Vector (Cross) Product

The Vector (Cross) Product

The second way to multiply vectors outputs a vector:

$$\vec{a} \times \vec{b} = |\vec{a}|\,|\vec{b}|\sin\theta\;\hat{n}$$

where $\theta$ ($0 \leq \theta \leq \pi$) is the angle between the vectors and $\hat{n}$ is the unit vector perpendicular to both $\vec{a}$ and $\vec{b}$, chosen so that $\vec{a}, \vec{b}, \hat{n}$ form a right-handed system: curl the right hand's fingers from $\vec{a}$ towards $\vec{b}$, and the thumb points along $\hat{n}$. If either vector is $\vec{0}$, define $\vec{a} \times \vec{b} = \vec{0}$.

Observations

1. $\vec{a} \times \vec{b}$ is a vector (the dot product was a scalar).
2. Parallel test: for nonzero vectors, $\vec{a} \times \vec{b} = \vec{0} \iff \vec{a} \parallel \vec{b}$ (then $\sin\theta = 0$). In particular $\vec{a} \times \vec{a} = \vec{0}$.
3. $\theta = \dfrac{\pi}{2}$ gives the maximum magnitude $|\vec{a}||\vec{b}|$.
4. The cycle: $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{i} = \hat{j}$; each self-cross is $\vec{0}$; and against the cycle the sign flips: $\hat{j} \times \hat{i} = -\hat{k}$, $\hat{k} \times \hat{j} = -\hat{i}$, $\hat{i} \times \hat{k} = -\hat{j}$.
5. Angle via sine: $\sin\theta = \dfrac{\left|\vec{a} \times \vec{b}\right|}{|\vec{a}||\vec{b}|}$.
6. NOT commutative: $\vec{a} \times \vec{b} = -\left(\vec{b} \times \vec{a}\right)$ — reversing the order reverses the thumb. It is distributive over addition, and scalars slide through: $\lambda\left(\vec{a} \times \vec{b}\right) = (\lambda\vec{a}) \times \vec{b} = \vec{a} \times (\lambda\vec{b})$.

Key Point (dot vs cross in one line): $\cos$ goes with the dot and gives a scalar; $\sin$ goes with the cross and gives a vector. Perpendicular vectors maximise the cross and kill the dot; parallel vectors do the exact opposite.

Areas, and the Determinant Formula

Areas

1. Triangle with adjacent sides $\vec{a}$ and $\vec{b}$:

$$\text{Area} = \frac{1}{2}\left|\vec{a} \times \vec{b}\right|$$

(base $|\vec{a}|$, height $|\vec{b}|\sin\theta$ — the cross product packages both).

2. Parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$:

$$\text{Area} = \left|\vec{a} \times \vec{b}\right|$$

For a triangle given by vertices $A, B, C$: form $\overrightarrow{AB}$ and $\overrightarrow{AC}$ first, then take half the magnitude of their cross product.

The determinant formula

Expanding $\left(a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\right) \times \left(b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\right)$ by distributivity and the cycle table collapses nine cross terms into the tidy determinant

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = \left(a_2b_3 - a_3b_2\right)\hat{i} - \left(a_1b_3 - a_3b_1\right)\hat{j} + \left(a_1b_2 - a_2b_1\right)\hat{k}$$

Key Point (the sign everyone forgets): the $\hat{j}$-component carries a minus sign from the cofactor expansion. Compute all three $2 \times 2$ minors, then flip the middle one — chanting "plus, MINUS, plus" while expanding prevents the most common cross-product error in existence.

A useful identity

Squaring the definition and using $\sin^2\theta = 1 - \cos^2\theta$:

$$\left|\vec{a} \times \vec{b}\right|^2 + \left(\vec{a} \cdot \vec{b}\right)^2 = |\vec{a}|^2\,|\vec{b}|^2$$

(Lagrange's identity) — it converts between dot and cross data without ever finding $\theta$ itself.

Solved Examples

Example 1: A straight determinant computation

Find $\vec{a} \times \vec{b}$ and its magnitude for $\vec{a} = 2\hat{i} + \hat{j} + 3\hat{k}$, $\vec{b} = 3\hat{i} + 5\hat{j} - 2\hat{k}$.

Solution:

1. Determinant: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 3 & 5 & -2 \end{vmatrix} = (-2 - 15)\hat{i} - (-4 - 9)\hat{j} + (10 - 3)\hat{k}$.
2. Simplify: $-17\hat{i} + 13\hat{j} + 7\hat{k}$.
3. Magnitude: $\sqrt{289 + 169 + 49} = \sqrt{507}$.

Answer: $\vec{a} \times \vec{b} = -17\hat{i} + 13\hat{j} + 7\hat{k}$, $\left|\vec{a} \times \vec{b}\right| = \sqrt{507}$ — note the minus applied to the $\hat{j}$ minor turning $(-4 - 9)$ into $+13$.

Example 2: A unit vector perpendicular to two vectors

Find a unit vector perpendicular to each of $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$, where $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.

Solution:

1. Build the two vectors: $\vec{a} + \vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}$, $\vec{a} - \vec{b} = -\hat{j} - 2\hat{k}$.
2. Cross them: $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 0 & -1 & -2 \end{vmatrix} = -2\hat{i} + 4\hat{j} - 2\hat{k}$.
3. Normalise: magnitude $= \sqrt{4 + 16 + 4} = 2\sqrt{6}$, giving $\dfrac{-1}{\sqrt{6}}\hat{i} + \dfrac{2}{\sqrt{6}}\hat{j} - \dfrac{1}{\sqrt{6}}\hat{k}$.

Answer: $\pm\dfrac{1}{\sqrt{6}}\left(-\hat{i} + 2\hat{j} - \hat{k}\right)$ — every plane has two perpendicular directions; the second comes from crossing in the opposite order.

Example 3: Area of a triangle from vertices

Find the area of the triangle with vertices $A(1, 1, 1)$, $B(1, 2, 3)$, $C(2, 3, 1)$.

Solution:

1. Side vectors from one vertex: $\overrightarrow{AB} = \hat{j} + 2\hat{k}$, $\overrightarrow{AC} = \hat{i} + 2\hat{j}$.
2. Cross: $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix} = -4\hat{i} + 2\hat{j} - \hat{k}$.
3. Halve the magnitude: $\left|\overrightarrow{AB} \times \overrightarrow{AC}\right| = \sqrt{16 + 4 + 1} = \sqrt{21}$.

Answer: area $= \dfrac{\sqrt{21}}{2}$ square units.

Example 4: Area of a parallelogram

Find the area of the parallelogram whose adjacent sides are $\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.

Solution:

1. Cross: $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix} = (1 + 4)\hat{i} - (3 - 4)\hat{j} + (-3 - 1)\hat{k} = 5\hat{i} + \hat{j} - 4\hat{k}$.
2. Magnitude: $\sqrt{25 + 1 + 16} = \sqrt{42}$.

Answer: area $= \sqrt{42}$ square units — no halving for a parallelogram.

Example 5: Completing a unit vector from two angles

A unit vector $\vec{a}$ makes angle $\dfrac{\pi}{3}$ with $\hat{i}$, angle $\dfrac{\pi}{4}$ with $\hat{j}$, and an acute angle $\theta$ with $\hat{k}$. Find $\theta$ and the components of $\vec{a}$.

Solution:

1. Direction cosines of a unit vector are its components: $a_1 = \cos\dfrac{\pi}{3} = \dfrac{1}{2}$, $a_2 = \cos\dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}$.
2. Use $l^2 + m^2 + n^2 = 1$: $\dfrac{1}{4} + \dfrac{1}{2} + \cos^2\theta = 1$, so $\cos^2\theta = \dfrac{1}{4}$ and (acute) $\cos\theta = \dfrac{1}{2}$.

Answer: $\theta = \dfrac{\pi}{3}$ and $\vec{a} = \dfrac{1}{2}\hat{i} + \dfrac{1}{\sqrt{2}}\hat{j} + \dfrac{1}{2}\hat{k}$.

Example 6: Parallel vectors fix the unknowns

Find $\lambda$ and $\mu$ if $\left(2\hat{i} + 6\hat{j} + 27\hat{k}\right) \times \left(\hat{i} + \lambda\hat{j} + \mu\hat{k}\right) = \vec{0}$.

Solution:

1. Zero cross means parallel: the components must be proportional: $\dfrac{1}{2} = \dfrac{\lambda}{6} = \dfrac{\mu}{27}$.
2. Solve: $\lambda = 3$, $\mu = \dfrac{27}{2}$.

Answer: $\lambda = 3, \ \mu = \dfrac{27}{2}$ — a cross-product equation solved with no determinant at all, by reading its geometric meaning.

Example 7: Another triangle area

Find the area of the triangle with vertices $A(1, 1, 2)$, $B(2, 3, 5)$, $C(1, 5, 5)$.

Solution:

1. Side vectors: $\overrightarrow{AB} = \hat{i} + 2\hat{j} + 3\hat{k}$, $\overrightarrow{AC} = 4\hat{j} + 3\hat{k}$.
2. Cross: $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix} = (6 - 12)\hat{i} - (3 - 0)\hat{j} + (4 - 0)\hat{k} = -6\hat{i} - 3\hat{j} + 4\hat{k}$.
3. Halve the magnitude: $\sqrt{36 + 9 + 16} = \sqrt{61}$.

Answer: area $= \dfrac{\sqrt{61}}{2}$ square units.

Example 8: Parallelogram with messier sides

Find the area of the parallelogram whose adjacent sides are $\hat{i} - \hat{j} + 3\hat{k}$ and $2\hat{i} - 7\hat{j} + \hat{k}$.

Solution:

1. Cross: $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix} = (-1 + 21)\hat{i} - (1 - 6)\hat{j} + (-7 + 2)\hat{k} = 20\hat{i} + 5\hat{j} - 5\hat{k}$.
2. Magnitude: $\sqrt{400 + 25 + 25} = \sqrt{450} = 15\sqrt{2}$.

Answer: area $= 15\sqrt{2}$ square units.