The Cross Product
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 2.4: “The Cross Product” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/2-4-the-cross-product |
| Textbook used in class | Stewart, Calculus, Section 12.4: “The Cross Product” (Examples 1, 2, 3) |
Opening Scenario
A wrench applies a force to a bolt. The torque (tendency to rotate) depends not just on the force and the arm length but also on the angle between them -- and it produces a rotation that is perpendicular to both the arm and the force. The cross product captures exactly this: from two vectors it produces a third vector perpendicular to both, with magnitude equal to the area of the parallelogram they span.
Quick Reference
Definition via the determinant: \[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = (a_2 b_3 - a_3 b_2)\,\mathbf{i} - (a_1 b_3 - a_3 b_1)\,\mathbf{j} + (a_1 b_2 - a_2 b_1)\,\mathbf{k}. \]
Key properties:
- $\mathbf{a} \times \mathbf{b}$ is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$.
- $\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})$ (anticommutative).
- $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta$, where $\theta \in [0, \pi]$.
- $|\mathbf{a} \times \mathbf{b}|$ equals the area of the parallelogram with sides $\mathbf{a}$ and $\mathbf{b}$.
- $\mathbf{a} \times \mathbf{b} = \mathbf{0}$ if and only if $\mathbf{a}$ and $\mathbf{b}$ are parallel (or one is zero).
Direction: use the right-hand rule -- curl the fingers of your right hand from $\mathbf{a}$ toward $\mathbf{b}$; your thumb points in the direction of $\mathbf{a}\times\mathbf{b}$.
Standard basis cross products: $\mathbf{i}\times\mathbf{j} = \mathbf{k}$, $\mathbf{j}\times\mathbf{k} = \mathbf{i}$, $\mathbf{k}\times\mathbf{i} = \mathbf{j}$ (and the reverse-order pairs are the negatives).
Key Concepts
1. Computing the Cross Product
Example 1. Find $\mathbf{a}\times\mathbf{b}$ for $\mathbf{a} = \langle 1, 3, 4\rangle$ and $\mathbf{b} = \langle 2, 7, -5\rangle$. (Stewart 12.4, Example 1.)
Expand along the first row: \[ \mathbf{a}\times\mathbf{b} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&3&4\\2&7&-5\end{vmatrix} \] \[ = \mathbf{i}(3\cdot(-5) - 4\cdot 7) - \mathbf{j}(1\cdot(-5) - 4\cdot 2) + \mathbf{k}(1\cdot 7 - 3\cdot 2) \] \[ = \mathbf{i}(-15 - 28) - \mathbf{j}(-5 - 8) + \mathbf{k}(7 - 6) \] \[ = -43\,\mathbf{i} + 13\,\mathbf{j} + 1\,\mathbf{k} = \langle -43, 13, 1\rangle. \]
Verify perpendicularity: $\langle -43,13,1\rangle\cdot\langle 1,3,4\rangle = -43 + 39 + 4 = 0$. Correct.
2. Finding a Normal to a Plane
If $\mathbf{a}$ and $\mathbf{b}$ lie in a plane, then $\mathbf{a}\times\mathbf{b}$ is a normal vector to that plane. This is one of the most frequent uses of the cross product.
Example 2. Find a vector perpendicular to the plane through $A = (1, 1, 0)$, $B = (-1, 0, 2)$, $C = (0, 2, 1)$.
\[ \overrightarrow{AB} = \langle -2, -1, 2\rangle, \quad \overrightarrow{AC} = \langle -1, 1, 1\rangle. \] \[ \overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-2&-1&2\\-1&1&1\end{vmatrix} = (-1-2)\mathbf{i} - (-2-(-2))\mathbf{j} + (-2-1)\mathbf{k} = \langle -3, 0, -3\rangle. \]
A normal to the plane is $\langle -3, 0, -3\rangle$, or simplified: $\langle 1, 0, 1\rangle$.
3. Area of a Parallelogram
Example 3. Find the area of the parallelogram with adjacent sides $\mathbf{a} = \langle 2, 1, 0\rangle$ and $\mathbf{b} = \langle 0, 3, 4\rangle$.
\[ \mathbf{a}\times\mathbf{b} = \langle (1)(4)-(0)(3), (0)(0)-(2)(4), (2)(3)-(1)(0)\rangle = \langle 4, -8, 6\rangle. \] \[ \text{Area} = |\mathbf{a}\times\mathbf{b}| = \sqrt{16 + 64 + 36} = \sqrt{116} = 2\sqrt{29}. \]
treating the cross product as commutative. $\mathbf{a}\times\mathbf{b} = -(\mathbf{b}\times\mathbf{a})$: the order matters. Swapping the two vectors flips the direction of the result. This is unlike the dot product, which is commutative. If you reverse the order, multiply the result by $-1$.
Common Errors Summary
| Error | Correction |
|---|---|
| Forgetting the minus sign on the $\mathbf{j}$ component | The cofactor expansion gives $+\mathbf{i}(\ldots) - \mathbf{j}(\ldots) + \mathbf{k}(\ldots)$ |
| Computing $\mathbf{a}\times\mathbf{a}$ as nonzero | $\mathbf{a}\times\mathbf{a} = \mathbf{0}$ for any vector; a vector crossed with itself is zero |
| Confusing cross product with dot product | Cross product gives a vector; dot product gives a scalar |
Common Misconceptions
the cross product is commutative, so $\mathbf{a} \times \mathbf{b} = \mathbf{b} \times \mathbf{a}$.
This is the concept-image-conflicts-definition error. The cross product is anticommutative: $\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})$. Swapping the two vectors reverses the direction of the result because the right-hand rule gives the opposite orientation. For example, $\mathbf{i} \times \mathbf{j} = \mathbf{k}$ but $\mathbf{j} \times \mathbf{i} = -\mathbf{k}$. Unlike the dot product, where order is irrelevant, order in the cross product determines the sign of the result.
Leveled Practice
Problem 1. Compute $\langle 1, -1, 2\rangle \times \langle 3, 1, 0\rangle$.
Show answer
$= \langle (-1)(0)-(2)(1), (2)(3)-(1)(0), (1)(1)-(-1)(3)\rangle = \langle -2, 6, 4\rangle$.
Problem 2. Find the area of the triangle with vertices $A = (1,0,0)$, $B = (0,2,0)$, $C = (0,0,3)$.
Show answer
$\overrightarrow{AB} = \langle -1,2,0\rangle$, $\overrightarrow{AC} = \langle -1,0,3\rangle$.
$\overrightarrow{AB}\times\overrightarrow{AC} = \langle 6, 3, 2\rangle$. Area of triangle $= \frac{1}{2}|\langle 6,3,2\rangle| = \frac{1}{2}\sqrt{49} = 7/2$.
Mastery Checklist
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