How the Cross Product Measures the Area of a Parallelogram

The dot product turns two vectors into a number. The cross product does something different: it produces a new vector perpendicular to both, whose length measures the area they span. It describes turning effects such as torque and the force on a moving charge in a magnetic field.

This lesson covers the definition and geometric meaning, areas of triangles and parallelograms, and testing collinearity.

**The vector product $\vec{a}\times \vec{b}$ is a vector of magnitude $|\vec{a}||\vec{b}|\sin \theta$, perpendicular to both $\vec{a}$ and $\vec{b}$, with direction given by the right-hand rule, and in components it is the determinant $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|$.

Unit vector products:**

- $\hat{i}\times \hat{j}=\hat{k}$, $\hat{j}\times \hat{k}=\hat{i}$ and $\hat{k}\times \hat{i}=\hat{j}$
- $\hat{j}\times \hat{i}=-\hat{k}$, and $\hat{i}\times \hat{i}=\vec{0}$

Properties:

- $\vec{b}\times \vec{a}=-(\vec{a}\times \vec{b})$ — the product is not commutative
- $\vec{a}\times (\vec{b}+\vec{c})=\vec{a}\times \vec{b}+\vec{a}\times \vec{c}$

Worked example. For $\vec{a}=2\hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+2\hat{k}$:

$$\vec{a}\times \vec{b}=\hat{i}(2-1)-\hat{j}(4+1)+\hat{k}(-2-1)=\hat{i}-5\hat{j}-3\hat{k}$$

Check. The dot product of $\hat{i}-5\hat{j}-3\hat{k}$ with $\vec{a}$ is $2-5+3=0$, and with $\vec{b}$ it is $1+5-6=0$ — the result is perpendicular to both.

An everyday example. Tightening a nut with a spanner produces a turning effect along the bolt, perpendicular to both the spanner and your push — the torque $\vec{r}\times \vec{F}$.

The substance. Order matters — reversing the two vectors reverses the direction of the product.

**A parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ has area $|\vec{a}\times \vec{b}|$, a triangle with two sides $\vec{a}$ and $\vec{b}$ has half that area, and a parallelogram with diagonals $\overrightarrow{d_1}$ and $\overrightarrow{d_2}$ has area $\frac{1}{2}|\overrightarrow{d_1}\times \overrightarrow{d_2}|$.

Worked example (parallelogram).** With adjacent sides $\vec{a}=2\hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+2\hat{k}$, the cross product is $\hat{i}-5\hat{j}-3\hat{k}$, so the area is

$$|\vec{a}\times \vec{b}|=\sqrt{1+25+9}=\sqrt{35}\approx 5.92\text{ square units}$$

Worked example (triangle). For $A(1,1,1)$, $B(1,2,3)$ and $C(2,3,1)$: $\overrightarrow{AB}=\hat{j}+2\hat{k}$ and $\overrightarrow{AC}=\hat{i}+2\hat{j}$.

$$\overrightarrow{AB}\times \overrightarrow{AC}=\hat{i}(0-4)-\hat{j}(0-2)+\hat{k}(0-1)=-4\hat{i}+2\hat{j}-\hat{k}$$

So the area is $\frac{1}{2}\sqrt{16+4+1}=\frac{\sqrt{21}}{2}\approx 2.29$ square units.

An everyday example. A triangular sunshade stretched between three hooks at different heights on a terrace has an area found exactly this way from the coordinates of the hooks.

The substance. This method works in three dimensions — the triangle does not need to lie in a coordinate plane, unlike the coordinate area formula.

**Two non-zero vectors are collinear exactly when their cross product is the zero vector, which happens when their components are proportional; three points A, B and C are collinear when $\overrightarrow{AB}\times \overrightarrow{AC}=\vec{0}$.

Why it works.** $|\vec{a}\times \vec{b}|=|\vec{a}||\vec{b}|\sin \theta$ is zero for non-zero vectors only when $\theta$ is $0°$ or $180°$.

Worked example. Are $\vec{a}=2\hat{i}-3\hat{j}+4\hat{k}$ and $\vec{b}=-4\hat{i}+6\hat{j}-8\hat{k}$ collinear?

$$\vec{a}\times \vec{b}=\hat{i}(24-24)-\hat{j}(-16+16)+\hat{k}(12-12)=\vec{0}$$

Yes — indeed $\vec{b}=-2\vec{a}$, so they lie along the same line in opposite directions.

Worked example 2. For $\hat{i}+2\hat{j}+\lambda \hat{k}$ and $3\hat{i}+6\hat{j}+9\hat{k}$ to be collinear, the ratios $\frac{1}{3}=\frac{2}{6}=\frac{\lambda }{9}$ give $\lambda =3$.

An everyday example. Checking that three lamp posts along a straight road really stand in line is testing whether the vectors between them are collinear.

The substance. Collinear vectors can point in opposite directions — a negative multiple still gives a zero cross product.