13.4 The Dot Product

In Section 13.3, we learned how add and subtract vectors and how to multiply vectors by scalars. In this section, we define a product of vectors. We begin with the following definition.

Definition

Given vectors $vector v equals ⟨ v sub 1, v sub 2 ⟩$ and $vector w equals ⟨ w sub 1, w sub 2 ⟩$ , the dot product of $\vec{v}$ and $\vec{w}$ is given by

$\vec{v} \cdot \vec{w} = ⟨ v_{1}, v_{2} ⟩ \cdot ⟨ w_{1}, w_{2} ⟩ = v_{1} w_{1} + v_{2} w_{2}$

For example, if $vector v equals ⟨ 3, 4 ⟩$ and $\vec{w} = ⟨ 1, - 2 ⟩$ ,then $\vec{v} \cdot \vec{w} = ⟨ 3, 4 ⟩ \cdot ⟨ 1, - 2 ⟩ = (3) (1) + (4) (- 2) = - 5$ .

Note that the dot product takes two vectors and produces a scalar. For that reason, the quantity $\vec{v} \cdot \vec{w}$ is often called the scalar product of $\vec{v}$ and $\vec{w}$ . The dot product enjoys the following properties.

Theorem

Properties of the Dot Product

Commutative Property: For all vectors $\vec{v}$ and $\vec{w}$ , $\vec{v} \cdot \vec{w} = \vec{w} \cdot \vec{v}$ .
Distributive Property: For all vectors $\vec{u}$ , $\vec{v}$ and $\vec{w}$ , $\vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w}$ .
Scalar Property: For all vectors $\vec{v}$ and $\vec{w}$ and scalars $k$ , $(k \vec{v}) \cdot \vec{w} = k (\vec{v} \cdot \vec{w}) = \vec{v} \cdot (k \vec{w})$ .
Relation to Magnitude: For all vectors $\vec{v}$ , $\vec{v} \cdot \vec{v} = {‖ \vec{v} ‖}^{2}$ .

Like most of the theorems involving vectors, the proof of Theorem 13.10 amounts to using the definition of the dot product and properties of real number arithmetic.

For example, to show the commutative property, let $vector v equals ⟨ v sub 1, v sub 2 ⟩$ and $vector w equals ⟨ w sub 1, w sub 2 ⟩$ . Then

$\begin{matrix} \vec{v} \cdot \vec{w} & = & ⟨ v_{1}, v_{2} ⟩ \cdot ⟨ w_{1}, w_{2} ⟩ \\ = & v_{1} w_{1} + v_{2} w_{2} & Definition of Dot Product \\ = & w_{1} v_{1} + w_{2} v_{2} & Commutativity of Real Number Multiplication \\ = & ⟨ w_{1}, w_{2} ⟩ \cdot ⟨ v_{1}, v_{2} ⟩ & Definition of Dot Product \\ = & \vec{w} \cdot \vec{v} \end{matrix}$

The distributive property is proved similarly and is left as an exercise.

For the scalar property, assume that $vector v equals ⟨ v sub 1, v sub 2 ⟩$ and $vector w equals ⟨ w sub 1, w sub 2 ⟩$ and $k$ is a scalar. Then

$\begin{matrix} (k \vec{v}) \cdot \vec{w} & = & (k ⟨ v_{1}, v_{2} ⟩) \cdot ⟨ w_{1}, w_{2} ⟩ \\ = & ⟨ k v_{1}, k v_{2} ⟩ \cdot ⟨ w_{1}, w_{2} ⟩ & Definition of Scalar Multiplication \\ = & (k v_{1}) (w_{1}) + (k v_{2}) (w_{2}) & Definition of Dot Product \\ = & k (v_{1} w_{1}) + k (v_{2} w_{2}) & Associativity of Real Number Multiplication \\ = & k (v_{1} w_{1} + v_{2} w_{2}) & Distributive Law of Real Numbers \\ = & k ⟨ v_{1}, v_{2} ⟩ \cdot ⟨ w_{1}, w_{2} ⟩ & Definition of Dot Product \\ = & k (\vec{v} \cdot \vec{w}) \end{matrix}$

We leave the proof of $k (\vec{v} \cdot \vec{w}) = \vec{v} \cdot (k \vec{w})$ as an exercise.

For the last property, we note that if $vector v equals ⟨ v sub 1, v sub 2 ⟩$ , then $\vec{v} \cdot \vec{v} = ⟨ v_{1}, v_{2} ⟩ \cdot ⟨ v_{1}, v_{2} ⟩ = v_{1}^{2} + v_{2}^{2} = {‖ \vec{v} ‖}^{2}$ , where the last equality comes courtesy of Definition 13.4.

The following example puts Theorem 13.10 to good use. As in Example 13.3.3, we work out the problem in great detail and encourage the reader to supply the justification for each step.

Example 1

Prove the identity: ${‖ \vec{v} - \vec{w} ‖}^{2} = {‖ \vec{v} ‖}^{2} - 2 (\vec{v} \cdot \vec{w}) + {‖ \vec{w} ‖}^{2}$ .

We begin by rewriting ${‖ \vec{v} - \vec{w} ‖}^{2}$ in terms of the dot product using Theorem 13.10.

$\begin{matrix} {‖ \vec{v} - \vec{w} ‖}^{2} & = & (\vec{v} - \vec{w}) \cdot (\vec{v} - \vec{w}) \\ = & (\vec{v} + [- \vec{w}]) \cdot (\vec{v} + [- \vec{w}]) \\ = & (\vec{v} + [- \vec{w}]) \cdot \vec{v} + (\vec{v} + [- \vec{w}]) \cdot [- \vec{w}] \\ = & \vec{v} \cdot (\vec{v} + [- \vec{w}]) + [- \vec{w}] \cdot (\vec{v} + [- \vec{w}]) \\ = & \vec{v} \cdot \vec{v} + \vec{v} \cdot [- \vec{w}] + [- \vec{w}] \cdot \vec{v} + [- \vec{w}] \cdot [- \vec{w}] \\ = & \vec{v} \cdot \vec{v} + \vec{v} \cdot [(- 1) \vec{w}] + [(- 1) \vec{w}] \cdot \vec{v} + [(- 1) \vec{w}] \cdot [(- 1) \vec{w}] \\ = & \vec{v} \cdot \vec{v} + (- 1) (\vec{v} \cdot \vec{w}) + (- 1) (\vec{w} \cdot \vec{v}) + [(- 1) (- 1)] (\vec{w} \cdot \vec{w}) \\ = & \vec{v} \cdot \vec{v} + (- 1) (\vec{v} \cdot \vec{w}) + (- 1) (\vec{v} \cdot \vec{w}) + \vec{w} \cdot \vec{w} \\ = & \vec{v} \cdot \vec{v} - 2 (\vec{v} \cdot \vec{w}) + \vec{w} \cdot \vec{w} \\ = & {‖ \vec{v} ‖}^{2} - 2 (\vec{v} \cdot \vec{w}) + {‖ \vec{w} ‖}^{2} \end{matrix}$

Hence, ${‖ \vec{v} - \vec{w} ‖}^{2} = {‖ \vec{v} ‖}^{2} - 2 (\vec{v} \cdot \vec{w}) + {‖ \vec{w} ‖}^{2}$ as required.

If we take a step back from the pedantry in Example 13.4.1, we see that the bulk of the work is needed to show that $(\vec{v} - \vec{w}) \cdot (\vec{v} - \vec{w}) = \vec{v} \cdot \vec{v} - 2 (\vec{v} \cdot \vec{w}) + \vec{w} \cdot \vec{w}$ . If this looks familiar, it should.

Since the dot product enjoys many of the same properties enjoyed by real numbers, the machinations required to expand $(\vec{v} - \vec{w}) \cdot (\vec{v} - \vec{w})$ for vectors $\vec{v}$ and $\vec{w}$ match those required to expand $(v - w) (v - w)$ for real numbers $v$ and $w$ , and hence we get similar looking results.

The identity verified in Example 13.4.1 plays a large role in the development of the geometric properties of the dot product, which we now explore.

Suppose $\vec{v}$ and $\vec{w}$ are two nonzero vectors. If we draw $\vec{v}$ and $\vec{w}$ with the same initial point, we define the angle between $\vec{v}$ and $\vec{w}$ to be the angle $θ$ determined by the rays containing the vectors $\vec{v}$ and $\vec{w}$ , as illustrated below. We require $0 less than or equal to θ less than or equal to π$ . (Think about why this is needed in the definition.)

Coordinate-plane figure. — **Figure 13.48**

$θ equals 0$

$0 less than θ less than π$

$θ = π$

The following theorem gives us some insight into the geometric role the dot product plays.

Theorem

Geometric Interpretation of Dot Product: If $\vec{v}$ and $\vec{w}$ are nonzero vectors then

$\vec{v} \cdot \vec{w} = ‖ \vec{v} ‖ ‖ \vec{w} ‖ \cos (θ),$

where $θ$ is the angle between $\vec{v}$ and $\vec{w}$ .

We prove Theorem 13.11 in cases. If $θ equals 0$ , then $\vec{v}$ and $\vec{w}$ have the same direction. It follows¹ that there is a real number $k greater than 0$ so that $\vec{w} = k \vec{v}$ . Hence, $\vec{v} \cdot \vec{w} = \vec{v} \cdot (k \vec{v}) = k (\vec{v} \cdot \vec{v}) = k {‖ \vec{v} ‖}^{2}$ .

Working from the other end of the equation, $‖ \vec{v} ‖ ‖ \vec{w} ‖ \cos (θ) = ‖ \vec{v} ‖ ‖ k \vec{v} ‖ \cos (0) = ‖ \vec{v} ‖ (| k | ‖ \vec{v} ‖) (1) = k {‖ \vec{v} ‖}^{2}$ , where $‖ k \vec{v} ‖ = | k | ‖ \vec{v} ‖$ courtesy of Theorem 13.8, and $| k | = k$ since $k greater than 0$ .

Hence, in the case $θ equals 0$ , we have shown $\vec{v} \cdot \vec{w} = k {‖ \vec{v} ‖}^{2}$ and $‖ \vec{v} ‖ ‖ \vec{w} ‖ \cos (θ) = k {‖ \vec{v} ‖}^{2}$ . Putting these two equations together shows that $\vec{v} \cdot \vec{w} = ‖ \vec{v} ‖ ‖ \vec{w} ‖ \cos (θ)$ holds in this case.

If $θ = π$ , $\vec{v}$ and $\vec{w}$ have the exact opposite directions, so there is a real number $k less than 0$ with $\vec{w} = k \vec{v}$ .

As before, we compute $\vec{v} \cdot \vec{w} = \vec{v} \cdot (k \vec{v}) = k (\vec{v} \cdot \vec{v}) = k {‖ \vec{v} ‖}^{2}$ . Since $k less than 0$ here, we have $| k | = - k$ . Hence, we find $‖ \vec{v} ‖ ‖ \vec{w} ‖ \cos (θ) = ‖ \vec{v} ‖ ‖ k \vec{v} ‖ \cos (π) = ‖ \vec{v} ‖ (| k | ‖ \vec{v} ‖) (- 1) = ‖ \vec{v} ‖ (- k) ‖ \vec{v} ‖ (- 1) = k {‖ \vec{v} ‖}^{2}$ .

Once again, both $\vec{v} \cdot \vec{w} = k {‖ \vec{v} ‖}^{2}$ and $‖ \vec{v} ‖ ‖ \vec{w} ‖ \cos (θ) = k {‖ \vec{v} ‖}^{2}$ , so $\vec{v} \cdot \vec{w} = ‖ \vec{v} ‖ ‖ \vec{w} ‖ \cos (θ)$ in this case.

Next, if $0 less than θ less than π$ , the vectors $\vec{v}$ , $\vec{w}$ and $\vec{v} - \vec{w}$ determine a triangle with side lengths $‖ \vec{v} ‖$ , $‖ \vec{w} ‖$ and $‖ \vec{v} - \vec{w} ‖$ , respectively, as seen in the diagram below.

The Law of Cosines yields ${‖ \vec{v} - \vec{w} ‖}^{2} = {‖ \vec{v} ‖}^{2} + {‖ \vec{w} ‖}^{2} - 2 ‖ \vec{v} ‖ ‖ \vec{w} ‖ \cos (θ)$ . From Example 13.4.1, we also have that ${‖ \vec{v} - \vec{w} ‖}^{2} = {‖ \vec{v} ‖}^{2} - 2 (\vec{v} \cdot \vec{w}) + {‖ \vec{w} ‖}^{2}$ .

Equating these two expressions for ${‖ \vec{v} - \vec{w} ‖}^{2}$ gives ${‖ \vec{v} ‖}^{2} + {‖ \vec{w} ‖}^{2} - 2 ‖ \vec{v} ‖ ‖ \vec{w} ‖ \cos (θ) = {‖ \vec{v} ‖}^{2} - 2 (\vec{v} \cdot \vec{w}) + {‖ \vec{w} ‖}^{2}$ which reduces to $- 2 ‖ \vec{v} ‖ ‖ \vec{w} ‖ \cos (θ) = - 2 (\vec{v} \cdot \vec{w})$ . Hence, $\vec{v} \cdot \vec{w} = ‖ \vec{v} ‖ ‖ \vec{w} ‖ \cos (θ)$ , as required.

An immediate consequence of Theorem 13.11 is the following.

Theorem

Let $\vec{v}$ and $\vec{w}$ be nonzero vectors and let $θ$ the angle between $\vec{v}$ and $\vec{w}$ . Then

$θ = \arccos (\frac{\vec{v} \cdot \vec{w}}{‖ \vec{v} ‖ ‖ \vec{w} ‖}) = \arccos (\hat{v} \cdot \hat{w})$

We obtain the formula in Theorem 13.12 by solving the equation given in Theorem 13.11 for $θ$ .

Since $\vec{v}$ and $\vec{w}$ are nonzero, so are $‖ \vec{v} ‖$ and $‖ \vec{w} ‖$ . Hence, we may divide both sides of $\vec{v} \cdot \vec{w} = ‖ \vec{v} ‖ ‖ \vec{w} ‖ \cos (θ)$ by $‖ \vec{v} ‖ ‖ \vec{w} ‖$ . Since $0 less than or equal to θ less than or equal to π$ by definition, the values of $θ$ exactly match the range of the arccosine function. Hence,

$\cos (θ) = \frac{\vec{v} \cdot \vec{w}}{‖ \vec{v} ‖ ‖ \vec{w} ‖} \Rightarrow θ = \arccos (\frac{\vec{v} \cdot \vec{w}}{‖ \vec{v} ‖ ‖ \vec{w} ‖}) .$

Using Theorem 13.10, we can rewrite

$\frac{\vec{v} \cdot \vec{w}}{‖ \vec{v} ‖ ‖ \vec{w} ‖} = (\frac{1}{‖ \vec{v} ‖} \vec{v}) \cdot (\frac{1}{‖ \vec{w} ‖} \vec{w}) = \hat{v} \cdot \hat{w},$

giving us the alternative formula listed in Theorem 13.12: $θ = \arccos (\hat{v} \cdot \hat{w})$ . We are overdue for an example.

Example 2

Find the angle between the following pairs of vectors. Graph each pair of vectors in standard position to check the reasonableness of your answer.

$\vec{v} = ⟨ 3, - 3 \sqrt{3} ⟩$ , and $\vec{w} = ⟨ - \sqrt{3}, 1 ⟩$
$vector v equals ⟨ 2, 2 ⟩$ , and $\vec{w} = ⟨ 5, - 5 ⟩$
$\vec{v} = ⟨ 3, - 4 ⟩$ , and $vector w equals ⟨ 2, 1 ⟩$

We use the formula $θ = \arccos (\frac{\vec{v} \cdot \vec{w}}{‖ \vec{v} ‖ ‖ \vec{w} ‖})$ from Theorem 13.12 in each case below.

We have $\vec{v} \cdot \vec{w} = ⟨ 3, - 3 \sqrt{3} ⟩ \cdot ⟨ - \sqrt{3}, 1 ⟩ = - 3 \sqrt{3} - 3 \sqrt{3} = - 6 \sqrt{3}$ . Computing lengths of vectors, we find $‖ \vec{v} ‖ = \sqrt{3^{2} + {(- 3 \sqrt{3})}^{2}} = \sqrt{36} = 6$ and $‖ \vec{w} ‖ = \sqrt{{(- \sqrt{3})}^{2} + 1^{2}} = \sqrt{4} = 2$ . Hence, we find $θ = \arccos (\frac{- 6 \sqrt{3}}{12}) = \arccos (- \frac{\sqrt{3}}{2}) = \frac{5 π}{6}$ . We check our answer geometrically by graphing this pair of vectors below on the left.
For $vector v equals ⟨ 2, 2 ⟩$ and $\vec{w} = ⟨ 5, - 5 ⟩$ , we find $\vec{v} \cdot \vec{w} = ⟨ 2, 2 ⟩ \cdot ⟨ 5, - 5 ⟩ = 10 - 10 = 0$ . Hence, it doesn't matter what $‖ \vec{v} ‖$ and $‖ \vec{w} ‖$ are, $θ = \arccos (\frac{\vec{v} \cdot \vec{w}}{‖ \vec{v} ‖ ‖ \vec{w} ‖}) = \arccos (0) = \frac{π}{2}$ . We check our answer geometrically by graphing this pair of vectors below on the right.
Figure 13.53
Figure 13.54
$\vec{v}$ and $\vec{w}$ from number
$\vec{v}$ and $\vec{w}$ from number
We find $\vec{v} \cdot \vec{w} = ⟨ 3, - 4 ⟩ \cdot ⟨ 2, 1 ⟩ = 6 - 4 = 2$ . Computing lengths, we find $‖ \vec{v} ‖ = \sqrt{3^{2} + {(- 4)}^{2}} = \sqrt{25} = 5$ and $\vec{w} = \sqrt{2^{2} + 1^{2}} = \sqrt{5}$ , so $θ = \arccos (\frac{2}{5 \sqrt{5}}) = \arccos (\frac{2 \sqrt{5}}{25})$ .
Since $\frac{2 \sqrt{5}}{25}$ isn't the cosine of one of the `common angles,' we leave our exact answer in terms of the arccosine function. For the purposes of checking our answer, however, we approximate $θ approximately equals 79.7 degrees$ .
Figure 13.55
$\vec{v}$ and $\vec{w}$ from number

A few remarks about Example 13.4.2 are in order. Note that for nonzero vectors $\vec{v}$ and $\vec{w}$ , the lengths $‖ \vec{v} ‖$ and $‖ \vec{w} ‖$ are always positive. Since Theorem 13.11 tells us that $\vec{v} \cdot \vec{w} = ‖ \vec{v} ‖ ‖ \vec{w} ‖ \cos (θ)$ , we know the sign of $\vec{v} \cdot \vec{w}$ is the same as the sign of $\cos (θ)$ .

Geometrically, if $vector v times vector w less than 0$ , then $cos ⁡ open parenthesis θ close parenthesis less than 0$ so $θ$ is an obtuse angle, demonstrated number above.

If $\vec{v} \cdot \vec{w} = 0$ , then $cos ⁡ open parenthesis θ close parenthesis equals 0$ so $θ = \frac{π}{2}$ as in number. In this case, the vectors $\vec{v}$ and $\vec{w}$ are called orthogonal. Geometrically, when orthogonal vectors are sketched with the same initial point, the lines containing the vectors are perpendicular. Hence, if $\vec{v}$ and $\vec{w}$ are orthogonal, we write $\vec{v} ⟂ \vec{w}$ .

Note there is no `zero product property' for the dot product. As with the vectors in number above, it is quite possible to have $\vec{v} \cdot \vec{w} = 0$ but neither $\vec{v}$ nor $\vec{w}$ be $vector 0$ .

Finally, if $vector v times vector w greater than 0$ , then $cos ⁡ open parenthesis θ close parenthesis greater than 0$ so $θ$ is an acute angle, as in the case of number above.

We summarize all of our observations in the schematic below.

$vector v times vector w less than 0$

$\vec{v} \cdot \vec{w} = 0$

$vector v times vector w greater than 0$

$θ$ is obtuse

$θ = \frac{π}{2} = 90^{\circ}$

$θ$ is acute

Of the three cases diagrammed above, the one which has the most mathematical significance moving forward is the orthogonal case. Hence, we state the corresponding theorem below.

Basically, Theorem 13.13 tells us that `the dot product detects orthogonality.' This is a helpful interpretation to keep in mind as you continue your study of vectors in later courses.

We have already argued one direction of Theorem 13.13, namely if $\vec{v} \cdot \vec{w} = 0$ then $\vec{v} ⟂ \vec{w}$ in the comments following Example 13.4.2.

To show the converse, we note if $\vec{v} ⟂ \vec{w}$ , then the angle between $\vec{v}$ and $\vec{w}$ , $θ = \frac{π}{2}$ . From Theorem 13.11, we have that $\vec{v} \cdot \vec{w} = ‖ \vec{v} ‖ ‖ \vec{w} ‖ \cos (\frac{π}{2}) = ‖ \vec{v} ‖ ‖ \vec{w} ‖ \cdot (0) = 0$ , as required.

We can use Theorem 13.13 in the following example to provide a different proof about the relationship between the slopes of perpendicular lines.²

Example 3

Let $L sub 1$ be the line $y = m_{1} x + b_{1}$ and let $L sub 2$ be the line $y = m_{2} x + b_{2}$ . Prove that $L sub 1$ is perpendicular to $L sub 2$ if and only if $m_{1} \cdot m_{2} = - 1$ .

Our strategy is to find two vectors: $vector v sub 1$ , which has the same direction as $L sub 1$ , and $vector v sub 2$ , which has the same direction as $L sub 2$ and show $vector v sub 1 ⟂ vector v sub 2$ if and only if $m_{1} m_{2} = - 1$ .

To that end, we substitute $x equals 0$ and $x equals 1$ into $y = m_{1} x + b_{1}$ to find two points which lie on $L sub 1$ , namely $P ⁢ open parenthesis 0, b sub 1 close parenthesis$ and $Q ⁢ open parenthesis 1, m sub 1 plus b sub 1 close parenthesis$ . We let $\vec{v_{1}} = \vec{P Q} = ⟨ 1 - 0, (m_{1} + b_{1}) - b_{1} ⟩ = ⟨ 1, m_{1} ⟩$ . Since $vector v sub 1$ is determined by two points on $L sub 1$ , it may be viewed as lying on $L sub 1$ , so $vector v sub 1$ has the same direction as $L sub 1$ .

Similarly, we get the vector $vector v sub 2 equals ⟨ 1, m sub 2 ⟩$ which has the same direction as the line $L sub 2$ . Hence, $L sub 1$ and $L sub 2$ are perpendicular if and only if $vector v sub 1 ⟂ vector v sub 2$ . According to Theorem 13.13, $vector v sub 1 ⟂ vector v sub 2$ if and only if $\vec{v_{1}} \cdot \vec{v_{2}} = 0$ .

Notice that $\vec{v_{1}} \cdot \vec{v_{2}} = ⟨ 1, m_{1} ⟩ \cdot ⟨ 1, m_{2} ⟩ = 1 + m_{1} m_{2}$ . Hence, $\vec{v_{1}} \cdot \vec{v_{2}} = 0$ if and only if $1 + m_{1} m_{2} = 0$ , which is true if and only if $m_{1} m_{2} = - 1$ , as required.

Vector Projections

While Theorem 13.13 certainly gives us some insight into what the dot product means geometrically, there is more to the story of the dot product. Consider the two nonzero vectors $\vec{v}$ and $\vec{w}$ drawn with a common initial point $O$ below. For the moment, assume that the angle between $\vec{v}$ and $\vec{w}$ , $θ$ , is acute.

We wish to develop a formula for the vector $\vec{p}$ , indicated below, which is called the orthogonal projection of $\vec{v}$ onto $\vec{w}$ . The vector $\vec{p}$ is obtained geometrically as follows: drop a perpendicular from the terminal point $T$ of $\vec{v}$ to the vector $\vec{w}$ and call the point of intersection $R$ . The vector $\vec{p}$ is then defined as $\vec{p} = \vec{O R}$ .

Like any vector, $\vec{p}$ is determined by its magnitude $‖ \vec{p} ‖$ and its direction $\hat{p}$ according to the formula $\vec{p} = ‖ \vec{p} ‖ \hat{p}$ . Since we want $\hat{p}$ to have the same direction as $\vec{w}$ , we have $\hat{p} = \hat{w}$ .

To determine $‖ \vec{p} ‖$ , we apply Definition B.1 to the right triangle $△ O R T$ . We find $\cos (θ) = \frac{‖ \vec{p} ‖}{‖ \vec{v} ‖}$ , or, equivalently, $‖ \vec{p} ‖ = ‖ \vec{v} ‖ \cos (θ)$ . Using Theorems 13.11 and 13.10, we get:

$‖ \vec{p} ‖ = ‖ \vec{v} ‖ \cos (θ) = \frac{‖ \vec{v} ‖ ‖ \vec{w} ‖ \cos (θ)}{‖ \vec{w} ‖} = \frac{\vec{v} \cdot \vec{w}}{‖ \vec{w} ‖} = \vec{v} \cdot (\frac{1}{‖ \vec{w} ‖} \vec{w}) = \vec{v} \cdot \hat{w} .$

Hence, $‖ \vec{p} ‖ = \vec{v} \cdot \hat{w}$ , and since $\hat{p} = \hat{w}$ , we have $\vec{p} = ‖ \vec{p} ‖ \hat{p} = (\vec{v} \cdot \hat{w}) \hat{w}$ .

Now suppose that the angle $θ$ between $\vec{v}$ and $\vec{w}$ is obtuse, and consider the diagram below.

In this case, we see that $\hat{p} = - \hat{w}$ and using the triangle $△ O R T$ , we find $‖ \vec{p} ‖ = ‖ \vec{v} ‖ \cos (θ^{'})$ . Since $θ + θ^{'} = π$ , it follows that $\cos (θ^{'}) = - \cos (θ)$ , which means $‖ \vec{p} ‖ = ‖ \vec{v} ‖ \cos (θ^{'}) = - ‖ \vec{v} ‖ \cos (θ)$ .

Rewriting this last equation in terms of $\vec{v}$ and $\vec{w}$ as before, we get $‖ \vec{p} ‖ = - (\vec{v} \cdot \hat{w})$ . Putting this together with $\hat{p} = - \hat{w}$ , we get $\vec{p} = ‖ \vec{p} ‖ \hat{p} = - (\vec{v} \cdot \hat{w}) (- \hat{w}) = (\vec{v} \cdot \hat{w}) \hat{w}$ in this case as well.

If the angle between $\vec{v}$ and $\vec{w}$ is $\frac{π}{2}$ then it is easy to show³ that $vector p equals vector 0$ . Since $\vec{v} ⟂ \vec{w}$ in this case, $\vec{v} \cdot \vec{w} = 0$ . It follows that $\vec{v} \cdot \hat{w} = 0$ and $\vec{p} = \vec{0} = 0 \hat{w} = (\vec{v} \cdot \hat{w}) \hat{w}$ in this case, too. We have motivated the following.

Definition

Let $\vec{v}$ and $\vec{w}$ be nonzero vectors.

The orthogonal projection of $\vec{v}$ onto $\vec{w}$ , denoted ${proj}_{\vec{w}} (\vec{v})$ is given by ${proj}_{\vec{w}} (\vec{v}) = (\vec{v} \cdot \hat{w}) \hat{w}$ .

Definition 13.8 gives us a good idea what the dot product does. The scalar $\vec{v} \cdot \hat{w}$ is a measure of how much of the vector $\vec{v}$ is in the direction of the vector $\vec{w}$ and is thus called the scalar projection of $\vec{v}$ onto $\vec{w}$ .

While the formula given in Definition 13.8 is theoretically appealing, because of the presence of the normalized unit vector $\hat{w}$ , computing the projection using the formula ${proj}_{\vec{w}} (\vec{v}) = (\vec{v} \cdot \hat{w}) \hat{w}$ can be messy. We present two other formulas that are often used in practice.

Theorem

Alternate Formulas for Vector Projections: If $\vec{v}$ and $\vec{w}$ are nonzero vectors then

${proj}_{\vec{w}} (\vec{v}) = (\vec{v} \cdot \hat{w}) \hat{w} = (\frac{\vec{v} \cdot \vec{w}}{‖ \vec{w} ‖^{2}}) \vec{w} = (\frac{\vec{v} \cdot \vec{w}}{\vec{w} \cdot \vec{w}}) \vec{w}$

The proof of Theorem 13.14, which we leave to the reader as an exercise, amounts to using the formula $\hat{w} = (\frac{1}{‖ \vec{w} ‖}) \vec{w}$ and properties of the dot product. It is time for an example.

Example 4

Let $vector v equals ⟨ 1, 8 ⟩$ and $\vec{w} = ⟨ - 1, 2 ⟩$ . Find $\vec{p} = {proj}_{\vec{w}} (\vec{v})$ . Check your answer geometrically.

We find $\vec{v} \cdot \vec{w} = ⟨ 1, 8 ⟩ \cdot ⟨ - 1, 2 ⟩ = (- 1) + 16 = 15$ and $\vec{w} \cdot \vec{w} = ⟨ - 1, 2 ⟩ \cdot ⟨ - 1, 2 ⟩ = 1 + 4 = 5$ . Hence,

$\vec{p} = \frac{\vec{v} \cdot \vec{w}}{\vec{w} \cdot \vec{w}} \vec{w} = \frac{15}{5} ⟨ - 1, 2 ⟩ = ⟨ - 3, 6 ⟩ .$

We plot $\vec{v}$ , $\vec{w}$ and $\vec{p}$ in standard position below on the left. We see $\vec{p}$ has the same direction as $\vec{w}$ , but we need to do more to show $\vec{p}$ in is indeed the orthogonal projection of $\vec{v}$ onto $\vec{w}$ .

Consider the vector $\vec{q}$ whose initial point is the terminal point of $\vec{p}$ and whose terminal point is the terminal point of $\vec{v}$ . From the definition of vector arithmetic, $\vec{p} + \vec{q} = \vec{v}$ , so that $\vec{q} = \vec{v} - \vec{p}$ .

Since $vector v equals ⟨ 1, 8 ⟩$ and $\vec{p} = ⟨ - 3, 6 ⟩$ , $\vec{q} = ⟨ 1, 8 ⟩ - ⟨ - 3, 6 ⟩ = ⟨ 4, 2 ⟩$ . To prove $\vec{q} ⟂ \vec{w}$ , we compute the dot product: $\vec{q} \cdot \vec{w} = ⟨ 4, 2 ⟩ \cdot ⟨ - 1, 2 ⟩ = (- 4) + 4 = 0$ . Hence, per Theorem 13.13, we know $\vec{q} ⟂ \vec{w}$ which completes our check.⁴

Figure 13.63

Figure 13.64

In Example 13.4.4 above, writing $\vec{v} = \vec{p} + \vec{q}$ is an example of what is called a vector decomposition of $\vec{v}$ . We generalize this result in the following theorem.

Theorem

Generalized Decomposition Theorem: Let $\vec{v}$ and $\vec{w}$ be nonzero vectors. There are unique vectors $\vec{p}$ and $\vec{q}$ such that $\vec{v} = \vec{p} + \vec{q}$ where $\vec{p} = k \vec{w}$ for some scalar $k$ , and $\vec{q} \cdot \vec{w} = 0$ .

If the vectors $\vec{p}$ and $\vec{q}$ in Theorem 13.15 are nonzero, then we can say $\vec{p}$ is `parallel'⁵ to $\vec{w}$ and $\vec{q}$ is `orthogonal' to $\vec{w}$ . In this case, the vector $\vec{p}$ is sometimes called the `vector component of $\vec{v}$ parallel to $\vec{w}$ ' and $\vec{q}$ is called the `vector component of $\vec{v}$ orthogonal to $\vec{w}$ .'

To prove Theorem 13.15, we take $\vec{p} = {proj}_{\vec{w}} (\vec{v})$ and $\vec{q} = \vec{v} - \vec{p}$ . Then $\vec{p}$ is, by definition, a scalar multiple of $\vec{w}$ . Next, we compute $\vec{q} \cdot \vec{w}$ .

$\begin{matrix} \vec{q} \cdot \vec{w} & = & (\vec{v} - \vec{p}) \cdot \vec{w} & Definition of \vec{q} . \\ = & \vec{v} \cdot \vec{w} - \vec{p} \cdot \vec{w} & Properties of Dot Product \\ = & \vec{v} \cdot \vec{w} - (\frac{\vec{v} \cdot \vec{w}}{\vec{w} \cdot \vec{w}} \vec{w}) \cdot \vec{w} & Since \vec{p} = {proj}_{\vec{w}} (\vec{v}) . \\ = & \vec{v} \cdot \vec{w} - (\frac{\vec{v} \cdot \vec{w}}{\vec{w} \cdot \vec{w}}) (\vec{w} \cdot \vec{w}) & Properties of Dot Product. \\ = & \vec{v} \cdot \vec{w} - \vec{v} \cdot \vec{w} \\ = & 0 . \end{matrix}$

Hence, $\vec{q} \cdot \vec{w} = 0$ , as required. At this point, we have shown that the vectors $\vec{p}$ and $\vec{q}$ guaranteed by Theorem 13.15 exist. Now we need to show that they are unique - that is, there is only one such way to decompose $\vec{v}$ in the manner described in Theorem 13.15.

Suppose $\vec{v} = \vec{p} + \vec{q} = {\vec{p}}^{'} + {\vec{q}}^{'}$ where the vectors ${\vec{p}}^{'}$ and ${\vec{q}}^{'}$ satisfy the same properties described in Theorem 13.15 as $\vec{p}$ and $\vec{q}$ . Then $\vec{p} - {\vec{p}}^{'} = {\vec{q}}^{'} - \vec{q}$ , so $\vec{w} \cdot (\vec{p} - {\vec{p}}^{'}) = \vec{w} \cdot ({\vec{q}}^{'} - \vec{q}) = \vec{w} \cdot {\vec{q}}^{'} - \vec{w} \cdot \vec{q} = 0 - 0 = 0$ . The long and short of this computation is that $\vec{w} \cdot (\vec{p} - {\vec{p}}^{'}) = 0$ .

Now there are scalars $k$ and $k^{'}$ so that $\vec{p} = k \vec{w}$ and ${\vec{p}}^{'} = k^{'} \vec{w}$ . This means $\vec{w} \cdot (\vec{p} - {\vec{p}}^{'}) = \vec{w} \cdot (k \vec{w} - k^{'} \vec{w}) = \vec{w} \cdot ([k - k^{'}] \vec{w}) = (k - k^{'}) (\vec{w} \cdot \vec{w}) = (k - k^{'}) {‖ \vec{w} ‖}^{2}$ .

Since $vector w not equal to vector 0$ , $‖ vector w ‖ squared not equal to 0$ , which means the only way $\vec{w} \cdot (\vec{p} - {\vec{p}}^{'}) = (k - k^{'}) {‖ \vec{w} ‖}^{2} = 0$ is for $k - k^{'} = 0$ , or $k = k^{'}$ . This means $\vec{p} = k \vec{w} = k^{'} \vec{w} = {\vec{p}}^{'}$ . Since ${\vec{q}}^{'} - \vec{q} = \vec{p} - {\vec{p}}^{'} = \vec{p} - \vec{p} = \vec{0}$ , it must be that ${\vec{q}}^{'} = \vec{q}$ as well.

Hence, we have shown there is only one way to write $\vec{v}$ as a sum of vectors as described in Theorem 13.15, so the decomposition listed there is unique.

We close this section with an application of the dot product. In Physics, if a constant force $F$ is exerted over a distance $d$ , the work $W$ done by the force is given by $W = F d$ . Here, the assumption is that the force is being applied in the direction of the motion. If the force applied is not in the direction of the motion, we can use the dot product to find the work done.

Consider the scenario sketched below in which the constant force $\vec{F}$ is applied to move an object from the point $P$ to the point $Q$ . Here the force is being applied at an angle $θ$ as opposed to being applied directly in the direction of the motion.

To find the work $W$ done in this scenario, we need to find how much of the force $\vec{F}$ is in the direction of the motion $\vec{P Q}$ . This is precisely what the dot product $\vec{F} \cdot \hat{P Q}$ represents.

Since the distance the object travels is $‖ \vec{P Q} ‖$ , we get $W = (\vec{F} \cdot \hat{P Q}) ‖ \vec{P Q} ‖$ . Since $\vec{P Q} = ‖ \vec{P Q} ‖ \hat{P Q}$ , we can simplify this formula as follows: $W = (\vec{F} \cdot \hat{P Q}) ‖ \vec{P Q} ‖ = \vec{F} \cdot (‖ \vec{P Q} ‖ \hat{P Q}) = \vec{F} \cdot \vec{P Q}$ .

Using Theorem 13.11, we can rewrite $W = \vec{F} \cdot \vec{P Q} = ‖ \vec{F} ‖ ‖ \vec{P Q} ‖ \cos (θ)$ , where $θ$ is the angle between the applied force $\vec{F}$ and the trajectory of the motion $\vec{P Q}$ . We have proved the following.

Theorem

Work as a Dot Product: Suppose a constant force $\vec{F}$ is applied along the vector $\vec{P Q}$ . The work $W$ done by $\vec{F}$ is given by

$W = \vec{F} \cdot \vec{P Q} = ‖ \vec{F} ‖ ‖ \vec{P Q} ‖ \cos (θ),$

where $θ$ is the angle between $\vec{F}$ and $\vec{P Q}$ .

We test out our formula for work in the following example.

Example 5

Taylor exerts a force of $10$ pounds to pull her wagon a distance of $50$ feet over level ground. If the handle of the wagon makes a $30 degrees$ angle with the horizontal, how much work did Taylor do pulling the wagon? Assume the force of $10$ pounds is exerted at a $30 degrees$ angle for the duration of the $50$ feet.

Figure 13.66

There are (at least) two ways to attack this problem. One way is to find the vectors $\vec{F}$ and $\vec{P Q}$ mentioned in Theorem 13.16 and compute $W = \vec{F} \cdot \vec{P Q}$ .

To do this, we assume the origin is at the point where the handle of the wagon meets the wagon and the positive $x$ -axis lies along the dashed line in the figure above.

To find the force vector $\vec{F}$ , we note the force in this situation is a constant 10 pounds, so $‖ vector F ‖ equals 10$ . Moreover, the force is being applied at a constant angle of $θ equals 30 degrees$ with respect to the positive $x$ -axis. Definition 13.4 gives us $\vec{F} = ∥ \vec{F} ∥ ⟨ \cos (θ), \sin (θ) ⟩ = 10 ⟨ \cos (30^{\circ}, \sin (30^{\circ}) ⟩ = ⟨ 5 \sqrt{3}, 5 ⟩$ .

Since the wagon is being pulled along 50 feet in the positive $x$ -direction, we find the displacement vector is $\vec{P Q} = 50 \hat{i} = 50 < 1, 0 > = < 50, 0 >$ .

Per Theorem 13.16, $W = \vec{F} \cdot \vec{P Q} = ⟨ 5 \sqrt{3}, 5 ⟩ \cdot ⟨ 50, 0 ⟩ = 250 \sqrt{3}$ . Since force is measured in pounds and distance is measured in feet, we get $W = 250 \sqrt{3}$ foot-pounds.

Alternatively, we can use the formula $W = ‖ \vec{F} ‖ ‖ \vec{P Q} ‖ \cos (θ)$ . With $‖ vector F ‖ equals 10$ pounds, $‖ vector P ⁢ Q ‖ equals 50$ feet and $θ equals 30 degrees$ , we get $W = (10 pounds) (50 feet) \cos (30^{\circ}) = 250 \sqrt{3}$ foot-pounds of work.

Exercises

In Exercises -, use the pair of vectors $\vec{v}$ and $\vec{w}$ to find the following quantities.

$\vec{v} \cdot \vec{w}$
The angle $θ$ (in degrees) between $\vec{v}$ and $\vec{w}$
${proj}_{\vec{w}} (\vec{v})$
$\vec{q} = \vec{v} - {proj}_{\vec{w}} (\vec{v})$ (Show that $\vec{q} \cdot \vec{w} = 0$ .)

$\vec{v} = ⟨ - 2, - 7 ⟩$ and $\vec{w} = ⟨ 5, - 9 ⟩$
$\vec{v} = ⟨ - 6, - 5 ⟩$ and $\vec{w} = ⟨ 10, - 12 ⟩$
$\vec{v} = ⟨ 1, \sqrt{3} ⟩$ and $\vec{w} = ⟨ 1, - \sqrt{3} ⟩$
$vector v equals ⟨ 3, 4 ⟩$ and $\vec{w} = ⟨ - 6, - 8 ⟩$
$\vec{v} = ⟨ - 2, 1 ⟩$ and $vector w equals ⟨ 3, 6 ⟩$
$\vec{v} = ⟨ - 3 \sqrt{3}, 3 ⟩$ and $\vec{w} = ⟨ - \sqrt{3}, - 1 ⟩$
$vector v equals ⟨ 1, 17 ⟩$ and $\vec{w} = ⟨ - 1, 0 ⟩$
$vector v equals ⟨ 3, 4 ⟩$ and $vector w equals ⟨ 5, 12 ⟩$
$\vec{v} = ⟨ - 4, - 2 ⟩$ and $\vec{w} = ⟨ 1, - 5 ⟩$
$\vec{v} = ⟨ - 5, 6 ⟩$ and $\vec{w} = ⟨ 4, - 7 ⟩$
$\vec{v} = ⟨ - 8, 3 ⟩$ and $vector w equals ⟨ 2, 6 ⟩$
$\vec{v} = ⟨ 34, - 91 ⟩$ and $vector w equals ⟨ 0, 1 ⟩$
$\vec{v} = 3 \hat{i} - \hat{j}$ and $vector w equals 4 j hat$
$\vec{v} = - 24 \hat{i} + 7 \hat{j}$ and $vector w equals 2 i hat$
$\vec{v} = \frac{3}{2} \hat{i} + \frac{3}{2} \hat{j}$ and $\vec{w} = \hat{i} - \hat{j}$
$\vec{v} = 5 \hat{i} + 12 \hat{j}$ and $\vec{w} = - 3 \hat{i} + 4 \hat{j}$
$\vec{v} = ⟨ \frac{1}{2}, \frac{\sqrt{3}}{2} ⟩$ and $\vec{w} = ⟨ - \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} ⟩$
$\vec{v} = ⟨ \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} ⟩$ and $\vec{w} = ⟨ \frac{1}{2}, - \frac{\sqrt{3}}{2} ⟩$
$\vec{v} = ⟨ \frac{\sqrt{3}}{2}, \frac{1}{2} ⟩$ and $\vec{w} = ⟨ - \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2} ⟩$
$\vec{v} = ⟨ \frac{1}{2}, - \frac{\sqrt{3}}{2} ⟩$ and $\vec{w} = ⟨ \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2} ⟩$
A force of $1500$ pounds is required to tow a trailer. Find the work done towing the trailer along a flat stretch of road $300$ feet. Assume the force is applied in the direction of the motion.
Find the work done lifting a $10$ pound book $3$ feet straight up into the air. Assume the force of gravity is acting straight downwards.
Suppose Taylor fills her wagon with rocks and must exert a force of 13 pounds to pull her wagon across the yard. If she maintains a $15 degrees$ angle between the handle of the wagon and the horizontal, compute how much work Taylor does pulling her wagon 25 feet. Round your answer to two decimal places.
In Exercise in Section 13.3, two drunken college students have filled an empty beer keg with rocks which they drag down the street by pulling on two attached ropes. The stronger of the two students pulls with a force of 100 pounds on a rope which makes a $13 degrees$ angle with the direction of motion. (In this case, the keg was being pulled due east and the student's heading was N $77 degrees$ E.) Find the work done by this student if the keg is dragged 42 feet.
Find the work done pushing a 200 pound barrel 10 feet up a $12.5 degrees$ incline. Ignore all forces acting on the barrel except gravity, which acts downwards. Round your answer to two decimal places.
HINT: Since you are working to overcome gravity only, the force being applied acts directly upwards. This means that the angle between the applied force in this case and the motion of the object is not the $12.5 degrees$ of the incline!
Prove the distributive property of the dot product in Theorem 13.10.
Finish the proof of the scalar property of the dot product in Theorem 13.10.
Show Theorem 13.15 reduces to Theorem 13.9 in the case $\vec{w} = \hat{i}$ .
Use the identity in Example 13.4.1 to prove the Parallelogram Law
${‖ \vec{v} ‖}^{2} + {‖ \vec{w} ‖}^{2} = \frac{1}{2} [{‖ \vec{v} + \vec{w} ‖}^{2} + {‖ \vec{v} - \vec{w} ‖}^{2}]$
We know that $| x + y | \leq | x | + | y |$ for all real numbers $x$ and $y$ by the Triangle Inequality established in Exercise in Section 1.3. We can now establish a Triangle Inequality for vectors. In this exercise, we prove that $‖ \vec{u} + \vec{v} ‖ \leq ‖ \vec{u} ‖ + ‖ \vec{v} ‖$ for all pairs of vectors $\vec{u}$ and $\vec{v}$ .
1. (Step 1) Show that ${‖ \vec{u} + \vec{v} ‖}^{2} = {‖ \vec{u} ‖}^{2} + 2 \vec{u} \cdot \vec{v} + {‖ \vec{v} ‖}^{2}$ .
2. (Step 2) Show that $| \vec{u} \cdot \vec{v} | \leq ‖ \vec{u} ‖ ‖ \vec{v} ‖$ . This is the celebrated Cauchy-Schwarz Inequality.⁶
  HINT: Start with $| \vec{u} \cdot \vec{v} | = | ‖ \vec{u} ‖ ‖ \vec{v} ‖ \cos (θ) |$ and use the fact that $vertical bar cos ⁡ open parenthesis θ close parenthesis vertical bar less than or equal to 1$ for all $θ$ .
3. (Step 3) Show:
  ${‖ \vec{u} + \vec{v} ‖}^{2} = {‖ \vec{u} ‖}^{2} + 2 \vec{u} \cdot \vec{v} + {‖ \vec{v} ‖}^{2} \leq {‖ \vec{u} ‖}^{2} + 2 | \vec{u} \cdot \vec{v} | + {‖ \vec{v} ‖}^{2} \leq {‖ \vec{u} ‖}^{2} + 2 ‖ \vec{u} ‖ ‖ \vec{v} ‖ + {‖ \vec{v} ‖}^{2} = {(‖ \vec{u} ‖ + ‖ \vec{v} ‖)}^{2} .$
4. (Step 4) Use Step 3 to show that $‖ \vec{u} + \vec{v} ‖ \leq ‖ \vec{u} ‖ + ‖ \vec{v} ‖$ for all pairs of vectors $\vec{u}$ and $\vec{v}$ .

Answers

$\vec{v} = ⟨ - 2, - 7 ⟩$ and $\vec{w} = ⟨ 5, - 9 ⟩$
$\vec{v} \cdot \vec{w} = 53$
$θ equals 45 degrees$
${proj}_{\vec{w}} (\vec{v}) = ⟨ \frac{5}{2}, - \frac{9}{2} ⟩$
$\vec{q} = ⟨ - \frac{9}{2}, - \frac{5}{2} ⟩$
$\vec{v} = ⟨ - 6, - 5 ⟩$ and $\vec{w} = ⟨ 10, - 12 ⟩$
$\vec{v} \cdot \vec{w} = 0$
$θ equals 90 degrees$
$proj sub vector w ⁢ open parenthesis vector v close parenthesis equals ⟨ 0, 0 ⟩$
$\vec{q} = ⟨ - 6, - 5 ⟩$
$\vec{v} = ⟨ 1, \sqrt{3} ⟩$ and $\vec{w} = ⟨ 1, - \sqrt{3} ⟩$
$\vec{v} \cdot \vec{w} = - 2$
$θ equals 120 degrees$
${proj}_{\vec{w}} (\vec{v}) = ⟨ - \frac{1}{2}, \frac{\sqrt{3}}{2} ⟩$
$\vec{q} = ⟨ \frac{3}{2}, \frac{\sqrt{3}}{2} ⟩$
$vector v equals ⟨ 3, 4 ⟩$ and $\vec{w} = ⟨ - 6, - 8 ⟩$
$\vec{v} \cdot \vec{w} = - 50$
$θ equals 180 degrees$
$proj sub vector w ⁢ open parenthesis vector v close parenthesis equals ⟨ 3, 4 ⟩$
$vector q equals ⟨ 0, 0 ⟩$
$\vec{v} = ⟨ - 2, 1 ⟩$ and $vector w equals ⟨ 3, 6 ⟩$
$\vec{v} \cdot \vec{w} = 0$
$θ equals 90 degrees$
$proj sub vector w ⁢ open parenthesis vector v close parenthesis equals ⟨ 0, 0 ⟩$
$\vec{q} = ⟨ - 2, 1 ⟩$
$\vec{v} = ⟨ - 3 \sqrt{3}, 3 ⟩$ and $\vec{w} = ⟨ - \sqrt{3}, - 1 ⟩$
$\vec{v} \cdot \vec{w} = 6$
$θ equals 60 degrees$
${proj}_{\vec{w}} (\vec{v}) = ⟨ - \frac{3 \sqrt{3}}{2}, - \frac{3}{2} ⟩$
$\vec{q} = ⟨ - \frac{3 \sqrt{3}}{2}, \frac{9}{2} ⟩$
$vector v equals ⟨ 1, 17 ⟩$ and $\vec{w} = ⟨ - 1, 0 ⟩$
$\vec{v} \cdot \vec{w} = - 1$
$θ approximately equals 93.37 degrees$
$proj sub vector w ⁢ open parenthesis vector v close parenthesis equals ⟨ 1, 0 ⟩$
$vector q equals ⟨ 0, 17 ⟩$
$vector v equals ⟨ 3, 4 ⟩$ and $vector w equals ⟨ 5, 12 ⟩$
$\vec{v} \cdot \vec{w} = 63$
$θ approximately equals 14.25 degrees$
${proj}_{\vec{w}} (\vec{v}) = ⟨ \frac{315}{169}, \frac{756}{169} ⟩$
$\vec{q} = ⟨ \frac{192}{169}, - \frac{80}{169} ⟩$
$\vec{v} = ⟨ - 4, - 2 ⟩$ and $\vec{w} = ⟨ 1, - 5 ⟩$
$\vec{v} \cdot \vec{w} = 6$
$θ approximately equals 74.74 degrees$
${proj}_{\vec{w}} (\vec{v}) = ⟨ \frac{3}{13}, - \frac{15}{13} ⟩$
$\vec{q} = ⟨ - \frac{55}{13}, - \frac{11}{13} ⟩$
$\vec{v} = ⟨ - 5, 6 ⟩$ and $\vec{w} = ⟨ 4, - 7 ⟩$
$\vec{v} \cdot \vec{w} = - 62$
$θ approximately equals 169.94 degrees$
${proj}_{\vec{w}} (\vec{v}) = ⟨ - \frac{248}{65}, \frac{434}{65} ⟩$
$\vec{q} = ⟨ - \frac{77}{65}, - \frac{44}{65} ⟩$
$\vec{v} = ⟨ - 8, 3 ⟩$ and $vector w equals ⟨ 2, 6 ⟩$
$\vec{v} \cdot \vec{w} = 2$
$θ approximately equals 87.88 degrees$
${proj}_{\vec{w}} (\vec{v}) = ⟨ \frac{1}{10}, \frac{3}{10} ⟩$
$\vec{q} = ⟨ - \frac{81}{10}, \frac{27}{10} ⟩$
$\vec{v} = ⟨ 34, - 91 ⟩$ and $vector w equals ⟨ 0, 1 ⟩$
$\vec{v} \cdot \vec{w} = - 91$
$θ approximately equals 159.51 degrees$
${proj}_{\vec{w}} (\vec{v}) = ⟨ 0, - 91 ⟩$
$vector q equals ⟨ 34, 0 ⟩$
$\vec{v} = 3 \hat{i} - \hat{j}$ and $vector w equals 4 j hat$
$\vec{v} \cdot \vec{w} = - 4$
$θ approximately equals 108.43 degrees$
${proj}_{\vec{w}} (\vec{v}) = ⟨ 0, - 1 ⟩$
$vector q equals ⟨ 3, 0 ⟩$
$\vec{v} = - 24 \hat{i} + 7 \hat{j}$ and $vector w equals 2 i hat$
$\vec{v} \cdot \vec{w} = - 48$
$θ approximately equals 163.74 degrees$
${proj}_{\vec{w}} (\vec{v}) = ⟨ - 24, 0 ⟩$
$vector q equals ⟨ 0, 7 ⟩$
$\vec{v} = \frac{3}{2} \hat{i} + \frac{3}{2} \hat{j}$ and $\vec{w} = \hat{i} - \hat{j}$
$\vec{v} \cdot \vec{w} = 0$
$θ equals 90 degrees$
$proj sub vector w ⁢ open parenthesis vector v close parenthesis equals ⟨ 0, 0 ⟩$
$\vec{q} = ⟨ \frac{3}{2}, \frac{3}{2} ⟩$
$\vec{v} = 5 \hat{i} + 12 \hat{j}$ and $\vec{w} = - 3 \hat{i} + 4 \hat{j}$
$\vec{v} \cdot \vec{w} = 33$
$θ approximately equals 59.49 degrees$
${proj}_{\vec{w}} (\vec{v}) = ⟨ - \frac{99}{25}, \frac{132}{25} ⟩$
$\vec{q} = ⟨ \frac{224}{25}, \frac{168}{25} ⟩$
$\vec{v} = ⟨ \frac{1}{2}, \frac{\sqrt{3}}{2} ⟩$ and $\vec{w} = ⟨ - \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} ⟩$
$\vec{v} \cdot \vec{w} = \frac{\sqrt{6} - \sqrt{2}}{4}$
$θ equals 75 degrees$
${proj}_{\vec{w}} (\vec{v}) = ⟨ \frac{1 - \sqrt{3}}{4}, \frac{\sqrt{3} - 1}{4} ⟩$
$\vec{q} = ⟨ \frac{1 + \sqrt{3}}{4}, \frac{1 + \sqrt{3}}{4} ⟩$
$\vec{v} = ⟨ \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} ⟩$ and $\vec{w} = ⟨ \frac{1}{2}, - \frac{\sqrt{3}}{2} ⟩$
$\vec{v} \cdot \vec{w} = \frac{\sqrt{2} - \sqrt{6}}{4}$
$θ equals 105 degrees$
${proj}_{\vec{w}} (\vec{v}) = ⟨ \frac{\sqrt{2} - \sqrt{6}}{8}, \frac{3 \sqrt{2} - \sqrt{6}}{8} ⟩$
$\vec{q} = ⟨ \frac{3 \sqrt{2} + \sqrt{6}}{8}, \frac{\sqrt{2} + \sqrt{6}}{8} ⟩$
$\vec{v} = ⟨ \frac{\sqrt{3}}{2}, \frac{1}{2} ⟩$ and $\vec{w} = ⟨ - \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2} ⟩$
$\vec{v} \cdot \vec{w} = - \frac{\sqrt{6} + \sqrt{2}}{4}$
$θ equals 165 degrees$
${proj}_{\vec{w}} (\vec{v}) = ⟨ \frac{\sqrt{3} + 1}{4}, \frac{\sqrt{3} + 1}{4} ⟩$
$\vec{q} = ⟨ \frac{\sqrt{3} - 1}{4}, \frac{1 - \sqrt{3}}{4} ⟩$
$\vec{v} = ⟨ \frac{1}{2}, - \frac{\sqrt{3}}{2} ⟩$ and $\vec{w} = ⟨ \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2} ⟩$
$\vec{v} \cdot \vec{w} = \frac{\sqrt{6} + \sqrt{2}}{4}$
$θ equals 15 degrees$
${proj}_{\vec{w}} (\vec{v}) = ⟨ \frac{\sqrt{3} + 1}{4}, - \frac{\sqrt{3} + 1}{4} ⟩$
$\vec{q} = ⟨ \frac{1 - \sqrt{3}}{4}, \frac{1 - \sqrt{3}}{4} ⟩$
$open parenthesis 1500 ⁢ pounds close parenthesis ⁢ open parenthesis 300 ⁢ feet close parenthesis ⁢ cos ⁡ open parenthesis 0 degrees close parenthesis equals 450, 000$ foot-pounds
$open parenthesis 10 ⁢ pounds close parenthesis ⁢ open parenthesis 3 ⁢ feet close parenthesis ⁢ cos ⁡ open parenthesis 0 degrees close parenthesis equals 30$ foot-pounds
$open parenthesis 13 ⁢ pounds close parenthesis ⁢ open parenthesis 25 ⁢ feet close parenthesis ⁢ cos ⁡ open parenthesis 15 degrees close parenthesis approximately equals 313.92$ foot-pounds
$open parenthesis 100 ⁢ pounds close parenthesis ⁢ open parenthesis 42 ⁢ feet close parenthesis ⁢ cos ⁡ open parenthesis 13 degrees close parenthesis approximately equals 4092.35$ foot-pounds
$open parenthesis 200 ⁢ pounds close parenthesis ⁢ open parenthesis 10 ⁢ feet close parenthesis ⁢ cos ⁡ open parenthesis 77.5 degrees close parenthesis approximately equals 432.88$ foot-pounds

Since $\vec{v} = ‖ \vec{v} ‖ \hat{v}$ and $\vec{w} = ‖ \vec{w} ‖ \hat{w}$ , if $\hat{v} = \hat{w}$ then $\vec{w} = ‖ \vec{w} ‖ \hat{v} = \frac{‖ \vec{w} ‖}{‖ \vec{v} ‖} (‖ \vec{v} ‖ \hat{v}) = \frac{‖ \vec{w} ‖}{‖ \vec{v} ‖} \vec{v}$ . In this case, $k = \frac{‖ \vec{w} ‖}{‖ \vec{v} ‖} > 0$ . ↩
See Exercise in Section A.5. ↩
In this case, the point $R$ coincides with the point $O$ , so $\vec{p} = \vec{O R} = \vec{O O} = \vec{0}$ . ↩
Note that, necessarily, $\vec{q} ⟂ \vec{p}$ as well! ↩
See Exercise in Section 13.3. ↩
It is also known by other names. Check out this site for details. ↩

Adapted from Precalculus, Preliminary 4th Edition (integrated calculus), by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.