an apparent effort to earn an appearance on the Destroyed in Seconds show, Caleb attempts a bicycle maneuver in which he jumps between two ramps whose elevated edges are located a distance of 1.8 meters apart. The ramps are angled at 35° and located at the same height. Determine the speed (in m/s and mi/hr) that Caleb must acquire to accomplish this stunt. (Given: 1.00 m/s = 2.24 mi/hr)

Problem 34:

Albert is South’s star punter for the varsity football team. His best hang time this past season was for a punt which he kicked at 74° above the horizontal. The punt had a 6.2 second hang time.

a. Determine the speed at which the ball was punted. b. Determine the horizontal distance which the ball traveled.

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Multivariable calculus, session 1: vectors.

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• AP Physics Problems

AP Physics 1: Vectors Practice Problems with Answers

All topics about vectors in AP Physics 1 exams are covered using the problem-solving method in this complete guide. It includes AP physics multiple-choice tests on vector addition and subtraction, dot and cross product, resultant vector, and so on. 

Solved Vector Problems:

Problem (1): Which of the following quantities are vectors in physics?

a. Electromotive force (emf). b. Electric current. c. Fluid pressure.                       d. gravitational field.

Solution : Those quantities that have both a direction and a magnitude are defined as vector quantities in physics, such as displacement, velocity, acceleration, force, and so on. Both electromotive force (emf) and electric current are not vectors since a number is sufficient to describe them completely. Pressure at the depth of a fluid is not a vector, either. The gravitational field at any point has both a direction and a magnitude.

Thus, the correct answer is d . 

Problem (2): Which of the following quantities are scalars in physics?

(a) Momentum (b) Displacement (c) Area (d) Average velocity

Solution : Many quantities in physics do not have any direction associated with them. Just a number or a unit fully describes them. Such quantities are called scalar quantities. Examples of scalar quantities are mass, time, area, temperature, emf, electric current, etc.

Thus, the correct answer is c . 

Be sure to read this article: Definition of a Vector in Physics . There you will find more problems on vectors.

Problem (3): The components of a vector are given as $A_x=5.3$ and $A_y=2.9$. What is the magnitude of this vector? (a) 3        (b) 6        (c) 4        (d) 5

Solution: the magnitude of a vector in component form is found using the Pythagorean formula as below: \[A=\sqrt{A_x^2+A_y^2}\] Substituting the numerical values of the components into the above equation, we have \[A=\sqrt{(5.3)^2+(2.9)^2}=\boxed{6}\] The correct answer is b .

Problem (4): We are given the components of a displacement vector as $d_x=23.5$ and $d_y=34.3$. What angle does this vector make with the positive $x$-axis? a. $34.4^\circ$            b. $67^\circ$ c. $55.6^\circ$            d. $17.3^\circ$

Solution : Once the components of a vector are known, we can find its direction from the $x$-axis by the following formula: \[\alpha=\tan^{-1}\left(\frac{d_y}{d_x}\right)\]

Substituting the components gives us \[\alpha=\tan^{-1}\left(\frac{34.3}{23.5}\right)=55.6^\circ\] Since all components are in the first quadrant, so this is the correct angle.

Thus, the correct answer is c .

More Problems about vector components are discussed in the article below. Vector practice problems  

Problem (5): The components of a velocity vector at a moment are given as $v_x=-9.8\,{\rm m/s}$ and $v_y=6.4\,{\rm m/s}$. The direction, from the $+x$-axis, and magnitude of this velocity vector (in $\rm m/s$) are closest to  a. $11.7\, , 147^\circ$            b. $12.7\, , 139^\circ$ c. $11.7\, , -33.15^\circ$    d. $11.7\, , 211.15^\circ$

Solution : Its magnitude is simply obtained using the Pythagorean theorem \begin{align*}v&=\sqrt{v_x^2+v_y^2}\\\\&=\sqrt{(9.8)^2+(6.4)^2}\\\\&=\boxed{11.7\,{\rm m/s}}\end{align*} The subtle point is finding its direction. Note that the components of this vector lie in the second quadrant. Using the equation in the previous problem, it gives us \begin{align*} \alpha&=\tan^{-1}\left(\frac{v_y}{v_x}\right) \\\\ &=\tan^{-1}\left(\frac{6.4}{-9.8}\right)\\\\&=-33.15^\circ\end{align*}

Pay attention to these notes when using this equation for finding the direction of a vector relative to the $x$-axis through the smallest angle (as shown in the figure  by $\alpha$):  Note (1): In the one and fourth quadrants, the formula gives the correct angle.  Note (2): In the second and third quadrants, add $180^\circ$ to the angle obtained from the formula. 

In this case, therefore, the correct angle is \[\beta=180^\circ+(-33.15^\circ)=\boxed{146.85^\circ}\] which can be rounded to $147^\circ$, Hence, t he correct answer is a. 

Problem (6): Two equal magnitude forces, forming an angle of $60^\circ$ with each other, act on a body. What is the ratio of their subtraction to their resultant? a. $\frac{\sqrt{3}}{3}$                      b. $\sqrt{3}$ c. $1$                                               d. $\frac{\sqrt{2}}{2}$     

Solution : This question aims to explore the concepts of vector subtraction and addition (the same resultant). Assume two vectors (say, force) $\vec{A}$ and $\vec{B}$. The magnitude of their subtraction,$\vec{C}=\vec{A}-\vec{B}$, is defined to be as \[ C=\sqrt{A^2+B^2-2AB\cos\theta}\] where $A$ and $B$ are the magnitudes of each vector separately, and $\theta$ is the angle between them. 

In this case, we're told that there are two equal magnitude forces, i.e., $F_1=F_2=F$. Substituting the values into the above, we will have \begin{align*} C&=\sqrt{F_1^2+F_2^2-2F_1 F_2 \cos\theta} \\\\ &=\sqrt{2F^2-2F^2 \cos 60^\circ}\\\\&=\boxed{F}\end{align*} where we used $\cos 60^\circ=\frac 12$. On the other side, the resultant vector is another name for the addition of vectors , $\vec{R}=\vec{A}+\vec{B}$. The magnitude of the resultant of two arbitrary vectors making an angle of $\theta$ is given by \[R=\sqrt{A^2+B^2+2AB\cos\theta}\] Thus, in this case, we have \begin{align*} R&=\sqrt{F_1^2+F_2^2+2F_1 F_2\cos\theta}\\\\ &=\sqrt{2F^2+2F^2 \cos 60^\circ}\\\\&=F\sqrt{3}\end{align*} Now their ratio is as \[\frac{C}{R}=\frac{F}{F\sqrt{3}}=\frac{1}{\sqrt{3}}\] By rationalizing the denominator, the ratio is obtained as $\frac{\sqrt{3}}{3}$.

Thus, the correct answer is a . 

More related problems on forces in the AP Physics 1 Exam: AP Physics 1 forces practice problems with MCQs .

Problem (7): The magnitudes of two displacement vectors are $d$ and $2d$, and the magnitude of the total displacement is also $d\sqrt{3}$. What is the angle between these two displacement vectors? a. $\frac{3\pi}{4}$                        b. $\frac{\pi}{4}$ c. $\frac{2\pi}{3}$                        d. $\frac{\pi}{2}$

Solution : Displacement is a vector quantity in physics. Here, we have two displacement vectors of magnitudes $d$ and $2d$. The total displacement vector is the same as the resultant (or net) vector that is obtained by adding the vectors. Recall that the magnitude of a resultant vector of two vectors $\vec{A}$ and $\vec{B}$ forming an angle of $\theta$ is given by \[R=\sqrt{A^2+B^2+2AB\cos\theta}\] where $A$ and $B$ are the magnitudes. Substituting the known data into it and solving for the unknown angle $\theta$, we will have \begin{gather*} R=\sqrt{A^2+B^2+2AB\cos\theta}\\\\ d\sqrt{3}=\sqrt{d^2+(2d)^2 +2(d)(2d)\cos\theta}\\\\ 3d^2 =d^2+4d^2+4d^2\cos\theta \\\\ -2d^2 =4d^2 \cos\theta \\\\ \Rightarrow \cos\theta =-\frac 12 \\\\ \Rightarrow \boxed{\theta=120^\circ=\frac{2\pi}{3}}\end{gather*} Where in the second equality, both sides were squared.

Hence, the correct answer is c . 

Problem (8): Use all the information provided by the graph below and find the magnitude and direction (with $+x$ axis) of vector $\vec{C}=\vec{A}-\vec{B}$.  (a) $6.1\quad, 215^\circ$        (b) $5.3\quad, 135^\circ$ (c) $4\quad, 25^\circ$                (d) $6.1\quad, -26.6^\circ$

Solution : To find the subtraction of two vectors, knowing their magnitude and direction, you must first resolve those into their components. Then, algebraically subtract them to find the result. 

The components of $\vec{A}$ and $\vec{B}$ are found to be \begin{align*} A_x&=A\sin 45^\circ\\&=2.9\times \frac{\sqrt{2}}2\\&=2 \\\\ A_y&=A\cos 45^\circ\\&=2.9\times \frac{\sqrt{2}}2\\&=2\\\\ B_x &=B\cos 27^\circ\\&=3 \\\\ B_y&=B\sin 27^\circ\\&=1.5\end{align*} Note that the vector $\vec{A}$ lies in the third quadrant, so all its components are toward the negative $x$ and $y$ axes. Hence, its correct components are $\vec{A}=(-2,-2)$. Now we must form the subtraction vector of two vectors $\vec{A}$ and $\vec{B}=(3,1.5)$, by subtracting the corresponding components as below \begin{align*} C_x&=A_x-B_x\\&=-2-3\\&=-5 \\\\ C_y&=A_y-B_y\\&=-2-1.5\\&=-3.5 \end{align*} Thus, these are the components of the subtraction vector whose magnitude is also found as \[C=\sqrt{C_x^2+C_y^2}=\sqrt{(-5)^2+(-3.5)^2}=\boxed{6.1}\] Given the components of a vector, one can use the following formula to find the angle that the vector makes with the positive $x$ direction \begin{align*} \alpha&=\tan^{-1}\left(\frac{C_y}{C_x}\right) \\\\&=\tan^{-1}\left(\frac{-3.5}{-5}\right) \\\\&=35^\circ \end{align*} But note that the components of $\vec{C}=(-5,-3.5)$ are in the third quadrant, so we must add $180^\circ$ to this angle to get the correct angle. Hence, \[\alpha= 180^\circ+(35^\circ)=\boxed{215^\circ}\] The correct answer is a.

Problem (9): Two vectors are given as below \begin{gather*} \vec{A}=-3\,\hat{i}+2\,\hat{j}+3\,\hat{k}\\ \vec{B}=\hat{i}+2\,\hat{k}\end{gather*} The dot product $\vec{A}\cdot \vec{B}$ equals  a. 2    b. 3    c. 4    d. 1

Solution : Assume the components of two vectors are given as $\vec{A}=(A_x, A_y , A_z)$ and $\vec{B}=(B_x,B_y,B_z)$. Their dot (scalar) product is defined as follows \[\vec{A}\cdot\vec{B}=A_xB_x+A_y B_y+A_z B_z\] In this case, we have \begin{align*}\vec{A}\cdot\vec{B}&=(-3)(1)+(2)(0)+(3)(2)\\&=\boxed{3}\end{align*} Note that the dot product of two vectors is just a number, not another vector.

Thus, the correct answer is b. 

Problem (10): What is the angle between the two vectors $\vec{A}=4\,\hat{i}+4\,\hat{j}$ and $\vec{B}=3\,\hat{i}-3\,\hat{j}$ in radians? (a) $\pi$                            (b) $\frac{\pi}{2}$         (c) $\frac{3\pi}{3}$             (d) $\frac{3\pi}{4}$

Solution: The angle between two vectors $\vec{A}$ and $\vec{B}$ is found using another definition of scalar (dot) product as below \[\cos\theta=\frac{\vec{A}\cdot \vec{B}}{|\vec{A}||\vec{B}|}\] On the other side, recall that \[\vec{A}\cdot\vec{B}=A_x B_x +A_y B_y\] Thus, we have \begin{align*} \cos\theta &=\frac{(4)(3)+(4)(-3)}{\sqrt{4^2+4^2}\sqrt{3^2+(-3)^2}}\\\\ &=\frac{12-12}{4\sqrt{2}\times 3\sqrt{2}}\\\\ &=0\end{align*} Take the inverse cosine of both sides above to find the desired angle \[\theta=\cos^{-1}(0)=\frac{\pi}{2}\] Hence the correct answer is b.

Problem (11): Two vectors $\vec{A}=10\,\hat{i}-6\,\hat{j}$ and $\vec{B}=-16\,\hat{j}$ are given. What angle does the vector $\vec{C}=\vec{A}-\vec{B}$ make with the positive $x$ axis?  a. $30^\circ$        b. $60^\circ$ c. $90^\circ$        d. $45^\circ$

Solution : We are given two vectors in components form as $\vec{A}=(10,-6)$ and $\vec{B}=(0,-16)$. First, construct the subtraction vector $\vec{C}$ as below \begin{align*} C_x &=A_x-B_x\\&=10-0\\&=10 \\\\ C_y&=A_y-B_y\\&=-6-(-16)\\&=10\end{align*} Thus, $\vec{C}=(10,10)$. The angle of a vector with the positive $x$ axis, provided its components are known, is obtained by the following formula \begin{align*}\theta&=\tan^{-1}\left(\frac{C_y}{C_x}\right)\\\\&=\tan^{-1}\left(\frac{10}{10}\right)\\\\ &=\tan^{-1}(1)\end{align*} We know that the angle whose tangent is $1$ is $\boxed{45^\circ}$.

The correct answer is d.  

Problem (12): The vector $\vec{A}=2\sqrt{3}\hat{i}+2\hat{j}$ is perpendicular to which of the vectors $\vec{B}=3\sqrt{3}\hat{i}-3\hat{j}$ and $\vec{C}=3\hat{i}-3\sqrt{3}\hat{j}$.

a. Only $B$                            b. Only $C$     c. Both $B$ and $C$            d. None of the above.

Solution : Two vectors $\vec{A}=A_x \hat{i}+A_y \hat{j}$ and $\vec{B}=B_x \hat{i}+B_y \hat{j}$ are perpendicular to each other when their scalar product is zero \begin{gather*} \vec{A}\cdot\vec{B}=0\\ A_x B_x +A_y B_y =0 \end{gather*} So, first check the two vectors $\vec{A}$ and $\vec{B}$ \begin{align*} \vec{A}\cdot\vec{B}&=(2\sqrt{3})(3\sqrt{3})+(2)(-3)\\ &\neq 0 \end{align*} Thus, these two vectors are not perpendicular. Now check $\vec{A}$ and $\vec{C}$. \begin{align*} \vec{A}\cdot\vec{C}&=(2\sqrt{3})(3)+(2)(-3\sqrt{3})\\ &=0 \end{align*} So, these two vectors are perpendicular.

Hence, the correct answer is b.  

Problem (13): What angle does vector $\hat{i}+\sqrt{3}\,\hat{j}$ make with vector $-\sqrt{3}\,\hat{i}$?  (a) zero                                (b) $\frac{\pi}{3}$ (c) $\frac{2\pi}{3}$                   (d)  $\frac{5\pi}{6}$

Solution : Consider two vectors $\vec{A}=(A_x,A_y)$ and $\vec{B}=(B_x,B_y)$. Scalar or dot product definition gives us the angle between these two vectors as \[\cos\theta=\frac{A_xB_x+A_y B_y}{AB}\] where $A$ and $B$ are the magnitudes of the vectors. 

In this problem, assume $\vec{A}=(1,\sqrt{3})$ and $\vec{B}=(-\sqrt{3},0)$. Their magnitudes are \begin{align*} A&=\sqrt{A_x^2+A_y^2}\\ &=\sqrt{1^2+(\sqrt{3})^2}\\&=2 \\\\ B&=\sqrt{(-\sqrt{3})^2+0^2}\\&=\sqrt{3}\end{align*} Therefore, the angle between them is calculated as \begin{align*}\cos\theta &=\frac{(1)(-\sqrt{3})+(0)(\sqrt{3})}{2\times \sqrt{3}}\\\\&=\frac{-1}{2}\end{align*} Taking the inverse cosine of both sides gives us the desired angle \[\theta=\cos^{-1}\left(\frac{-1}{2}\right)=\frac{2\pi}{3}\] Thus, the correct angle is (c).

Problem (14): Which of the following vectors is perpendicular to the vector $\vec{a}=-2\,\hat{i}+3\,\hat{j}$?  a. $3\,\hat{i}+3\,\hat{j}$                b. $-\hat{i}+5\,\hat{j}$ c. $-3\,\hat{i}+2\,\hat{j}$             d. $3\,\hat{i}+2\,\hat{j}$

Solution : When the dot product of two vectors becomes zero, those vectors are perpendicular (the angle between them is $90^\circ$), to each other. Having the components of a vector, we can write the dot product between them as below \[ \vec{a}\cdot\vec{b}=a_x b_x+a_y b_y\] In this problem, we must check each choice separately.

a.  This is false since \[(-2)(3)+(3)(3)=3\neq 0\] b.  False \[(-2)(-1)+(3)(5)=18\neq 0\] c.  False \[(-2)(-3)+(3)(2)=12 \neq 0\] d.  Correct \[(-2)(3)+(3)(2)=0 \] Hence, the correct answer is (d).

Problem (15): Consider the vector $\vec{A}=0.5\,\hat{i}-\frac 23\,\hat{j}$. What is the magnitude of the vector $6\vec{A}$?  a. -1        b. +1        c. 5        d. $\sqrt{7}$

Solution : The purpose of this problem is to explore the concept of multiplying a vector by a scalar. If a vector with components $\vec{A}=(A_x,A_y)$ is given, then the vector $\vec{B}=k\vec{A}$, where $k$ is some number, is constructed as below \[\boxed{\vec{B}=(kA_x,kA_y)}\] Thus, in this case, we will have \[6\vec{A}=6(0.5,-\frac 23)=(3,-4)\] Its magnitude is also determined using the Pythagorean theorem \[\sqrt{3^2+(-4)^2}=5\] The correct answer is c . 

Problem (16): Two vectors are given as below: \begin{gather*} \vec{A}=3\,\hat{i}+2\,\hat{j}-\hat{k}\\\\\vec{B}=-2\,\hat{i}+4\,\hat{k}\end{gather*} What is the magnitude of the cross product $\vec{A}\times\vec{B}$? 

Solution : There are two methods to solve cross-product problems in the AP Physics exams. One is using the definition of cross product as below which only gives us its magnitude \[|\vec{A}\times\vec{B}|=AB\sin\theta\] and the next, using the determinants, which is a bit difficult but, in turn, gives the vector itself. We chose the first method. 

In the above $|\cdots|$ denotes the magnitude of the cross product.

To use the definition of the cross product, you must know the angle $\theta$ between the given vectors. In the previous problems, you learned how to find the angle between two arbitrary vectors using the dot product. So, that angle is obtained as \begin{align*} \cos\theta&=\frac{A_xB_x+A_yB_y+A_zB_z}{AB}\\\\&=\frac{(3)(-2)+(2)(0)+(-1)(4)}{\sqrt{14}\sqrt{20}}\\\\ &=\frac{-10}{\sqrt{14\times 20}}\end{align*} Taking the inverse  cosine of both sides, get \[\boxed{\alpha=126.7^\circ}\] where $A$ and $B$ are the magnitudes of the given vectors. Now that the angle is known, we can simply use the cross-product definition to find its magnitude as \begin{align*} |\vec{A}\times\vec{B}|&=AB\sin\theta\\&=\sqrt{14}\sqrt{20}\sin 126.7^\circ\\&=\boxed{13.4}\end{align*} 

Problem (17): Three vectors are shown in the figure below. The number next to each vector and between two adjacent vectors represent the magnitude and angle, respectively. Use this information to find the magnitude and direction of the following cross products: (a) $\vec{A}\times\vec{B}$, (b) $\vec{B}\times \vec{C}$. 

Solution : The cross product of two vectors $\vec{A}$ and $\vec{B}$ is another vector at the right angle (or, perpendicular) to both. The magnitude of this vector is calculated using the formula \[|\vec{A}\times\vec{B}|=AB\sin\theta\] where $\theta$ is the acute angle (the smallest) between the two vectors. 

The direction of the cross product is found using the right-hand rule. According to this rule, point the fingers of your right hand along the first vector $\vec{A}$ and turn those to the next vector $\vec{B}$. In this way, your thumb is directed along the direction of $\vec{A}\times \vec{B}$. 

(a) In this case, $\theta=42^\circ$. Using the equation for the cross product above, the magnitude is \begin{align*} |\vec{A}\times\vec{B}|&=AB\sin\theta \\ &=(3)(6) \sin 42^\circ\\&=12 \end{align*} According to the right-hand rule prescription, the direction is into the page. 

(b) The angle is $77^\circ+42^\circ=119^\circ$, so \begin{align*} |\vec{B}\times\vec{C}|&=BC\sin\theta \\ &=(3)(7) \sin 119^\circ\\&=18.3 \end{align*} Its direction is also out of the page . 

Refer to the page below to practice more problems on the right-hand rule.

Right-hand rule: example problems

Problem (18): Five equal-magnitude forces apply to an object. If the magnitude of each force is $F$, find the resultant force vector.  a. 1F                b. 2F c. 3F                d. 5F

Solution : Always, the best method to find the resultant vector of a couple of vectors is to decompose all vectors along the horizontal and vertical directions, then use the rules of vector addition. Of five equal-magnitude forces, two of them are directed at an angle of $30^\circ$. So, resolve those into their components. 

The force vector that is in the first quadrant has the following components \begin{gather*} F_x=F\cos 30^\circ=F\frac{\sqrt{3}}2\\\\ F_y=F\sin 30^\circ=F\left(\frac 12\right)\end{gather*}

Similarly, the force in the second quadrant has the same components, but with a small difference. The  $x$-component of this vector is to the left, so its correct component is $-F\frac{\sqrt{3}}2$. 

Now that all vectors have been decomposed along the $x$ and $y$ axes, in each direction, algebraically add them to find the component of the resultant vector in that direction. In the $x$-direction, we have \[F\frac{\sqrt{3}}2+F-F\frac{\sqrt{3}}2-F=0\] In the $y$-direction, \[F+F\frac 12+F\frac 12=2F\] Therefore, the resultant vector $\vec{R}$ has the following components \[\vec{R}=(0,2F)\] Its magnitude is also \begin{align*}R&=\sqrt{R_x^2+R_y^2}\\&=\sqrt{0^2+(2F)^2}\\&=\boxed{2F}\end{align*} Thus, the correct answer is b . 

Problem (19): What is the resultant of the forces shown in the figure below? a. 28        b. $12\sqrt{2}$        c. $15\sqrt{2}$        d. 20

Solution : First of all, resolve the angled vector into its components along the positive $x$ and $y$ axes. Recall that a vector of magnitude $A$ that makes an angle of $\theta$ with the positive $x$ axis has a component of $A\cos\theta$ along the $x$ axis, and a component of $A\sin\theta$ along the $y$ axis. 

In this problem, the tilted vector has a magnitude of $5\sqrt{2}$ and an angle of $45^\circ$ with the horizontal. So, its components are \begin{align*} A_x&=5\sqrt{2}\cos 45^\circ\\&=5\sqrt{2}\times \sqrt{2}/2 \\&=5 \\\\ A_y&=5\sqrt{2}\sin 45^\circ \\&=5\end{align*}

Now we have several vectors in each direction of the coordinate system. The resultant vector of a couple of vectors is defined as their vector addition. To find it, we proceed as below: Stage (I): Add vectors lying in each direction. Along $+x$: $16.5+5=21.5\,{\rm N}$ Along $-x$: $5.5\,{\rm N}$ Along $+y$: $11.5+5=16.5\,{\rm N}$ Along $-y$: $4.5\,{\rm N}$

Stage (II): Along each direction, subtract the vectors in the negative direction from the positive one. This gives the components of the resultant (net) vector along that direction. \begin{gather*} R_x = 21.5-5.5=16\,{\rm N} \\ R_y=16.5-4.5=12\,{\rm N}\end{gather*} where we called the resultant's components as $R_x$ and $R_y$. Thus, the resultant vector is written as \[\vec{R}=16\,\hat{i}+12\,\hat{j}\] The magnitude of a vector is also found using the Pythagorean theorem \begin{align*} R&=\sqrt{R_x^2 +R_y^2}\\\\&=\sqrt{16^2+12^2}\\\\&=\boxed{20\,{\rm N}}\end{align*} Hence, the correct answer is d.  

Problem (20): What is the magnitude and direction of the resultant vector of the following vectors shown in the diagram below? (Take $\cos 53^\circ=0.6$ and $\sin 53^\circ=0.8$).  a. $2\,{\rm N}$ due west        b.$2\,{\rm N}$ due east c. $1\,{\rm N}$ due east        d. $1\,{\rm N}$ due west

Solution : There are two vectors directed due east and north, respectively. The $10\,{\rm N}$ vector is also directed $53^\circ$ west of south. Resolve this tilted vector into its components along the $x$ and $y$ directions. \begin{align*} A_x&=A\sin\theta \\&=10\,\sin 53^\circ\\&=8 \\\\ A_y&=A\cos\theta\\&=10\,\cos 53^\circ\\&=6\end{align*}

Now, along each direction, there are two vectors in opposite directions. So subtract them to find the net vector in that direction. \begin{gather*} R_x=10-8=2\,{\rm N}\\\\ R_y=6-6=0\end{gather*} These are the components of the resultant vector whose magnitude gets as \[R=\sqrt{R_x^2+R_y^2}=\sqrt{2^2+0^2}=2\,{\rm N}\] The angle that a vector makes with the positive $x$-axis is also found by \[\alpha=\tan^{-1}\left(\frac{R_y}{R_x}\right)\] So, substituting the components gives us \[\alpha=\tan^{-1}\left(\frac{0}{2}\right)=0^\circ\] Hence, the resultant vector lies toward the east and has a magnitude of $2\,{\rm N}$.

The correct answer is b.  

Problem (21): As shown in the free-body diagram below, a body is subjected to three forces. The body is in equilibrium condition. Find the magnitude and direction (the angle $\theta$) of the vector $\vec{F}$

Solution : ''Equilibrium'' means that the resultant or net force on the object is zero. Three forces are acting on the body. First, resolve the tilted force vectors, find their vector addition (resultant vector), and then set it to zero.  The vector $\vec{F}$ is resolved into its components as below \begin{gather*} F_x=F\cos\theta \\F_y=F\sin\theta\end{gather*} Similarly, the components of $40\sqrt{3}-{\rm N}$ vector are \begin{align*} A_x&=A\cos 60^\circ\\&=40\sqrt{3}\left(\frac{1}{2}\right)\\&=20\sqrt{3}\,{\rm N}\\\\ A_y&=A\sin 60^\circ\\&=40\sqrt{3}\times \frac{\sqrt{3}}{2}\\&=60\,{\rm N}\end{align*} The algebraic addition of vectors along the $x$ direction gives the $x$ component of the resultant vector \[R_x=F\cos\theta-20\sqrt{3}\] Similarly, the $y$ component of the resultant vector is also found as \[R_y=F\sin\theta+60-80\] Since the object is in equilibrium, the net force on it must be zero. Thus, equate those components to zero \begin{gather*} R_x=0 \Rightarrow F\cos\theta=20\sqrt{3} \\\\\ R_y=0 \Rightarrow F\sin\theta=20\end{gather*} By dividing them, one of the unknowns, $F$, is removed. This way, the other unknown, $\theta$, is found. \begin{gather*} \frac{F\cos\theta}{F\sin\theta}=\frac{20\sqrt{3}}{20}\\\\ \Rightarrow \cot\theta =\sqrt{3} \\\\ \Rightarrow\quad  \boxed{\theta=30^\circ}\end{gather*} Now, substitute the obtained angle into one of the equations $R_x=0$ or $R_y=0$ and solve for $F$. \begin{gather*}R_x=F\cos \theta-20\sqrt{3}=0\\ F\cos 30^\circ=20\sqrt{3} \\\Rightarrow \boxed{F=40\,{\rm N}}\end{gather*} where we set $\cos 30^\circ=\frac{\sqrt{3}}2$. 

Problem (22): In the figure below, the rope $OB$ is horizontal, and the tension force in the rope $OA$ equals $60\sqrt{2}$. The system is in equilibrium. What is the mass of the hanging object?  a. $\sqrt{2}$     b. $6$        c. $3\sqrt{2}$    d. $3$

Solution : This setup is in equilibrium, so the net or resultant force on the body must be zero. Three forces are acting on the block. Two tension forces in the ropes and one weight force. 

The tension force in the rope $OA$, $T_1$, is directed at an angle of $180^\circ-135^\circ=45^\circ$ with the horizontal as shown in the free-body diagram below. Resolve this tension into its components, then find the net force vector resulting from these three forces.

The components of the tension force $T_1$ are found to be \begin{align*} T_{1x}&=T_1\cos 45^\circ\\&=60\sqrt{2}\times \frac{\sqrt{2}}2\\&=60\,{\rm N}\\\\T_{1y}&=T_1\sin 45^\circ\\&=60\sqrt{2}\times \frac{\sqrt{2}}2\\&=60\,{\rm N}\end{align*} The object does not move vertically, so the net force in this direction must be zero, i.e., $T_{1y}=W$. Therefore, we have \[T_{1y}=W \Rightarrow W=60\,{\rm N}\] Using the definition of weight as $W=mg$, the mass of the object is \[m=\frac{W}{g}=\frac{60}{10}=6\,{\rm kg}\] Hence, t he correct answer is b . (Find the tension in the rope $OB$.)

Problem (23): If the resultant of the following three vectors is zero, find $a$ and $b$, respectively. \begin{gather*} \vec{A}=3\,\hat{i}+2\,\hat{j}\\ \vec{B}=-5\,\hat{i}+3\,\hat{j}\\\vec{C}=a\,\hat{i}+b\,\hat{j}\end{gather*}

Solution : In all the AP Physics exams, the resultant vector means the addition of vectors. We call it $\vec{R}$. So, we must find the addition of the three above vectors, set it to zero, and solve the obtained equations for the unknowns. \[\vec{R}=\vec{A}+\vec{B}+\vec{C}=0\] We summarize each vector as below for simplicity \[\vec{A}=(3,2) \, , \vec{B}=(-5,3) \, , \vec{C}=(a,b)\] By adding the corresponding components with each other, we will have \begin{align*} R_x &=3+(-5)+a\\R_y &= 2+3+b \end{align*} These are the components of the resultant vector, setting these to zero yields \begin{gather*} R_x=0 \Rightarrow \boxed{a=2} \\ R_y=0 \Rightarrow \boxed{b=-5} \end{gather*} 

Author : Dr. Ali Nemati Page Published : 10/19/2021  

© 2015 All rights reserved. by Physexams.com

2.16: Dot Product: Problem Solving

Chapter 0: physics basics, chapter 1: an introduction to statics, chapter 2: force vectors, chapter 3: equilibrium of a particle, chapter 4: force system resultants, chapter 5: equilibrium of a rigid body, chapter 6: structural analysis, chapter 7: internal forces, chapter 8: friction, chapter 9: center of gravity and centroid, chapter 10: moment of inertia, chapter 11: virtual work, chapter 12: kinematics of a particle, chapter 13: kinetics of a particle: force and acceleration, chapter 14: kinetics of a particle: impulse and momentum, chapter 15: planar kinematics of a rigid body, chapter 16: 3-dimensional kinetics of a rigid body, chapter 17: concept of stress, chapter 18: stress and strain - axial loading.

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Consider a rod fixed to a wall, which can be pulled by a chain by applying a force at one of its ends. The position of the rod is defined using a three-dimensional coordinate system.

The angle theta between the force vector and the rod, and the projection of force along the rod needs to be determined.

First, the position vectors for the two ends of the rod are defined. Then the position vector along the rod is determined.

The next step determines the magnitude of the position vector r AB and the force vector.

Now, the dot product of the position vector with the force vector is determined by multiplying the components of the two vectors. Angle theta is then estimated as the inverse cosine function of the ratio of the dot product and the product of magnitudes of the two vectors.

The projection of the force along the rod can be determined as the product of the magnitude of force and the cosine of theta.

The dot product is a powerful tool in problem-solving involving vectors, given that the dot product of two vectors is the product of their magnitudes and the cosine of the angle between them measured anti-clockwise. Solving problems involving the dot product requires understanding its properties and developing a step-by-step process to solve them. Here are the main steps to follow when solving any general problem involving the dot product:

Identify the problem: Start by reading the problem and identifying the question that needs to be answered. This will enable you to determine the purpose and direction for solving the problem.

Define the vectors: List the given vectors and represent them in the Cartesian or component form.

Decide which operation to use: The dot product is appropriate when the problem involves finding the angle between two vectors, calculating the component of a vector along a given direction, testing orthogonality, or finding the projection of one vector onto another vector. Ensure that the problem requires the use of the dot product before proceeding.

Calculate the dot product: Multiply the corresponding components of the two vectors and sum their products. This gives the value of their dot product.

Verify the solution: Check your solution to ensure that it satisfies the given conditions in the problem. Be sure to round off the answer appropriately and include the correct units where necessary.

The angle between two vectors can be obtained from the inverse cosine of the dot product of the two vectors divided by the product of the magnitudes of the two vectors. The dot product can also be employed to find the component of a vector along a given direction by projecting it onto a unit vector in the desired direction. This technique is particularly useful for decomposing complex vector problems into simpler components. Additionally, the dot product can be used to test orthogonality between two vectors. If their dot product is zero, the vectors are orthogonal, meaning they are perpendicular to each other. Lastly, the projection of one vector onto another can be found using the dot product by multiplying the magnitude of the first vector by the cosine of the angle between the two vectors.

• Hibbeler, R.C. (2016). Engineering Mechanics ‒ Statics and Dynamics. Hoboken, New Jersey: Pearson Prentice Hall. pp 71- 76.

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• \begin{pmatrix}1&0&3\end{pmatrix}+\begin{pmatrix}-1&4&2\end{pmatrix}
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• \begin{pmatrix}1&2&3\end{pmatrix}\times\begin{pmatrix}1&5&7\end{pmatrix}
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• What are vectors in math?
• In math, a vector is an object that has both a magnitude and a direction. Vectors are often represented by directed line segments, with an initial point and a terminal point. The length of the line segment represents the magnitude of the vector, and the arrowhead pointing in a specific direction represents the direction of the vector.
• What are the types of vectors?
• The common types of vectors are cartesian vectors, column vectors, row vectors, unit vectors, and position vectors.
• How do you add two vectors?
• To add two vectors, add the corresponding components from each vector. Example: the sum of (1,3) and (2,4) is (1+2,3+4), which is (3,7)

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2.1 Scalars and Vectors

Learning objectives.

By the end of this section, you will be able to:

• Describe the difference between vector and scalar quantities.
• Identify the magnitude and direction of a vector.
• Explain the effect of multiplying a vector quantity by a scalar.
• Describe how one-dimensional vector quantities are added or subtracted.
• Explain the geometric construction for the addition or subtraction of vectors in a plane.
• Distinguish between a vector equation and a scalar equation.

Many familiar physical quantities can be specified completely by giving a single number and the appropriate unit. For example, “a class period lasts 50 min” or “the gas tank in my car holds 65 L” or “the distance between two posts is 100 m.” A physical quantity that can be specified completely in this manner is called a scalar quantity . Scalar is a synonym of “number.” Time, mass, distance, length, volume, temperature, and energy are examples of scalar quantities.

Scalar quantities that have the same physical units can be added or subtracted according to the usual rules of algebra for numbers. For example, a class ending 10 min earlier than 50 min lasts 50 min − 10 min = 40 min 50 min − 10 min = 40 min . Similarly, a 60-cal serving of corn followed by a 200-cal serving of donuts gives 60 cal + 200 cal = 260 cal 60 cal + 200 cal = 260 cal of energy. When we multiply a scalar quantity by a number, we obtain the same scalar quantity but with a larger (or smaller) value. For example, if yesterday’s breakfast had 200 cal of energy and today’s breakfast has four times as much energy as it had yesterday, then today’s breakfast has 4 ( 200 cal ) = 800 cal 4 ( 200 cal ) = 800 cal of energy. Two scalar quantities can also be multiplied or divided by each other to form a derived scalar quantity. For example, if a train covers a distance of 100 km in 1.0 h, its speed is 100.0 km/1.0 h = 27.8 m/s, where the speed is a derived scalar quantity obtained by dividing distance by time.

Many physical quantities, however, cannot be described completely by just a single number of physical units. For example, when the U.S. Coast Guard dispatches a ship or a helicopter for a rescue mission, the rescue team must know not only the distance to the distress signal, but also the direction from which the signal is coming so they can get to its origin as quickly as possible. Physical quantities specified completely by giving a number of units (magnitude) and a direction are called vector quantities . Examples of vector quantities include displacement, velocity, position, force, and torque. In the language of mathematics, physical vector quantities are represented by mathematical objects called vectors ( Figure 2.2 ). We can add or subtract two vectors, and we can multiply a vector by a scalar or by another vector, but we cannot divide by a vector. The operation of division by a vector is not defined.

Let’s examine vector algebra using a graphical method to be aware of basic terms and to develop a qualitative understanding. In practice, however, when it comes to solving physics problems, we use analytical methods, which we’ll see in the next section. Analytical methods are more simple computationally and more accurate than graphical methods. From now on, to distinguish between a vector and a scalar quantity, we adopt the common convention that a letter in bold type with an arrow above it denotes a vector, and a letter without an arrow denotes a scalar. For example, a distance of 2.0 km, which is a scalar quantity, is denoted by d = 2.0 km, whereas a displacement of 2.0 km in some direction, which is a vector quantity, is denoted by d → d → .

Suppose you tell a friend on a camping trip that you have discovered a terrific fishing hole 6 km from your tent. It is unlikely your friend would be able to find the hole easily unless you also communicate the direction in which it can be found with respect to your campsite. You may say, for example, “Walk about 6 km northeast from my tent.” The key concept here is that you have to give not one but two pieces of information—namely, the distance or magnitude (6 km) and the direction (northeast).

Displacement is a general term used to describe a change in position , such as during a trip from the tent to the fishing hole. Displacement is an example of a vector quantity. If you walk from the tent (location A ) to the hole (location B ), as shown in Figure 2.3 , the vector D → D → , representing your displacement , is drawn as the arrow that originates at point A and ends at point B . The arrowhead marks the end of the vector. The direction of the displacement vector D → D → is the direction of the arrow. The length of the arrow represents the magnitude D of vector D → D → . Here, D = 6 km. Since the magnitude of a vector is its length, which is a positive number, the magnitude is also indicated by placing the absolute value notation around the symbol that denotes the vector; so, we can write equivalently that D ≡ | D → | D ≡ | D → | . To solve a vector problem graphically, we need to draw the vector D → D → to scale. For example, if we assume 1 unit of distance (1 km) is represented in the drawing by a line segment of length u = 2 cm, then the total displacement in this example is represented by a vector of length d = 6 u = 6 ( 2 cm ) = 12 cm d = 6 u = 6 ( 2 cm ) = 12 cm , as shown in Figure 2.4 . Notice that here, to avoid confusion, we used D = 6 km D = 6 km to denote the magnitude of the actual displacement and d = 12 cm to denote the length of its representation in the drawing.

Suppose your friend walks from the campsite at A to the fishing pond at B and then walks back: from the fishing pond at B to the campsite at A . The magnitude of the displacement vector D → A B D → A B from A to B is the same as the magnitude of the displacement vector D → B A D → B A from B to A (it equals 6 km in both cases), so we can write D A B = D B A D A B = D B A . However, vector D → A B D → A B is not equal to vector D → B A D → B A because these two vectors have different directions: D → A B ≠ D → B A D → A B ≠ D → B A . In Figure 2.3 , vector D → B A D → B A would be represented by a vector with an origin at point B and an end at point A , indicating vector D → B A D → B A points to the southwest, which is exactly 180 ° 180 ° opposite to the direction of vector D → A B D → A B . We say that vector D → B A D → B A is antiparallel to vector D → A B D → A B and write D → A B = − D → B A D → A B = − D → B A , where the minus sign indicates the antiparallel direction.

Two vectors that have identical directions are said to be parallel vectors —meaning, they are parallel to each other. Two parallel vectors A → A → and B → B → are equal, denoted by A → = B → A → = B → , if and only if they have equal magnitudes | A → | = | B → | | A → | = | B → | . Two vectors with directions perpendicular to each other are said to be orthogonal vectors . These relations between vectors are illustrated in Figure 2.5 .

Check Your Understanding 2.1

Two motorboats named Alice and Bob are moving on a lake. Given the information about their velocity vectors in each of the following situations, indicate whether their velocity vectors are equal or otherwise. (a) Alice moves north at 6 knots and Bob moves west at 6 knots. (b) Alice moves west at 6 knots and Bob moves west at 3 knots. (c) Alice moves northeast at 6 knots and Bob moves south at 3 knots. (d) Alice moves northeast at 6 knots and Bob moves southwest at 6 knots. (e) Alice moves northeast at 2 knots and Bob moves closer to the shore northeast at 2 knots.

Algebra of Vectors in One Dimension

Vectors can be multiplied by scalars, added to other vectors, or subtracted from other vectors. We can illustrate these vector concepts using an example of the fishing trip seen in Figure 2.6 .

Suppose your friend departs from point A (the campsite) and walks in the direction to point B (the fishing pond), but, along the way, stops to rest at some point C located three-quarters of the distance between A and B , beginning from point A ( Figure 2.6 (a)). What is his displacement vector D → A C D → A C when he reaches point C ? We know that if he walks all the way to B , his displacement vector relative to A is D → A B D → A B , which has magnitude D A B = 6 km D A B = 6 km and a direction of northeast. If he walks only a 0.75 fraction of the total distance, maintaining the northeasterly direction, at point C he must be 0.75 D A B = 4.5 km 0.75 D A B = 4.5 km away from the campsite at A . So, his displacement vector at the rest point C has magnitude D A C = 4.5 km = 0.75 D A B D A C = 4.5 km = 0.75 D A B and is parallel to the displacement vector D → A B D → A B . All of this can be stated succinctly in the form of the following vector equation :

In a vector equation, both sides of the equation are vectors. The previous equation is an example of a vector multiplied by a positive scalar (number) α = 0.75 α = 0.75 . The result, D → A C D → A C , of such a multiplication is a new vector with a direction parallel to the direction of the original vector D → A B D → A B .

In general, when a vector A → A → is multiplied by a positive scalar α α , the result is a new vector B → B → that is parallel to A → A → :

The magnitude | B → | | B → | of this new vector is obtained by multiplying the magnitude | A → | | A → | of the original vector, as expressed by the scalar equation :

In a scalar equation, both sides of the equation are numbers. Equation 2.2 is a scalar equation because the magnitudes of vectors are scalar quantities (and positive numbers). If the scalar α α is negative in the vector equation Equation 2.1 , then the magnitude | B → | | B → | of the new vector is still given by Equation 2.2 , but the direction of the new vector B → B → is antiparallel to the direction of A → A → . These principles are illustrated in Figure 2.7 (a) by two examples where the length of vector A → A → is 1.5 units. When α = 2 α = 2 , the new vector B → = 2 A → B → = 2 A → has length B = 2 A = 3.0 units B = 2 A = 3.0 units (twice as long as the original vector) and is parallel to the original vector. When α = −2 α = −2 , the new vector C → = −2 A → C → = −2 A → has length C = | − 2 | A = 3.0 units C = | − 2 | A = 3.0 units (twice as long as the original vector) and is antiparallel to the original vector.

Now suppose your fishing buddy departs from point A (the campsite), walking in the direction to point B (the fishing hole), but he realizes he lost his tackle box when he stopped to rest at point C (located three-quarters of the distance between A and B , beginning from point A ). So, he turns back and retraces his steps in the direction toward the campsite and finds the box lying on the path at some point D only 1.2 km away from point C (see Figure 2.6 (b)). What is his displacement vector D → A D D → A D when he finds the box at point D ? What is his displacement vector D → D B D → D B from point D to the hole? We have already established that at rest point C his displacement vector is D → A C = 0.75 D → A B D → A C = 0.75 D → A B . Starting at point C , he walks southwest (toward the campsite), which means his new displacement vector D → C D D → C D from point C to point D is antiparallel to D → A B D → A B . Its magnitude | D → C D | | D → C D | is D C D = 1.2 km = 0.2 D A B D C D = 1.2 km = 0.2 D A B , so his second displacement vector is D → C D = −0.2 D → A B D → C D = −0.2 D → A B . His total displacement D → A D D → A D relative to the campsite is the vector sum of the two displacement vectors: vector D → A C D → A C (from the campsite to the rest point) and vector D → C D D → C D (from the rest point to the point where he finds his box):

The vector sum of two (or more) vectors is called the resultant vector or, for short, the resultant . When the vectors on the right-hand-side of Equation 2.3 are known, we can find the resultant D → A D D → A D as follows:

When your friend finally reaches the pond at B , his displacement vector D → A B D → A B from point A is the vector sum of his displacement vector D → A D D → A D from point A to point D and his displacement vector D → D B D → D B from point D to the fishing hole: D → A B = D → A D + D → D B D → A B = D → A D + D → D B (see Figure 2.6 (c)). This means his displacement vector D → D B D → D B is the difference of two vectors :

Notice that a difference of two vectors is nothing more than a vector sum of two vectors because the second term in Equation 2.5 is vector − D → A D − D → A D (which is antiparallel to D → A D ) D → A D ) . When we substitute Equation 2.4 into Equation 2.5 , we obtain the second displacement vector:

This result means your friend walked D D B = 0.45 D A B = 0.45 ( 6.0 km ) = 2.7 km D D B = 0.45 D A B = 0.45 ( 6.0 km ) = 2.7 km from the point where he finds his tackle box to the fishing hole.

When vectors A → A → and B → B → lie along a line (that is, in one dimension), such as in the camping example, their resultant R → = A → + B → R → = A → + B → and their difference D → = A → − B → D → = A → − B → both lie along the same direction. We can illustrate the addition or subtraction of vectors by drawing the corresponding vectors to scale in one dimension, as shown in Figure 2.7 .

To illustrate the resultant when A → A → and B → B → are two parallel vectors, we draw them along one line by placing the origin of one vector at the end of the other vector in head-to-tail fashion (see Figure 2.7 (b)). The magnitude of this resultant is the sum of their magnitudes: R = A + B . The direction of the resultant is parallel to both vectors. When vector A → A → is antiparallel to vector B → B → , we draw them along one line in either head-to-head fashion ( Figure 2.7 (c)) or tail-to-tail fashion. The magnitude of the vector difference, then, is the absolute value D = | A − B | D = | A − B | of the difference of their magnitudes. The direction of the difference vector D → D → is parallel to the direction of the longer vector.

In general, in one dimension—as well as in higher dimensions, such as in a plane or in space—we can add any number of vectors and we can do so in any order because the addition of vectors is commutative ,

and associative ,

Moreover, multiplication by a scalar is distributive :

We used the distributive property in Equation 2.4 and Equation 2.6 .

When adding many vectors in one dimension, it is convenient to use the concept of a unit vector . A unit vector, which is denoted by a letter symbol with a hat, such as u ^ u ^ , has a magnitude of one and does not have any physical unit so that | u ^ | ≡ u = 1 | u ^ | ≡ u = 1 . The only role of a unit vector is to specify direction. For example, instead of saying vector D → A B D → A B has a magnitude of 6.0 km and a direction of northeast, we can introduce a unit vector u ^ u ^ that points to the northeast and say succinctly that D → A B = ( 6.0 km ) u ^ D → A B = ( 6.0 km ) u ^ . Then the southwesterly direction is simply given by the unit vector − u ^ − u ^ . In this way, the displacement of 6.0 km in the southwesterly direction is expressed by the vector

Example 2.1

A ladybug walker.

The total displacement D → D → is the resultant of all its displacement vectors.

In this calculation, we use the distributive law given by Equation 2.9 . The result reads that the total displacement vector points away from the 100-cm mark (initial landing site) toward the end of the meter stick that touches the wall. The end that touches the wall is marked 0 cm, so the final position of the ladybug is at the (100 – 32)cm = 68-cm mark.

Check Your Understanding 2.2

A cave diver enters a long underwater tunnel. When her displacement with respect to the entry point is 20 m, she accidentally drops her camera, but she doesn’t notice it missing until she is some 6 m farther into the tunnel. She swims back 10 m but cannot find the camera, so she decides to end the dive. How far from the entry point is she? Taking the positive direction out of the tunnel, what is her displacement vector relative to the entry point?

Algebra of Vectors in Two Dimensions

When vectors lie in a plane—that is, when they are in two dimensions—they can be multiplied by scalars, added to other vectors, or subtracted from other vectors in accordance with the general laws expressed by Equation 2.1 , Equation 2.2 , Equation 2.7 , and Equation 2.8 . However, the addition rule for two vectors in a plane becomes more complicated than the rule for vector addition in one dimension. We have to use the laws of geometry to construct resultant vectors, followed by trigonometry to find vector magnitudes and directions. This geometric approach is commonly used in navigation ( Figure 2.9 ). In this section, we need to have at hand two rulers, a triangle, a protractor, a pencil, and an eraser for drawing vectors to scale by geometric constructions.

For a geometric construction of the sum of two vectors in a plane, we follow the parallelogram rule . Suppose two vectors A → A → and B → B → are at the arbitrary positions shown in Figure 2.10 . Translate either one of them in parallel to the beginning of the other vector, so that after the translation, both vectors have their origins at the same point. Now, at the end of vector A → A → we draw a line parallel to vector B → B → and at the end of vector B → B → we draw a line parallel to vector A → A → (the dashed lines in Figure 2.10 ). In this way, we obtain a parallelogram. From the origin of the two vectors we draw a diagonal that is the resultant R → R → of the two vectors: R → = A → + B → R → = A → + B → ( Figure 2.10 (a)). The other diagonal of this parallelogram is the vector difference of the two vectors D → = A → − B → D → = A → − B → , as shown in Figure 2.10 (b). Notice that the end of the difference vector is placed at the end of vector A → A → .

It follows from the parallelogram rule that neither the magnitude of the resultant vector nor the magnitude of the difference vector can be expressed as a simple sum or difference of magnitudes A and B , because the length of a diagonal cannot be expressed as a simple sum of side lengths. When using a geometric construction to find magnitudes | R → | | R → | and | D → | | D → | , we have to use trigonometry laws for triangles, which may lead to complicated algebra. There are two ways to circumvent this algebraic complexity. One way is to use the method of components, which we examine in the next section. The other way is to draw the vectors to scale, as is done in navigation, and read approximate vector lengths and angles (directions) from the graphs. In this section we examine the second approach.

If we need to add three or more vectors, we repeat the parallelogram rule for the pairs of vectors until we find the resultant of all of the resultants. For three vectors, for example, we first find the resultant of vector 1 and vector 2, and then we find the resultant of this resultant and vector 3. The order in which we select the pairs of vectors does not matter because the operation of vector addition is commutative and associative (see Equation 2.7 and Equation 2.8 ). Before we state a general rule that follows from repetitive applications of the parallelogram rule, let’s look at the following example.

Suppose you plan a vacation trip in Florida. Departing from Tallahassee, the state capital, you plan to visit your uncle Joe in Jacksonville, see your cousin Vinny in Daytona Beach, stop for a little fun in Orlando, see a circus performance in Tampa, and visit the University of Florida in Gainesville. Your route may be represented by five displacement vectors A → , A → , B → B → , C → C → , D → D → , and E → E → , which are indicated by the red vectors in Figure 2.11 . What is your total displacement when you reach Gainesville? The total displacement is the vector sum of all five displacement vectors, which may be found by using the parallelogram rule four times. Alternatively, recall that the displacement vector has its beginning at the initial position (Tallahassee) and its end at the final position (Gainesville), so the total displacement vector can be drawn directly as an arrow connecting Tallahassee with Gainesville (see the green vector in Figure 2.11 ). When we use the parallelogram rule four times, the resultant R → R → we obtain is exactly this green vector connecting Tallahassee with Gainesville: R → = A → + B → + C → + D → + E → R → = A → + B → + C → + D → + E → .

Drawing the resultant vector of many vectors can be generalized by using the following tail-to-head geometric construction . Suppose we want to draw the resultant vector R → R → of four vectors A → A → , B → B → , C → C → , and D → D → ( Figure 2.12 (a)). We select any one of the vectors as the first vector and make a parallel translation of a second vector to a position where the origin (“tail”) of the second vector coincides with the end (“head”) of the first vector. Then, we select a third vector and make a parallel translation of the third vector to a position where the origin of the third vector coincides with the end of the second vector. We repeat this procedure until all the vectors are in a head-to-tail arrangement like the one shown in Figure 2.12 . We draw the resultant vector R → R → by connecting the origin (“tail”) of the first vector with the end (“head”) of the last vector. The end of the resultant vector is at the end of the last vector. Because the addition of vectors is associative and commutative, we obtain the same resultant vector regardless of which vector we choose to be first, second, third, or fourth in this construction.

Example 2.2

Geometric construction of the resultant.

For (c), we can start with vector −3 B → −3 B → and draw the remaining vectors tail-to-head as shown in Figure 2.15 . In vector addition, the order in which we draw the vectors is unimportant, but drawing the vectors to scale is very important. Next, we draw vector S → S → from the origin of the first vector to the end of the last vector and place the arrowhead at the end of S → S → . We use a ruler to measure the length of S → S → , and find that its magnitude is S = 36.9 cm. We use a protractor and find that its direction angle is θ S = 52.9 ° θ S = 52.9 ° . This solution is shown in Figure 2.15 .

Check Your Understanding 2.3

Using the three displacement vectors A → A → , B → B → , and F → F → in Figure 2.13 , choose a convenient scale, and use a ruler and a protractor to find vector G → G → given by the vector equation G → = A → + 2 B → − F → G → = A → + 2 B → − F → .

Interactive

Observe the addition of vectors in a plane by visiting this vector calculator , and by engaging the Phet simulation below.

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HUA Problem Solving Team Has Yet to Convene After 2 Weeks

Two weeks after the Harvard Undergraduate Association began the process of forming a problem solving team and indefinitely postponed all student referenda — including the College-wide referendum on divestment from Israel — the team still has not set a time for its first meeting.

On April 12, two HUA officers filed a motion to create a problem solving team to “solve a dispute” over referenda guidelines, after the HUA approved a petition from the Palestine Solidarity Committee for a referendum on whether Harvard should divest from institutions with ties to Israel’s settlements and its war on Gaza. The creation of the team led to a delay in all HUA-held referenda.

The HUA’s announcement of the postponement of all referenda came under intense fire from the PSC, who protested the decision in a rally on April 13 and chanted outside the Smith Center: “Hey hey, ho ho, the HUA should let us vote.”

The process to start forming a problem-solving team began after the HUA received a petition appearing to parody the PSC petition from a group calling themselves “Are Harvard Students, Students Against Hate?”

Seven students were selected to serve on the problem solving team last Friday, according to a communication to the team obtained by The Crimson. A day before the team was finalized, the HUA had sent out an email to the undergraduate student body inviting students to apply to be randomly selected for the team.

Now, with Harvard’s recent suspension of the PSC and pro-Palestine organizers currently encamped in Harvard Yard to protest the suspension, the problem solving team may face additional complications in deciding the fate of a referendum initiated by a suspended student organization.

More than 500 Harvard affiliates gathered in Harvard Yard for an emergency protest of the PSC’s suspension at noon on Wednesday, as a smaller group of protesters began putting up the encampment.

On Tuesday, former HUA co-treasurer and Crimson Editorial editor Josh A. Kaplan ’26 circulated a “charge sheet” document to the problem solving team.

It is unclear why the communication came from Kaplan, whose term as HUA co-treasurer ended on April 20 and who did not respond to a request for comment.

According to the document, the team's recommendations should be sent to HUA’s executive officers, headed by newly elected co-presidents Ashley C. Adirika ’26 and Jonathan Haileselassie ’26.

According to the document, HUA executive officers believed a potential interpretation of the HUA’s constitution indicates “a referendum cannot be held if the referendum asks for student feedback on Harvard issues that are outside the Harvard Undergraduate Association’s scope of governance.”

The document also states that “the HUA has received multiple questions addressing a variety of campus and global issues,” including “the usage of Harvard facilities, the names of Harvard College buildings, the Harvard University endowment, and other Harvard University schools.”

Under a heading titled “Your Charge,” the problem solving team is asked to provide recommendations to the HUA’s executive team on how to solve the constitutional dispute and “clearly define” the HUA’s constitutional process for all future referenda.

“The intent of this investigation is to clearly define the Constitutional process for the HUA Referendum process for future referendum,” the document says. “This Problem-Solving Team is directed to focus on the referendum process, not the content of the proposed referendum.”

—Staff writer Cam N. Srivastava can be reached at [email protected] . Follow him on X @camsrivastava .

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Climate advocates want to solve their 'biggest problem' in the US: Turning out voters

ABC News embedded with an environmental group canvassing in Pittsburgh.

ABC News is taking a look at solutions for issues related to climate change and the environment with the series "The Power of Us: People, The Climate, and Our Future."

In battleground states across the country, environmental activists like Dr. Emily Church are canvassing on behalf of an organization called the Environmental Voter Project in an effort to turn out people who care the most about climate change -- but who haven't shown up for past elections.

During a recent effort in Pittsburgh, Church, a biology professor who leads local canvasses for the project, recalled to ABC News how she used to lobby lawmakers directly to take action on climate change, but they told her voters don't care about the issue.

She said she's now trying to prove them wrong.

"The people who prioritize climate and the environment need to show up," Church said. "That's how we're going to get anything done."

The Environmental Voter Project, or EVP, is targeting very specific individuals: registered voters who list climate change as their No. 1 issue but who are unlikely to cast ballots in November's election based on their voting history.

"Our biggest problem in the climate movement right now [is] we don't have enough voting power," EVP founder and executive Nathaniel Stinnett said.

EVP takes a targeted approach to door knocking, Stinnett explained. Using polling, the group first determines which registered voters in a particular area, like Pittsburgh, would rank climate as their top voting issue. They then cross-reference profiles with voting records to find people who have not come out to the polls recently or regularly.

By Stinnett's accounting, the group has been successful across general elections, primaries and even in local races.

"We've sometimes increased turnout by as much as 1.8 percentage points in general elections, 3.6 points in primaries and 5.7 points in local elections," he said, noting that while 1.8 percentage points might sound small, it could determine an election. Pennsylvania, for example, was only won by President Joe Biden in 2020 by 1.17%.

For the canvassing effort in Pittsburgh, Stinnett said EVP targeted people who didn't vote in the 2020 election or elections in the years since. He added that they identified 22,135 voters in the city who are highly likely to rank climate as their top priority but unlikely to vote in November.

The group claims nonpartisanship but acknowledges that right now it's Democrats working on climate change almost exclusively. One of their hopes is to bring more Republicans to the table, too.

"We want to scare the bejesus out of as many politicians as possible, no matter what side of the aisle they're on, until they think, 'You know what, the only way I can win elections is if I start recognizing the biggest crisis,'" Stinnett said.

MORE:Coastal US cities are sinking as sea levels continue to rise, new research shows

Over time, climate change has become a more salient voting issue. In 2010, only a slim majority of Americans agreed that global warming was occurring, according to polling by the Yale Program on Climate Change Communication . Now, 72% of Americans agree.

But climate is currently not one of the biggest motivators for people this election cycle, surveys have indicated -- though climate advocates hope to change the electorate by encouraging turnout of climate-concerned voters.

According to a February poll by the Wall Street Journal , registered voters listed immigration (20%), the economy (14%), abortion (8%) and democracy (8%) as their top issues. Climate change ranked 11th, with 2% of voters choosing it as their top issue.

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More broadly, Gallup's tracking of what Americans say is the country's most important problem over time shows climate, pollution and the environment at 2% in March, far below economic issues and immigration.

Polling has also shown that in addition to a partisan divide on the issue, a generational shift may be at play.

"Young voters in general tend to be more Democratic, and that is kind of tied up inextricably with their belief that climate is really important," said Nathaniel Rakich, a senior editor and senior elections analyst at 538. "So if Republicans don't want to basically be losing this upcoming electorate by large margins for decades to come, they're going to have to eat into that Democratic support by at least proposing some solutions and addressing climate change."

Even the Biden administration, which has prioritized fighting climate change, is being pushed by progressives to do more on the issue.

Twenty-one activists with the environmental advocacy group Sunrise Movement were arrested outside of Biden campaign headquarters in Wilmington, Delaware, in February. That group and other advocates have additional demonstrations planned in the run-up to the November election.

"I think there were some missteps by the administration -- permitting the Willow project in Alaska was a step backwards. That was unfortunate," Evergreen Action Executive Director Lena Moffitt said, referring to a large-scale oil drilling initiative backed by Alaska lawmakers and others in the state for its economic value, but which environmentalists criticized as undercutting the White House's climate goals.

MORE: Biden admin approves Willow Project despite oil drilling concerns

"We know that we need to move away from fossil fuels and, at the same time, the administration is doing a lot to hasten that move away from fossil fuels," Moffitt said.

The choice for voters in November, on the issue of climate, is stark. President Biden has spoken urgently of the dangers of not slowing climate change and has pushed renewable energy solutions, backed electric vehicle infrastructure and created a new Climate Corps to train and expand the environmental workforce.

Biden last week finalized new protections against oil and gas production for some13 million acres of land in Alaska and, through the Environmental Protection Agency, has imposed aggressive emissions standards for vehicles to cut future greenhouse gases.

Meanwhile, former President Donald Trump, who has long questioned climate science, without evidence, has opposed Biden's clean energy policies and promised to roll them back -- arguing they are a drag on the economy and make the U.S. less competitive and independent.

MORE: EPA sets new emissions standards for heavy-duty vehicles in effort to fight climate change

"The fact is President Biden has done more to address climate change than any president in U.S. history. And there's a lot more to be done," Moffitt said. "Scientists have said that we still can avoid the worst of the worst of the climate crisis. But what we do in these next few years is essential to which path we choose."

Stinnett agreed, telling ABC News that too often Americans have been told to focus on their own individual habits rather than government policy.

"[Politicians say,] 'Hey, don't pay attention to that coal-fired power plant back there. Instead, it's all your fault for having a plastic water bottle in your hand.' And we bought it. We bought it hook, line and sinker," he said. "In truth, it is far more of a political and a systemic problem that needs political and systemic solutions."

In Pittsburgh, Church said that despite the difficulty in getting new, environmentally minded voters to the polls, she thinks the challenge is worth it.

"The science is very clear. So we know what we need to do," she said, "it's just a matter of getting it done."

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The Annenberg School of Journalism at the University of Southern California seeks a highly qualified adjunct instructor to teach  JOUR 499: Problem Solving with AI. This course dives into real-time case studies that examine how AI tools are opening up new frontiers in newsrooms, writers’ rooms and boardrooms. 

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The USC Annenberg School for Communication and Journalism is among the nation’s leading institutions devoted to the study of communication, journalism and public relations.  With an enrollment of 2,400 undergraduate and graduate students, USC Annenberg scholars, both students and faculty, are defining these fields for the 21st century and beyond.

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SevenRooms CEO Joel Montaniel Is Solving the Hospitality Industry’s Data Problem

Sevenrooms ceo joel montaniel speaks about how his crm company helps restaurants redefine their relationship with customers..

Joel Montaniel ’s SevenRooms is more than just another restaurant reservation system. “If you ask a restaurant, ‘What are your top 10, your top 25, and your top 50 customers?’ There are 15 million restaurants on the planet and 14.95 million of them can’t answer that question today,” Montaniel tells Observer. So, SevenRooms set out to solve that problem—by offering a suite of tools that empower restaurants and other types of hospitality operators to take full control and make the best use of their customer data. 

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“Basically what Shopify has been doing for retail is what we’re doing for restaurants,” Montaniel explained. “We help restaurants do three things: make money, build relationships with their customers, and then streamline their operations.” Montaniel, a former investment banker and real estate investor, founded SevenRooms in 2011 with his childhood friend Kinesh Patel and former Credit Suisse (CS) colleague Allison Page . Inspired by the   former Vanity Fair editor Graydon Carter ‘s “seven rooms theory,” the company aspires to create spaces where patrons feel at home, regardless of the venue’s exclusivity. From high-end establishments to smaller independent eateries, SevenRooms caters to a diverse range of hospitality operators, boasting an impressive clientele of more than 11,000 hospitality operators around the world.

Montaniel was named on Observer’s 2023 Nightlife + Dining Power List. We interviewed him earlier this month about SevenRooms’ founding story, his journey of finding the right business model, and the broader landscape of the hospitality industry amid high inflation and shifting consumer behavior. The following conversation has been edited for length and clarity.

Observer: How did you meet your co-founders and how did the idea of SevenRooms come about? 

Joel Montaniel: There are three of us. One of them, our CTO, Kinesh, I grew up with in Texas, so we’ve known each other since we were 10 years old. My other co-founder, Alison, and I met at our first jobs out of school—we were investment banking analysts and shared a cubicle wall.

At first, our idea was to solve a consumer problem and try to make it easier for consumers to go to these exclusive places. We launched a website in 2009 and it failed miserably. W hat we learned was that there actually was a bigger problem to solve. When we looked at the systems that hospitality operators were using, they didn’t have any customer data. So, we switched over from thinking about the consumer to thinking about the restaurant or the operator. We found that by solving that problem and giving them the technology that can help them understand their customers, that could then translate into an amazing customer experience, which is what we were trying to solve in the first place.

Before diving into the SevenRooms we know today, in retrospect, why do you think your original idea “failed miserably?”

I think we weren’t solving a real problem at that point in time, and I don’t think the world was ready for something like that. I remember talking to a restaurant and they told us point blank, “We sell dinner here, we don’t sell the right to have dinner.” 

We learned that we shouldn’t be solving things that we think are cool. We should work on things that actually create real value based on problems that we’ve experienced ourselves or problems that the industry talked to us about.

How would you describe SevenRooms’s core products and services? 

The way we think about ourselves is basically what Shopify is doing for retail is what we’re doing for restaurants. At the center is CRM—that is guest data. Then, there’s operations marketing, and commerce that sit on top of it. We work with restaurants, hotels and entertainment venues. We help them do three things: make money, build relationships with their customers, and streamline their operations.

What was the traditional approach to achieving those three objectives and what innovation and value does SevenRooms provide?

The traditional approach would have been to use a reservation system. In the U.S., that means using an OpenTable or a Resy system, for example. Those systems are really good at bringing customers in. But what hospitality operators haven’t really made as much investment in is the guest database and marketing tools. All of the guest information would be stored in their heads. 

If you ask a restaurant, “What are your top 10, your top 25, and your top 50 customers?” There are 15 million restaurants on the planet and 14.95 million of them can’t answer that question today, because restaurants don’t have the customer data, they’re not really doing retention marketing. 

On your website, it says the company’s name was inspired by the “seven rooms theory.” Can you explain what that is and why it’s a fitting name for your company?

We believe technology is more than just transactional, and so we wanted to find a name that is about helping companies build relationships with people. W e stumbled across this theory by the former Vanity Fair editor Graydon Carter called “the seven rooms theory.” It says that, in New York, there are seven interconnected rooms, each one more exclusive than the one before it. It suggests that when you think you’re at the top spot and the best place, there’s always another room that you don’t have access to that you haven’t discovered yet.

It hit the nail on the head for us where, you know, the seventh room is not about exclusivity; it’s about the place you go to where you feel most at home. For one of us, it might be a coffee shop, and for another one of us, it might be a bar. That captures the essence of what we want to do, which is we’re in business to help restaurants and hospitality operators make their guests feel at home. 

What types of hospitality operators does SevenRooms work with?

At the very beginning, we were working with more higher-end restaurants and nightclubs. These were places that want to make people feel special. One of the very first customers we had was Tao Group Hospitality in New York. Another really early customer was LDV Hospitality, which owns Scarpetta and other places. They were super helpful in helping us understand the technology and what customer information mattered.

But today, we are in every size and dollar amount. We have over 11,000 restaurants using the platform.  We power 80 percent of Las Vegas today, so all of MGM, Wynn, Cosmo and Venetian. We have everything from big hotel groups like Marriott to small, independent restaurants. And we have everything from inexpensive to $200-plus per person. As an example, 40 percent of our customers are under $50 per person and 9 percent are under $25. 

Covid-19 and high inflation for the past two years have changed consumer behavior in many ways. What are some of the emerging trends you’ve observed in hospitality?

Hospitality has been resilient overall. Sales are roughly flat, but they’re not going down. People are going out a little bit more, but they’re spending a little bit less.

We will continue to see that trend, which I think is largely driven by an older demographic that has a lot of disposable income. They’re spending money on travel and they’re going out to eat. Then you have a younger generation that is all about the experiences, all about Instagram and TikTok. The reason I think restaurants are really taking off is that they’re the new kind of social currency, and people are posting about food, restaurants and even dinner reservations on social media. I’m finding that food and experiences are central to both the younger generation and the older generation. 

• SEE ALSO : FCC Chair Jessica Rosenworcel Supports TikTok Ban, Calls Lack of Oversight ‘Stunning’

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