Vector (cross) product

The vector product (or cross product) takes two vectors and produces a new vector , one that is perpendicular to both inputs and whose magnitude equals the area of the parallelogram they span. It is the key tool for finding normals, computing areas in 3D, and constructing perpendiculars.

Definition

Given $\vec{a}, \vec{b} \in \mathbb{R}^3$, the cross product $\vec{a} \times \vec{b}$ is the unique vector satisfying:

1. Magnitude: $|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta$, where $\theta \in [0, \pi]$ is the angle between $\vec{a}$ and $\vec{b}$.
2. Direction: perpendicular to both $\vec{a}$ and $\vec{b}$, with sense given by the right-hand rule , curl the fingers from $\vec{a}$ to $\vec{b}$; the thumb points in the direction of $\vec{a}\times\vec{b}$.

Determinant formula

In components, $\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = (a_2 b_3 - a_3 b_2)\hat{i} - (a_1 b_3 - a_3 b_1)\hat{j} + (a_1 b_2 - a_2 b_1)\hat{k}.$

This is the operational form you'll use most.

Properties

• Anti-commutative: $\vec{a}\times\vec{b} = -\vec{b}\times\vec{a}$.
• Distributive over addition: $\vec{a}\times(\vec{b} + \vec{c}) = \vec{a}\times\vec{b} + \vec{a}\times\vec{c}$.
• Scalar factors: $(\lambda\vec{a})\times\vec{b} = \lambda(\vec{a}\times\vec{b}) = \vec{a}\times(\lambda\vec{b})$.
• $\vec{a}\times\vec{a} = \vec{0}$ (sine of zero).
• Parallel vectors: $\vec{a}\times\vec{b} = \vec{0} \iff \vec{a} \parallel \vec{b}$ (one is a scalar multiple of the other, including $\vec{0}$).
• Basis cross products: $\hat{i}\times\hat{j} = \hat{k}$, $\hat{j}\times\hat{k} = \hat{i}$, $\hat{k}\times\hat{i} = \hat{j}$ (cyclic). Reversing the order flips the sign.

Area interpretation

The magnitude $|\vec{a}\times\vec{b}|$ equals the area of the parallelogram with sides $\vec{a}$ and $\vec{b}$ (base $|\vec{a}|$, height $|\vec{b}|\sin\theta$).

The area of a triangle with two sides $\vec{a}$ and $\vec{b}$ is $\dfrac{1}{2}|\vec{a}\times\vec{b}|$.

The area of a triangle with vertices $A$, $B$, $C$ is $\dfrac{1}{2}|\vec{AB}\times\vec{AC}|$.

Finding a perpendicular

To find a vector perpendicular to two given vectors $\vec{a}$ and $\vec{b}$, just compute $\vec{a}\times\vec{b}$. To get a unit perpendicular, divide by the magnitude: $\hat{n} = \frac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}.$

This is exactly how you find the normal to a plane spanned by two vectors.

Worked examples

Example 1. $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$. Find $\vec{a}\times\vec{b}$.

$\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 1 & -1 \end{vmatrix} = \hat{i}(2(-1) - 3(1)) - \hat{j}(1(-1) - 3(2)) + \hat{k}(1(1) - 2(2)) = \hat{i}(-5) - \hat{j}(-7) + \hat{k}(-3) = -5\hat{i} + 7\hat{j} - 3\hat{k}$.

Example 2. Find the area of the parallelogram with adjacent sides $\vec{a} = \hat{i} + \hat{j} - \hat{k}$, $\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$.

$\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\1 & 1 & -1\\2 & -1 & 1\end{vmatrix} = \hat{i}(1 - 1) - \hat{j}(1 + 2) + \hat{k}(-1 - 2) = -3\hat{j} - 3\hat{k}$. Area $= \sqrt{9 + 9} = 3\sqrt 2$.

Example 3. Area of the triangle with vertices $A = (1, 1, 1)$, $B = (4, 5, 6)$, $C = (0, 2, 3)$.

$\vec{AB} = (3, 4, 5)$, $\vec{AC} = (-1, 1, 2)$. $\vec{AB}\times\vec{AC} = \begin{vmatrix}\hat i & \hat j & \hat k\\3 & 4 & 5\\-1 & 1 & 2\end{vmatrix} = \hat i(8 - 5) - \hat j(6 + 5) + \hat k(3 + 4) = 3\hat i - 11\hat j + 7\hat k$. Magnitude $= \sqrt{9 + 121 + 49} = \sqrt{179}$. Area $= \sqrt{179}/2$.

Example 4. Find a unit vector perpendicular to both $\hat{i} + \hat{j}$ and $\hat{j} + \hat{k}$.

Cross product: $\begin{vmatrix}\hat i & \hat j & \hat k\\1 & 1 & 0\\0 & 1 & 1\end{vmatrix} = \hat i(1) - \hat j(1) + \hat k(1) = \hat i - \hat j + \hat k$. Magnitude $\sqrt 3$. Unit vector: $\dfrac{1}{\sqrt 3}(\hat i - \hat j + \hat k)$.

Example 5. Show that $|\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2$.

By definition: $|\vec{a}\times\vec{b}|^2 = |\vec a|^2|\vec b|^2 \sin^2\theta$ and $(\vec{a}\cdot\vec{b})^2 = |\vec a|^2|\vec b|^2\cos^2\theta$. Add: $|\vec a|^2|\vec b|^2(\sin^2\theta + \cos^2\theta) = |\vec a|^2|\vec b|^2$.

Example 6. Find the sine of the angle between $\vec{a} = 2\hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.

$\vec{a}\times\vec{b} = \begin{vmatrix}\hat i & \hat j & \hat k\\2 & 1 & 1\\1 & -1 & 1\end{vmatrix} = \hat i(1 + 1) - \hat j(2 - 1) + \hat k(-2 - 1) = 2\hat i - \hat j - 3\hat k$. Magnitude $\sqrt{14}$. $|\vec a| = \sqrt 6$, $|\vec b| = \sqrt 3$. $\sin\theta = \dfrac{\sqrt{14}}{\sqrt{18}} = \dfrac{\sqrt{14}}{3\sqrt 2} = \dfrac{\sqrt 7}{3}$.

Try it yourself

1. Find $\vec{a}\times\vec{b}$ if $\vec{a} = \hat{i} + \hat{j}$, $\vec{b} = \hat{j} + \hat{k}$.
2. Find the area of the parallelogram with sides $\vec{a} = 3\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{b} = \hat{i} - 3\hat{j} + 4\hat{k}$.
3. Area of the triangle with vertices $(0, 0, 0)$, $(1, 0, 0)$, $(0, 1, 0)$.
4. Find a unit vector perpendicular to $\hat{i} - \hat{j}$ and $\hat{i} + \hat{k}$.
5. Show that $\vec{a}\times\vec{a} = \vec{0}$.
6. Compute $(\hat{i} + 2\hat{j})\times(3\hat{i} - \hat{j})$.
7. If $\vec{a}\times\vec{b} = \vec{0}$ and neither vector is zero, what is the relationship between $\vec{a}$ and $\vec{b}$?
8. Find $\lambda$ if $(2\hat i + \lambda \hat j + \hat k) \times (\hat i + \hat j) = \vec{0}$.
9. Verify $\hat{i}\times(\hat{j}\times\hat{k}) = \hat{i}\times\hat{i} = \vec{0}$.
10. Find the area of the triangle with vertices $A = (1, 2, 3)$, $B = (2, -1, 4)$, $C = (4, 5, -1)$.
11. Show that $\vec{a}\times\vec{b} = -\vec{b}\times\vec{a}$.
12. Find a vector of magnitude $5$ perpendicular to both $\hat{i} + 2\hat{j}$ and $3\hat{i} - \hat{k}$.
13. Find $|\vec{a}\times\vec{b}|$ if $|\vec{a}| = 4$, $|\vec{b}| = 3$, angle $\theta = 30^\circ$.
14. Show that the diagonals of a rhombus are perpendicular using cross products.

Pitfalls and tricks

• Cross product is a vector in 3D; it doesn't exist in 2D in the same form.
• Right-hand rule determines orientation; reversing the order flips the sign.
• Don't confuse $\vec{a}\times\vec{b}$ with $\vec{a}\cdot\vec{b}$. The cross is a vector; the dot is a scalar.
• For area, take the magnitude. Just $|\vec{a}\times\vec{b}|$ for parallelogram, halved for triangle.
• The determinant formula is mechanical , memorise it.