The Dot Product
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 2.3: “The Dot Product” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/2-3-the-dot-product |
| Textbook used in class | Stewart, Calculus, Section 12.3: “The Dot Product” (Examples 1, 2, 3, 4) |
Opening Scenario
Work done by a force is $W = F \cdot d$ when force and displacement are in the same direction. When they are not, only the component of force along the displacement contributes. The dot product computes exactly that: $\mathbf{F} \cdot \mathbf{d} = |\mathbf{F}||\mathbf{d}|\cos\theta$, where $\theta$ is the angle between the vectors. This single operation encodes both the magnitudes and the angle between two vectors.
Quick Reference
Component formula: $\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$.
Geometric formula: $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$, where $\theta \in [0, \pi]$ is the angle between them.
The dot product is a scalar, not a vector.
Key consequences:
- $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2$.
- $\mathbf{a} \perp \mathbf{b} \iff \mathbf{a} \cdot \mathbf{b} = 0$ (for nonzero vectors).
- $\cos\theta = \dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$.
Properties:
- Commutative: $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$.
- Distributive: $\mathbf{a}\cdot(\mathbf{b}+\mathbf{c}) = \mathbf{a}\cdot\mathbf{b} + \mathbf{a}\cdot\mathbf{c}$.
- $(c\mathbf{a})\cdot\mathbf{b} = c(\mathbf{a}\cdot\mathbf{b})$.
Key Concepts
1. Computing the Dot Product
Example 1. Compute $\langle 2, -1, 3\rangle \cdot \langle -1, 4, 2\rangle$. (Stewart 12.3, Example 1.)
\[ = (2)(-1) + (-1)(4) + (3)(2) = -2 - 4 + 6 = 0. \]
Since the dot product is $0$, the two vectors are perpendicular.
2. Finding the Angle Between Two Vectors
Example 2. Find the angle between $\mathbf{a} = \langle 1, 2, -1\rangle$ and $\mathbf{b} = \langle 3, -1, 2\rangle$. (Stewart 12.3, Example 2.)
\[ \mathbf{a}\cdot\mathbf{b} = 3 - 2 - 2 = -1, \quad |\mathbf{a}| = \sqrt{6}, \quad |\mathbf{b}| = \sqrt{14}. \] \[ \cos\theta = \frac{-1}{\sqrt{6}\sqrt{14}} = \frac{-1}{\sqrt{84}}. \] \[ \theta = \arccos\left(\frac{-1}{\sqrt{84}}\right) \approx 96.3^\circ. \]
Since $\theta > 90^\circ$, the vectors point into “opposite hemispheres.”
3. Testing for Perpendicularity
Perpendicular (orthogonal) vectors satisfy $\mathbf{a}\cdot\mathbf{b} = 0$. The zero dot product is the algebraic definition of perpendicularity in any dimension.
Example 3. Is $\mathbf{u} = \langle 2, 3, -1\rangle$ perpendicular to $\mathbf{v} = \langle 1, 0, 2\rangle$?
$\mathbf{u}\cdot\mathbf{v} = 2 + 0 - 2 = 0$: yes, they are perpendicular.
thinking the dot product is a vector. The dot product of two vectors is a scalar (a number). Adding or subtracting it as if it were a vector is a type error. The cross product (next section) is the vector product. When a formula yields $\mathbf{a}\cdot\mathbf{b}$, the result is a number, not a direction.
Common Errors Summary
| Error | Correction |
|---|---|
| Computing $\mathbf{a}\cdot\mathbf{b}$ as $a_1 b_1 + a_2 b_2 + a_3 b_3$ but multiplying component pairs incorrectly | Carefully pair $a_1$ with $b_1$, $a_2$ with $b_2$, $a_3$ with $b_3$ |
| Confusing $\mathbf{a}\cdot\mathbf{a} = |\mathbf{a}|^2$ with $\mathbf{a}\cdot\mathbf{a} = |\mathbf{a}|$ | The dot of $\mathbf{a}$ with itself equals the square of the magnitude |
| Using the formula $\cos\theta = \mathbf{a}\cdot\mathbf{b}$ without dividing by the magnitudes | Always divide: $\cos\theta = (\mathbf{a}\cdot\mathbf{b})/(|\mathbf{a}||\mathbf{b}|)$ |
Common Misconceptions
the dot product of two vectors is a vector.
This is the concept-image-conflicts-definition error. The dot product $\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$ is a scalar (a single real number), not a vector. Adding or subtracting the result as if it were a vector is a type error. For example, $\langle 2, -1, 3\rangle \cdot \langle -1, 4, 2\rangle = -2 - 4 + 6 = 0$; the answer $0$ is a number, not the zero vector $\langle 0, 0, 0\rangle$. The cross product is the operation that produces a vector output.
Leveled Practice
Problem 1. Find the angle between $\mathbf{a} = \langle 1, 0, 1\rangle$ and $\mathbf{b} = \langle 0, 1, 1\rangle$.
Show answer
$\mathbf{a}\cdot\mathbf{b} = 1$, $|\mathbf{a}| = \sqrt{2}$, $|\mathbf{b}| = \sqrt{2}$. $\cos\theta = 1/2$, so $\theta = \pi/3 = 60^\circ$.
Problem 2. For which value of $c$ is $\langle c, 3, -2\rangle$ perpendicular to $\langle 2, -1, c\rangle$?
Show answer
$\langle c,3,-2\rangle\cdot\langle 2,-1,c\rangle = 2c - 3 - 2c = -3 \neq 0$. There is no value of $c$ that makes them perpendicular.
Wait, let me recheck: $2c + (3)(-1) + (-2)(c) = 2c - 3 - 2c = -3$. So no, they cannot be made perpendicular.
Answer: No value of $c$ works; the dot product is $-3$ regardless of $c$.