Vector product

The vector product (cross product) is an operation between two vectors in space that gives a new vector perpendicular to both of them.

What is the vector product

The vector product is defined between two vectors, and only in space. Unlike the scalar product, the result of a vector product is again a vector. The vector product of the vectors $\overrightarrow{\mathbf{u}}$ and $\overrightarrow{\mathbf{v}}$ is a vector $\overrightarrow{\mathbf{w}}$ with the following properties:

$$|\overrightarrow{\mathbf{w}}|=|\overrightarrow{\mathbf{u}}|\cdot |\overrightarrow{\mathbf{v}}|\cdot \sin \alpha ,$$

where α is the angle between the vectors $\overrightarrow{\mathbf{u}}$ and $\overrightarrow{\mathbf{v}}$. The vector $\overrightarrow{\mathbf{w}}$ is also perpendicular to both $\overrightarrow{\mathbf{u}}$ and $\overrightarrow{\mathbf{v}}$. Its direction follows the right-hand rule. To tell the vector product from the scalar product, we use the sign × for the vector product: $\overrightarrow{\mathbf{u}}\times \overrightarrow{\mathbf{v}}$.

The picture shows one possible vector product $\overrightarrow{\mathbf{u}}\times \overrightarrow{\mathbf{v}}$. The result is a vector $\overrightarrow{\mathbf{w}}$ perpendicular to both vectors. We are working in space, so the vectors $\overrightarrow{\mathbf{u}}$ and $\overrightarrow{\mathbf{v}}$ lie in one plane and the vector $\overrightarrow{\mathbf{w}}$ is perpendicular to that plane.

If one vector is a linear combination of the other, then their vector product is the zero vector. This covers the cases where both vectors lie on one line. If at least one vector is the zero vector, the product is also zero.

How to find the coordinates of the resulting vector

So far we can work out the magnitude and direction of the resulting vector. It is more practical, though, to know its coordinates directly. The formula itself looks like this:

$$u\times v=(u_2{v}_3-v_2{u}_3,u_3{v}_1-v_3{u}_1,u_1{v}_2-v_1{u}_2)$$

This formula is rather hard to remember, so here is a memory aid:

With this table we can multiply vectors with ease. For example, take the vectors $\overrightarrow{\mathbf{u}}=(4,0,0)$ and $\overrightarrow{\mathbf{v}}=(0,5,0)$. These vectors lie entirely along the x-axis and the y-axis respectively. Their vector product is

$$\begin{array}{cccccccc}\overrightarrow{\mathbf{u}}& 0& & 0& & 4& & 0\\ \overrightarrow{\mathbf{v}}& 5& & 0& & 0& & 5\\ \overrightarrow{\mathbf{u}}\times \overrightarrow{\mathbf{v}}& & 0& & 0& & 20\end{array}$$

The resulting vector $\overrightarrow{\mathbf{w}}$ has coordinates (0, 0, 20). We can check that the result is right using the first definition, which says that

$$|w|=|u|\cdot |v|\cdot \sin \alpha ,$$

Our vector $\overrightarrow{\mathbf{w}}$ has magnitude 20. By the formula we must therefore have

$$20=|u|\cdot |v|\cdot \sin \alpha$$

Substituting:

$$20=4\cdot 5\cdot \sin {90}^{\circ }=20\cdot 1=20.$$

What is it good for

Clearly we can use the vector product to find a vector perpendicular to two other vectors. We can check that this really works using the scalar product: the result must be zero. Checking the previous example:

$$\begin{array}{rcl}\overrightarrow{\mathbf{u}}\cdot \overrightarrow{\mathbf{w}}& =& 4\cdot 0+0+0\cdot 20=0\\ \overrightarrow{\mathbf{v}}\cdot \overrightarrow{\mathbf{w}}& =& 0+5\cdot 0+0\cdot 20=0\end{array}$$

Another use is finding the area of a parallelogram in space. If we have a parallelogram ABCD in space and vectors $\overrightarrow{\mathbf{u}}$ and $\overrightarrow{\mathbf{v}}$ formed by the sides AB and AD, then the area of the parallelogram is $S=|\overrightarrow{\mathbf{u}}\times \overrightarrow{\mathbf{v}}|$. If we have a triangle ABC in space and vectors $\overrightarrow{\mathbf{u}}$ and $\overrightarrow{\mathbf{v}}$ formed by the sides AB and AC, then its area is $T=\frac{1}{2}|\overrightarrow{\mathbf{u}}\times \overrightarrow{\mathbf{v}}|$.

For example, find the area of the triangle in space with vertices A(1, 0, 3), B(5, 6, 8) and C(3, 5, 4).

First we have to find the vectors $\overrightarrow{\mathbf{u}}$ and $\overrightarrow{\mathbf{v}}$.

$$\begin{array}{rcl}u& =& B-A=(4,6,5)\\ v& =& C-A=(2,5,1)\end{array}$$

Now we work out their product:

$$\overrightarrow{\mathbf{w}}=\overrightarrow{\mathbf{u}}\times \overrightarrow{\mathbf{v}}=(-19,6,8)$$

Then its magnitude:

$$|\overrightarrow{\mathbf{w}}|=\sqrt{(-19{)}^2+6^2+8^2}=\sqrt{461}\approx 21.47$$

And finally we divide the result by two:

$$T=\frac{21.47}{2}=10.735$$