The Dot Product

The dot product is a way to “multiply” two vectors, and the answer is a plain number (a scalar), not another vector. That number tells you how much two vectors point in the same direction. It’s the quickest way to find the angle between two vectors, to test whether they’re perpendicular, and to work out the work done by a force.

Key ideas

The geometric definition

Place $\vec{a}$ and $\vec{b}$ tail to tail, and let $\theta$ be the angle between them, with $0^\circ \le \theta \le 180^\circ$. The dot product is

$$\vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}| \cos\theta$$

Read it as “a dot b”. Since $|\vec{a}|$ and $|\vec{b}|$ are positive (for non-zero vectors), the sign of the dot product comes from $\cos\theta$:

Angle	$\cos\theta$	Dot product
acute (or $0^\circ$), $0^\circ \le \theta \lt 90^\circ$	positive	$\vec{a} \cdot \vec{b} \gt 0$
right, $\theta = 90^\circ$	$0$	$\vec{a} \cdot \vec{b} = 0$
obtuse, $90^\circ \lt \theta \le 180^\circ$	negative	$\vec{a} \cdot \vec{b} \lt 0$

The sign of $\vec{a} \cdot \vec{b}$ tells you whether the angle between the vectors is acute, right, or obtuse.

The component formula

When the vectors are in Cartesian form (written with square brackets, like the rest of this unit; some books use $(a_1, a_2)$ or $\langle a_1, a_2 \rangle$), multiply matching components and add:

$$\begin{aligned} \text{2-D:}\quad [a_1, a_2] \cdot [b_1, b_2] &= a_1 b_1 + a_2 b_2 \\ \text{3-D:}\quad [a_1, a_2, a_3] \cdot [b_1, b_2, b_3] &= a_1 b_1 + a_2 b_2 + a_3 b_3 \end{aligned}$$

For example, $[2, -1, 3] \cdot [4, 0, 5] = 8 + 0 + 15 = 23$.

Why do the two formulas agree? Draw $\vec{a}$, $\vec{b}$, and $\vec{a} - \vec{b}$ as a triangle. The cosine law gives $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2|\vec{a}||\vec{b}|\cos\theta$. Writing each magnitude with components and expanding, almost everything cancels and you’re left with $a_1 b_1 + a_2 b_2 + a_3 b_3 = |\vec{a}||\vec{b}|\cos\theta$.

The angle between two vectors

Put the two formulas together and solve for $\cos\theta$:

$$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|}$$

Then use $\cos^{-1}$ (calculator in degrees). Because $\cos^{-1}$ always gives an angle from $0^\circ$ to $180^\circ$, you get the right angle every time, acute or obtuse.

Perpendicular (orthogonal) vectors

Two non-zero vectors are perpendicular, also called orthogonal, exactly when their dot product is zero:

$$\vec{a} \perp \vec{b} \quad \Longleftrightarrow \quad \vec{a} \cdot \vec{b} = 0$$

This is the fastest perpendicular test there is. For example, $[3, -2, 1] \cdot [1, 2, 1] = 3 - 4 + 1 = 0$, so these vectors meet at $90^\circ$.

Finding a vector perpendicular to a given one.

• In 2-D, swap the components and change one sign: $[a, b]$ is perpendicular to $[-b, a]$ and to $[b, -a]$. Check: $a(-b) + b(a) = 0$.
• In 3-D there are infinitely many perpendicular directions. Choose two components and solve for the third, or set one component to $0$ and use the 2-D trick on the other two. For $[1, 2, 3]$, setting the middle component to $0$ gives $[3, 0, -1]$: check $3 + 0 - 3 = 0$.

Properties

These all follow from the component formula, so you can check any of them by expanding.

Property	Rule
Commutative	$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$
Distributive	$\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}$
Scalar multiples	$(k\vec{a}) \cdot \vec{b} = k(\vec{a} \cdot \vec{b}) = \vec{a} \cdot (k\vec{b})$
Dot with itself	$\vec{a} \cdot \vec{a} = \lvert\vec{a}\rvert^2$
Zero vector	$\vec{a} \cdot \vec{0} = 0$

The rule $\vec{a} \cdot \vec{a} = |\vec{a}|^2$ comes from $\theta = 0^\circ$: $|\vec{a}||\vec{a}|\cos 0^\circ = |\vec{a}|^2$. It’s very useful for finding magnitudes, as in Practice question 5.

The dot product is not associative, and the question doesn’t even make sense. To be associative you’d need $(\vec{a} \cdot \vec{b}) \cdot \vec{c} = \vec{a} \cdot (\vec{b} \cdot \vec{c})$. But $\vec{a} \cdot \vec{b}$ is a number, and you can’t dot a number with a vector. The closest you can write is $(\vec{a} \cdot \vec{b})\vec{c}$, a scalar multiple of $\vec{c}$, and that’s usually different from $\vec{a}(\vec{b} \cdot \vec{c})$, a multiple of $\vec{a}$. For $\vec{a} = [1, 2]$, $\vec{b} = [3, -1]$, $\vec{c} = [2, 2]$: $(\vec{a} \cdot \vec{b})\vec{c} = 1[2, 2] = [2, 2]$, but $\vec{a}(\vec{b} \cdot \vec{c}) = 4[1, 2] = [4, 8]$.

Work

When a constant force $\vec{F}$ moves an object through a displacement $\vec{d}$, the work done is

$$W = \vec{F} \cdot \vec{d} = |\vec{F}|\,|\vec{d}| \cos\theta$$

With force in newtons (N) and distance in metres (m), work is in joules (J). Only the part of the force that points along the motion does work: pull straight ahead and $\cos 0^\circ = 1$ (all of the force counts); pull at right angles to the motion and $\cos 90^\circ = 0$ (no work at all).

Worked examples

Example 1: Using the geometric definition

$|\vec{a}| = 5$ and $|\vec{b}| = 8$. Find $\vec{a} \cdot \vec{b}$ if the angle between them is (a) $60^\circ$ and (b) $120^\circ$.

Solution.

(a) $\vec{a} \cdot \vec{b} = (5)(8)\cos 60^\circ = 40 \times \dfrac{1}{2} = 20$

(b) $\vec{a} \cdot \vec{b} = (5)(8)\cos 120^\circ = 40 \times \left(-\dfrac{1}{2}\right) = -20$

The obtuse angle gives a negative dot product, just as the table predicts.

Example 2: The angle between two vectors in 3-D

Find the angle between $\vec{a} = [2, -1, 3]$ and $\vec{b} = [4, 2, -1]$, to one decimal place.

Solution. Find the dot product and both magnitudes:

$$\begin{aligned} \vec{a} \cdot \vec{b} &= 2(4) + (-1)(2) + 3(-1) = 8 - 2 - 3 = 3 \\ |\vec{a}| &= \sqrt{4 + 1 + 9} = \sqrt{14} \\ |\vec{b}| &= \sqrt{16 + 4 + 1} = \sqrt{21} \end{aligned}$$

Then

$$\cos\theta = \frac{3}{\sqrt{14}\,\sqrt{21}} = \frac{3}{\sqrt{294}} \approx 0.1750 \qquad\Rightarrow\qquad \theta = \cos^{-1}(0.1750) \approx 79.9^\circ$$

The dot product is small and positive, so the angle is acute but close to $90^\circ$. That matches.

Example 3: Making vectors perpendicular

(a) Find $k$ so that $\vec{u} = [2, k, 5]$ and $\vec{v} = [k, 3, -4]$ are perpendicular.

(b) Find two different vectors perpendicular to $[2, -5]$.

Solution.

(a) Perpendicular means the dot product is $0$:

$$\begin{aligned} 2k + 3k - 20 &= 0 \\ 5k &= 20 \\ k &= 4 \end{aligned}$$

Check: $[2, 4, 5] \cdot [4, 3, -4] = 8 + 12 - 20 = 0$. ✓

(b) Swap the components and change one sign: $[5, 2]$ works, since $2(5) + (-5)(2) = 0$. Any multiple also works, such as $[-5, -2]$ or $[10, 4]$.

Example 4: Work done pulling a sled

You pull a sled $50$ m across flat snow with a force of $80$ N along a rope held at $30^\circ$ above the horizontal.

(a) How much work do you do?

(b) How much more work would the same force do if you lowered the rope to $15^\circ$?

Solution. The displacement is horizontal, so the angle between $\vec{F}$ and $\vec{d}$ is the rope’s angle.

(a)

$$W = |\vec{F}||\vec{d}|\cos\theta = (80)(50)\cos 30^\circ = 4000 \times \frac{\sqrt{3}}{2} = 2000\sqrt{3} \approx 3464 \text{ J}$$

(b) $W = (80)(50)\cos 15^\circ \approx 3864$ J, which is about $400$ J more.

A lower rope sends more of your pull along the direction of motion. (Pulling at $0^\circ$ would give the full $4000$ J, but then your hand would be on the ground.)

Common mistakes

Writing the answer as a vector. The dot product of two vectors is a number. $[1, 2, 3] \cdot [4, 5, 6] = 4 + 10 + 18 = 32$, not $[4, 10, 18]$.

Using an angle that isn’t tail to tail. The angle in $|\vec{a}||\vec{b}|\cos\theta$ is between vectors that start at the same point. In triangle $ABC$, the angle at $B$ is between $\overrightarrow{BA}$ and $\overrightarrow{BC}$. If you use $\overrightarrow{AB}$ and $\overrightarrow{BC}$ (head to tail), you get $180^\circ$ minus the angle you want.

Calculator in radians. This course gives angles between vectors in degrees. If $\cos^{-1}(0.175)$ comes out as $1.395$, your calculator is in radian mode.

Thinking a zero dot product means a zero vector. $\vec{a} \cdot \vec{b} = 0$ usually means the vectors are perpendicular, not that one of them is $\vec{0}$. For example, $[3, 1] \cdot [-1, 3] = 0$.

Trying to “associate” dot products. $(\vec{a} \cdot \vec{b}) \cdot \vec{c}$ is meaningless, because $\vec{a} \cdot \vec{b}$ is a number. And $(\vec{a} \cdot \vec{b})\vec{c}$ is generally not equal to $\vec{a}(\vec{b} \cdot \vec{c})$: they’re multiples of different vectors.

Using the whole force for work. Work uses only the component of the force along the motion, $|\vec{F}|\cos\theta$. Multiplying $|\vec{F}|$ by the distance overestimates the work whenever the force is at an angle.

Practice

1. (Warm-up) $|\vec{u}| = 6$, $|\vec{v}| = 4$, and the angle between them is $45^\circ$. Find $\vec{u} \cdot \vec{v}$, exactly and to two decimal places.

Solution

$$\vec{u} \cdot \vec{v} = (6)(4)\cos 45^\circ = 24 \times \frac{\sqrt{2}}{2} = 12\sqrt{2} \approx 16.97$$

2. (Warm-up) Find each dot product.

• (a) $[3, -2] \cdot [5, 4]$
• (b) $[1, 0, -4] \cdot [2, 7, -1]$

Solution

(a) $3(5) + (-2)(4) = 15 - 8 = 7$

(b) $1(2) + 0(7) + (-4)(-1) = 2 + 0 + 4 = 6$

3. (Core) Find the angle between $\vec{a} = [1, 2, 2]$ and $\vec{b} = [3, 0, -4]$, to one decimal place.

Solution

$\vec{a} \cdot \vec{b} = 3 + 0 - 8 = -5$, $|\vec{a}| = \sqrt{1 + 4 + 4} = 3$, and $|\vec{b}| = \sqrt{9 + 0 + 16} = 5$.

$$\cos\theta = \frac{-5}{(3)(5)} = -\frac{1}{3} \qquad\Rightarrow\qquad \theta \approx 109.5^\circ$$

The negative dot product means the angle is obtuse. ✓

4. (Core) Find all values of $k$ for which $[k, k, 3]$ and $[k, -5, 2]$ are perpendicular.

Solution

Set the dot product to zero:

$$\begin{aligned} k^2 - 5k + 6 &= 0 \\ (k - 2)(k - 3) &= 0 \\ k = 2 \quad &\text{or} \quad k = 3 \end{aligned}$$

Check $k = 2$: $[2, 2, 3] \cdot [2, -5, 2] = 4 - 10 + 6 = 0$. ✓ Check $k = 3$: $[3, 3, 3] \cdot [3, -5, 2] = 9 - 15 + 6 = 0$. ✓

5. (Core) $|\vec{a}| = 3$, $|\vec{b}| = 5$, and $\vec{a} \cdot \vec{b} = -6$. Use the properties of the dot product to find:

• (a) $(\vec{a} + 2\vec{b}) \cdot (\vec{a} - \vec{b})$
• (b) $|\vec{a} + \vec{b}|$

Solution

(a) Expand like a product of binomials (the distributive and commutative properties allow this):

$$\begin{aligned} (\vec{a} + 2\vec{b}) \cdot (\vec{a} - \vec{b}) &= \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} + 2\vec{b} \cdot \vec{a} - 2\vec{b} \cdot \vec{b} \\ &= |\vec{a}|^2 + \vec{a} \cdot \vec{b} - 2|\vec{b}|^2 \\ &= 9 + (-6) - 2(25) \\ &= -47 \end{aligned}$$

(b) Use $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$:

$$|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2\,\vec{a} \cdot \vec{b} + |\vec{b}|^2 = 9 - 12 + 25 = 22$$

So $|\vec{a} + \vec{b}| = \sqrt{22} \approx 4.69$.

6. (Core) A constant force $\vec{F} = [25, 10, -5]$ (in newtons) moves an object in a straight line from $A(1, 2, 0)$ to $B(9, 5, 4)$ (in metres). Find the work done.

Solution

The displacement is $\vec{d} = \overrightarrow{AB} = [9 - 1, 5 - 2, 4 - 0] = [8, 3, 4]$.

$$W = \vec{F} \cdot \vec{d} = 25(8) + 10(3) + (-5)(4) = 200 + 30 - 20 = 210 \text{ J}$$

7. (Core) A triangle has vertices $A(2, 1, -1)$, $B(4, 3, 0)$, and $C(1, 5, 1)$. Find the angle at $A$, to one decimal place.

Solution

The angle at $A$ is between $\overrightarrow{AB}$ and $\overrightarrow{AC}$ (both start at $A$):

$$\overrightarrow{AB} = [2, 2, 1], \qquad \overrightarrow{AC} = [-1, 4, 2]$$

$\overrightarrow{AB} \cdot \overrightarrow{AC} = -2 + 8 + 2 = 8$, $|\overrightarrow{AB}| = \sqrt{4 + 4 + 1} = 3$, $|\overrightarrow{AC}| = \sqrt{1 + 16 + 4} = \sqrt{21}$.

$$\cos A = \frac{8}{3\sqrt{21}} \approx 0.5819 \qquad\Rightarrow\qquad A \approx 54.4^\circ$$

8. (Challenge) A rhombus has all four sides equal. Its sides from one corner are $\vec{a}$ and $\vec{b}$, with $|\vec{a}| = |\vec{b}|$, so its diagonals are $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$. Use the dot product to prove that the diagonals of a rhombus are perpendicular.

Solution

$$\begin{aligned} (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) &= \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b} \\ &= |\vec{a}|^2 - |\vec{b}|^2 && \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \\ &= 0 && |\vec{a}| = |\vec{b}| \end{aligned}$$

The dot product of the diagonals is $0$, so they’re perpendicular. (For a rectangle that isn’t a square, $|\vec{a}| \ne |\vec{b}|$, so its diagonals are not perpendicular.)

9. (Challenge) Find a vector that is perpendicular to both $\vec{a} = [1, 2, 0]$ and $\vec{b} = [0, 1, 3]$.

Solution

Let the vector be $[x, y, z]$. It needs a zero dot product with each:

$$x + 2y = 0 \qquad\text{and}\qquad y + 3z = 0$$

That’s two equations in three unknowns, so pick one value. Let $z = 1$. Then $y = -3$, and $x = -2y = 6$. One answer is $[6, -3, 1]$ (any non-zero multiple also works).

Check: $[6, -3, 1] \cdot [1, 2, 0] = 6 - 6 + 0 = 0$ ✓ and $[6, -3, 1] \cdot [0, 1, 3] = 0 - 3 + 3 = 0$ ✓

The cross product gives a vector like this in one step.