9.3 The Dot Product

Components

We have seen that it can be useful to resolve a vector into horizontal and vertical components. We can also break a vector into components that point in other directions.

Imagine the following experiment: Delbert holds a ball at shoulder height and then drops it, so that it falls to the ground. Francine holds a ball at shoulder height on an inclined ramp, then releases it so that it rolls downhill. Which ball will reach the ground first?

Although gravity causes both balls to speed up, the free-falling ball will reach the ground first. The force of gravity pulls straight down, the same direction as the motion of the free-falling ball, but the rolling ball must move at an angle to the pull of gravity, along the surface of the ramp. Only part of the gravitational force accelerates the rolling ball, and the rest of the force is counteracted by the surface of the ramp. What fraction of the gravitational force causes the ball to roll?

In figure (a), the gravitational force $F$ is resolved into the sum of two vectors, $F = u + v$ , where $v$ points down the ramp, and $u$ is perpendicular to the ramp. The magnitude of $v$ is called the component of $F$ in the direction of motion, and is denoted by ${comp}_{v} F$ . This is the portion of the gravitational force that moves the ball. From figure (b), we see that $comp sub v F equals ∥ F ∥ cos ⁡ open parenthesis θ close parenthesis$ , where $θ$ is the angle between $F$ and $v$ .

With a little geometry, you can verify that in this example the angle $θ$ is the complement of the angle of inclination of the ramp, $α$ . (Think of similar triangles.) Now suppose that we increase the angle of inclination. As $α$ increases, $θ$ decreases, $\cos (θ)$ increases, and hence ${comp}_{v} F$ increases. This result agrees with our experience: as the ramp gets steeper, the ball rolls faster.

Example 1

Delbert uses a sheet of plywood as a ramp for his wheelbarrow. The ramp is inclined at an angle of $15 degrees$ to the horizontal. The plywood can support a maximum weight of 200 pounds. What weight will the ramp support?

The component of the weight perpendicular to the ramp cannot be more than 200 pounds. If the weight of the wheelbarrow is $w$ , and $v$ is a vector perpendicular to the ramp, as shown below, then the angle between $v$ and $w$ is $15 degrees$ , by similar triangles. Thus, the component of $w$ in the direction of $w$ is

$\begin{aligned} {comp}_{v} w = ∥ w ∥ \cos (15^{\circ}) & = 200 \\ ∥ w ∥ (0.9659) & = 200 \\ ∥ w ∥ & = 207.06 \end{aligned}$

The ramp can support no more than about 207 pounds.

Delbert pushes a lawn mower whose handle makes a $40 degrees$ angle with the horizontal. If he applies a force of 30 pounds in the direction of the handle, what is the component of the force in the horizontal direction?

23 pounds

Coordinate Form for Components

In the examples above, we computed the component of a force $F$ in the direction of a vector $v$ by knowing the angle between $F$ and $v$ . If the vectors are given in coordinate form (that is, $v = a i + b j$ ), we may not know the angle between them. Can we compute the component of a vector $w$ in the direction of $v$ , in terms of the coordinates of $v$ and $w$ ?

Suppose $v = a i + b j$ and $w = c i + d j$ , as shown below. We will need to compute the cosine of $θ$ in terms of $a, b, c,$ and $d$ .

First verify that

$\begin{aligned} \cos (α) & = \frac{a}{∥ v ∥} and \sin (α) = \frac{b}{∥ v ∥} \\ \cos (β) & = \frac{c}{∥ w ∥} and \sin (β) = \frac{d}{∥ w ∥} \end{aligned}$

Because $θ = β = α$ , we use the subtraction formula for cosine.

$\begin{aligned} \cos (θ) & = \cos (α) \cos (β) + \sin (α) \sin (β) \\ = \frac{a}{∥ v ∥} \frac{c}{∥ w ∥} + \frac{b}{∥ v ∥} \frac{d}{∥ w ∥} \\ = \frac{1}{∥ v ∥ ∥ w ∥} (a c + b d) \end{aligned}$

And finally,

${comp}_{v} w = ∥ w ∥ \cos (θ) = \frac{a c + b d}{∥ v ∥}$

Now we have a fromula for the component of a vector $w$ in the direction of a vector $v$ .

Example 2

Compute the component of $w = 6 i + 2 j$ in the direction of $v = 4 i + 3 j$ .
Find the vector component of $w$ in the direction of $v$ .
Find the vector component of $w$ perpendicular to $v$ .

We use the formula derived above.
${comp}_{v} w = \frac{a c + b d}{∥ v ∥} = \frac{4 (6) + 3 (2)}{\sqrt{4^{2} + 3^{2}}} = \frac{30}{5} = 6$
Remember that the component of $w$ in the direction of $v$ is a scalar; it is the length of the projection of $w$ onto $v$ .
We would like to find a vector $q$ of length 6 in the direction of $v$ , as shown at right. Because $∥ v ∥ equals 5$ , we scale the components of $v$ by $\frac{6}{5}$ to get
$q = \frac{6}{5} (4 i + 3 j) = \frac{24}{5} i + \frac{18}{5} j$
Because $w = p + q$ , where $p$ is perpendicular to $q$ , we have
$p = w - q = (6 i + 2 j) - (\frac{24}{5} i + \frac{18}{5} j) = \frac{6}{5} i - \frac{8}{5} j$

Write the vector $w = 6 i + 2 j$ as the sum of two components, one parallel to $v = i + j$ and the other perpendicular to $v$ .

$w = (4 i + 4 j) + (2 i - 2 j)$

The Dot Product

The expression $a c + b d$ , which we encountered above as part of the formula for ${comp}_{v} w$ , is quite useful and is given a name; it is called the dot product of the vectors $v = a i + b j$ and $w = c i + d j$ .

It is easy to remember the formula for the dot product if we think of adding the product of the $i$ -components and the product of the $j$ -components of the two vectors.

In the examples above, you can see that the dot product of two vectors is a scalar. For this reason, the dot product is also called the scalar product.

Compute the dot product of $v = 6 i + 2 j$ and $w = - 2 i + 3 j$ .

$minus 6$

We can now write the formula for ${comp}_{v} w$ using the dot product.

${comp}_{v} w = ∥ w ∥ \cos (θ) = \frac{a c + b d}{∥ v ∥} = \frac{v \cdot w}{∥ v ∥}$

We have derived an alternate formula for a component of a vector.

Example 4

Compute the component of $v = 5 i - 3 j$ in the direction of $w = 4 i + j$ .

Using the formula above, we have ${comp}_{w} v = \frac{v \cdot w}{∥ w ∥}$ . We first compute $v \cdot w$ and $∥ w ∥$ .

$v \cdot w = 5 (4) + (- 3) (1) = 17 and ∥ w ∥ = \sqrt{4^{2} + 1^{2}} = \sqrt{17}$

Thus,

${comp}_{w} v = \frac{v \cdot w}{∥ w ∥} = \frac{17}{\sqrt{17}} = \sqrt{17}$

The length of the projection of $v$ in the direction of $w$ is $the square root of 17$ units, as shown at right.

Compute the component of $u = 2 i + 3 j$ in the direction of $v = 6 i + 5 j$ .

$\frac{27}{\sqrt{61}}$

Geometric Meaning of the Dot Product

An even more important relationship, which gives geometric meaning to the dot product, follows from the formula for a component. We now have two ways to compute the component of $w$ in the direction of $v$ :

${comp}_{v} w = ∥ w ∥ \cos θ and {comp}_{v} w = \frac{v \cdot w}{∥ v ∥}$

Equating these two expressions, we find

$∥ w ∥ \cos (θ) = \frac{v \cdot w}{∥ v ∥}$

or $∥ v ∥ ∥ w ∥ \cos θ = v \cdot w$ . This is a geometric formula for the dot product.

The dot product is a way of multiplying two vectors that depends on the angle between them.

If $θ equals 0 degrees$ , so that $v$ and $w$ point in the same direction, then $cos ⁡ open parenthesis θ close parenthesis equals 1$ and $v \cdot w$ is just the product of their lengths, $∥ v ∥ ∥ w ∥$ .
If $v$ and $w$ are perpendicular, then $cos ⁡ open parenthesis θ close parenthesis equals 0$ , so $v \cdot w = 0$ . (Two vectors $v$ and $w$ are said to be orthogonal if their dot product is zero.)
If $θ$ is between $0 degrees$ and $90 degrees$ , the dot product multiplies the length of $v$ times the component of $w$ in the direction of $v$ .

Example 5

Show that the vectors $v = 2 i + 6 j$ and $w = - 9 i + 3 j$ are orthogonal.

We compute the dot product of the vectors.

$v \cdot w = 2 (- 9) + 6 (3) = 0$

Thus, $∥ v ∥ ∥ w ∥ cos ⁡ open parenthesis θ close parenthesis equals 0$ , so $v$ and $w$ are orthogonal. Because neither $∥ v ∥ equals 0$ or $∥ w ∥ equals 0$ , it must be the case that $cos ⁡ open parenthesis θ close parenthesis equals 0$ , so $θ equals 90 degrees$ or $270 degrees$ .

Show that the vectors $v = a i + b j$ and $w = - b i + a j$ are orthogonal.
Find a vector $w$ perpendicular to $v = - 3 i - 5 j$ .

$u \cdot v = - a b + a b = 0$
$5 i minus 3 j$

Using the dot product, we can find the angle between two vectors.

Example 6

Find the angle between the vectors $u = - 3 i + 2 j$ and $v = 5 i + 3 j$ .

We first compute $u \cdot v$ and the magnitude of each vector.

$\begin{aligned} u \cdot v & = - 3 (5) + 2 (3) = - 9 \\ ∥ u ∥ & = \sqrt{(- 3)^{2} + 2^{2}} = \sqrt{13} \\ ∥ v ∥ & = \sqrt{5^{2} + 3^{2}} = \sqrt{34} \end{aligned}$

Then we apply the formula for $\cos (θ)$ .

$\begin{aligned} \cos (θ) & = \frac{v \cdot w}{∥ v ∥ ∥ w ∥} = \frac{- 9}{\sqrt{13} \sqrt{34}} = - 0.4281 \\ θ & = \cos^{- 1} (- 0.4281) = {115.35}^{\circ} \end{aligned}$

The angle between the vectors is $115.35 degrees$ .

Find the angle between the vectors $u = 4 i - 6 j$ and $v = 2 i + 8 j$ .

$132.27 degrees$

Review the following skills you will need for this section.

Section 9.3 Summary

Vocabulary

Dot product
Scalar product
Orthogonal

Concepts

Note
Dot Product (Coordinate Formula)
The dot product of two vectors $v = v_{1} i + v_{2} j$ and $w = w_{1} i + w_{2} j$ is the scalar
$v \cdot w = v_{1} w_{1} + v_{2} w_{2}$
The dot product is a way of multiplying two vectors that depends on the angle between them.
Note
Dot Product (Geometric Formula)
The dot product of two vectors $v$ and $w$ is the scalar
$v \cdot w = ∥ v ∥ ∥ w ∥ \cos (θ)$
where $θ$ is the angle between the vectors.
The component of a vector $w$ in the direction of vector $v$ is the length of the vector projection of $w$ onto $w$ .
Note
Component of a Vector
The component of $w$ in the direction of $v$ is the scalar
${comp}_{v} w = \frac{v \cdot w}{∥ v ∥}$
Note
Angle Between Two Vectors
The angle $θ$ between two vectors $v$ and $w$ is given by
$\cos (θ) = \frac{v \cdot w}{∥ v ∥ ∥ w ∥}$
Two vectors $v$ and $w$ are orthogonal if $v \cdot w = 0$

Study Questions

If $u$ and $v$ have the same direction, what is ${comp}_{u} v$ ?
If $u$ is perpendicular to $v$ , what is ${comp}_{u} v$ ?
What is the angle between $u$ and $v$ that makes their dot product as large as possible?
What does the dot product of two unit vectors tell you?

Skills

Find the component of $w$ in the direction of $v$ #1–6, 37–40
Compute the dot product #11–22, 27–36
Find the angle between two vectors #23–26
Resolve a vector into components in given directions #7–10, 41–42

Homework 9-3

For Problems 1–6, find the component of $w$ in the direction of $v$ .

$w = 5 i + 9 j,$ $v = 3 i + 2 j$

$\frac{33}{\sqrt{13}}$

$w = 7 i + 4 j,$ $v = 2 i + 3 j$

$w = - 6 i + 5 j,$ $v = i + j$

$\frac{- 1}{\sqrt{2}}$

$w = 10 i - 14 j, v = i + j$

$w = 4 i - 3 j,$ $v = - i + 2 j$

$- 2 \sqrt{5}$

$w = - 2 i - 3 j, v = i - 2 j$

For Problems 7–10,

Resolve $w$ into two components, one parallel to $v$ and the other orthogonal to $v$ .
Sketch both vectors and the vector components.

$w = 8 i + 4 j,$ $v = 2 i + 3 j$

$w = (\frac{56}{13} i + \frac{84}{13} j) + (\frac{48}{13} i - \frac{32}{13} j)$

$w = - 3 i + 7 j,$ $v = 4 i + 2 j$

$w = 6 i - 2 j,$ $v = i - j$

$w = (4 i - 4 j) + (2 i + 2 j)$

$w = - 5 i + 3 j,$ $v = - i - 3 j$

For Problems 11–18, compute the dot product $u \cdot v$ .

$u = 3 i + 7 j,$ $v = - 2 i + 4 j$

$22$

$u = - 1.3 i + 5.6 j,$ $v = 3 i - 5 j$

$u = 3 i - 4 j,$ $v = 20 i + 15 j$

$0$

$u = 2 i + j,$ $v = 6 i + 3 j$

$u$ has magnitude 3 and direction $27 degrees$ , and $v$ has magnitude 8 and direction $minus 33 degrees$ .

$12$

$u$ has magnitude $the square root of 7$ and direction $112 degrees$ , and $v$ has magnitude $the square root of 14$ and direction $157 degrees$ .

$minus 318.2$

For Problems 19–22, decide whether the pair of vectors is orthogonal.

$2 i plus 3 j$ and $- 3 i - 2 j$

not orthogonal

$- 5 i + 7 j$ and $7 i plus 5 j$

$4 i plus 6 j$ and $- 15 i + 10 j$

orthogonal

$3 i minus 4 j$ and $- 3 i + 4 j$

For Problems 23–26, find the angle between the vectors.

$3 i plus 5 j$ and $2 i plus 4 j$

$4.4 degrees$

$i minus 2 j$ and $- 2 i - 3 j$

$4 i minus 8 j$ and $6 i plus 4 j$

$97.1 degrees$

$- 6 i + 8 j$ and $18 i minus 24 j$

For Problems 27–30, find a value of $k$ so that $v$ is orthogonal to $w$ .

$w = 8 i - 3 j,$ $v = 3 i + k j$

$8$

$w = 2 i + 7 j,$ $v = k i + 4 j$

$w = - 2 i - 5 j,$ $v = k i + 4 j$

$minus 10$

$w = 5 i + 3 j,$ $v = - 2 i + k j$

For Problems 31–36, evaluate the expression for the vectors

$u = 2 i + 5 j, v = - 3 i + 4 j, w = 3 i - 2 j$

$w \cdot (u + v)$

$minus 21$

$w \cdot u + w \cdot v$

$(u \cdot v) w$

$42 i minus 28 j$

$(u \cdot v) (u \cdot w)$

$(u + v) \cdot (u - v)$

$4$

$\frac{w \cdot v}{w \cdot w} w$

Gary pulls a loaded wagon along a flat road. The handle of the wagon makes an angle of $50 degrees$ to the horizontal. If Gary pulls with a force of 60 pounds, find the component of the force in the direction of motion.

38.57 lbs

Wassily is trying to topple a statue by pulling on a rope tied to the statue's upraised arm. The rope is making a $35 degrees$ angle from horizontal. If Wassily is pulling on the rope with a force of 250 pounds, find the component of the force in the horizontal direction.

An SUV weighing 6200 pounds is parked on a hill with slope $12 degrees$ . Find the force needed to keep the SUV from rolling down the hill.

1289 lbs

Steve's boat is headed due north, and the sail points at an angle of $15 degrees$ east of north. The wind is blowing in the direction $60 degrees$ west of south, but because of the difference in air pressure between the front and back surfaces of the sail, the boat experiences a force of 400 pounds in the direction the sail is facing. Find the component of the force in the direction of the boat's motion.

Find unit vectors $u$ and $v$ in the directions of $i + j$ and $i - j$ .
Show that $u$ and $v$ are orthogonal.
Find the components of $w = 3 i + 8 j$ in the directions of $u$ and $v$ .
Sketch the vectors $u, v$ and $w$ , and show the components of $w$ .

$\frac{1}{\sqrt{2}} i + \frac{1}{\sqrt{2}} j and \frac{- 1}{\sqrt{2}} i + \frac{1}{\sqrt{2}} j$
$u \cdot v = 0$
$\frac{11}{\sqrt{2}}$ and $\frac{5}{\sqrt{2}}$

Find unit vectors $u$ and $v$ in the directions of $30 degrees$ and $120 degrees$ .
Show that $u$ and $v$ are orthogonal.
Find the components of $w = - 4 i + 4 j$ in the directions of $u$ and $v$ .
Sketch the vectors $u, v$ and $w$ , and show the components of $w$ .

For Problems 43–48, let $u = a i + b j$ and $v = c i + d j$ .

Show that $v \cdot v = ∥ v ∥^{2}$ .

$v \cdot v = c^{2} + d^{2}$

If $∥ u ∥ equals 1$ , show that ${comp}_{u} v = u \cdot v$ .

Show that $k u \cdot v = k (u \cdot v) = u \cdot k v$ .

$k u \cdot v = k a c + k b d = k (a c + b d) = (a k c + b k d)$

Prove the distributive law: $u \cdot (v + w) = u \cdot v + u \cdot w$ .

Show that $(u - v) \cdot (u + v) = ∥ u ∥^{2} - ∥ v ∥^{2}$ .

$\begin{aligned} (u - v) \cdot (u + v) & = (a - c) (a + c) + (b - d) (b + d) \\ = (a^{2} + b^{2}) - (c^{2} + d^{2}) \end{aligned}$

If $∥ u ∥ equals ∥ v ∥$ , show that $u + v$ is perpendicular to $u - v$ .

Show that the component of $v = a i + b j$ in the direction of $i$ is $a$ , and the component of $v$ in the direction of $j$ is $b$ .

$\frac{a \cdot 1 + b \cdot 0}{1} = a$ and $\frac{a \cdot 0 + b \cdot 1}{1} = b$

Show that the dot product $u \cdot v$ gives the length of $v$ times the component of $u$ in the direction of $v$ .

Start from the geometric definition $v \cdot w = ∥ v ∥ ∥ w ∥ \cos (θ)$ and show that $i \cdot i = 1, j \cdot j = 1$ and $i \cdot j = 0$ .
Use part (a) and Problems 45 and 46 to derive the coordinate definition of $v \cdot w$ .

Both $i \cdot i = 1$ and $j \cdot j = 1$ because $1 \cdot 1 \cos 0 = 1$ ; $i \cdot j = 1 \cdot 1 \cos 90^{\circ} = 0$
$(a i + b j) \cdot (c i + d j) = a c (1) + a d (0) + b c (0) + b d (1) = a c + b d$

Show that $(v + w) \cdot (v + w) = ∥ v ∥^{2} + ∥ w ∥^{2} + 2 (v \cdot w)$
Use part (a) to prove the triangle inequality:
$∥ v + w ∥ \leq ∥ v ∥ + ∥ w ∥$

Use the dot product to show that $∥ u - v ∥^{2} = ∥ u ∥^{2} + ∥ v ∥^{2} - 2 ∥ u ∥ ∥ v ∥ \cos (θ)$ .
Use the figure at right to explain why part (a) proves the law of cosines.

$∥ u - v ∥^{2} = u \cdot u - 2 u \cdot v + v \cdot v = ∥ u ∥^{2} + ∥ v ∥^{2} - 2 ∥ u ∥ ∥ v ∥ \cos θ$
Let $a = ∥ u ∥, b = ∥ v ∥, c = ∥ u - v ∥$ , and $C = θ$

If $u$ is a unit vector, and the angle between $u$ and $i$ is $α$ , show that $u = \cos (α) i + \sin (α) j$ .
Suppose $u$ and $v$ are unit vectors, as shown in the figure at right. Use the dot product $u \cdot v$ to prove that
$\cos (β - α) = \cos (β) \cos (α) + \sin (β) \sin (α)$

Trigonometry by Katherine Yoshiwara (yoshiwarabooks.org), GNU Free Documentation License 1.2 or later. Adapted for the XYZ HTML edition with the authors' permission (recorded 2026-07-04). License: GFDL-1.2-or-later.