Module 2: The Relationship Between Angles, Magnitudes, and the Cross Product

As with the dot product, the cross product offers an interesting relation about the arrangement of two vectors:

$\|\vec{a} \times \vec{b}\| = \|\vec{a} \| \| \vec{b} \| \sin (\theta)$

where $\| \, \|$ is the norm, and $\theta$ is the angle between vectors $\vec{a}$ and $\vec{b}$.

Whereas the dot product (with cosine) gave us a zero value when the vectors are perpendicular, the cross product (with sine) will give us the zero vector when both vectors are parallel! However, this is often not the best use of the cross product (it is usually easy to see just by looking at them if vectors are parallel). Moreover, this is not the best way to find the angle between given vectors either, since the computation in the dot product formula is often simpler.

Instead we offer another application of this result: Examine the diagram below to find a parallelogram with sides labeled as lengths of vectors. Then the area of the parallelogram will be the value of $\|\vec{a} \times \vec{b}\|$!

The height of the parallelogram is $h\:=\:\left|\vec{b}\right|\sin\left(\theta\right)$, so the area of the parallelogram is $|\vec{a}||\vec{b}|\sin(\theta) = |\vec{a} \times \vec{b}|$. 

For example: The area of the parallelogram with sides given by:

$\vec{a} = \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix} \mbox{ and } \vec{b} = \begin{pmatrix} 0 \\ 3 \\ -2 \end{pmatrix}$

will be

$\|\vec{a} \times \vec{b} \| = \left\| \begin{pmatrix} -8 \\ -2 \\ -3 \end{pmatrix} \right\| = \sqrt{64+4+9} = \sqrt{77}.$

Another interesting geometric relation between vectors and their cross product is the following: Let us compute $\vec{a} \cdot (\vec{a} \times \vec{b})$. That is, take two vectors and compute their cross product. Then take the dot product of the result with one of the two initial vectors:

$\vec{a} \cdot (\vec{a} \times \vec{b}) = \begin{pmatrix} a_x \\ a_y \\ a_z \end{pmatrix} \cdot \begin{pmatrix} a_yb_z - a_zb_y \\ a_zb_x - a_xb_z \\ a_xb_y -a_yb_x \end{pmatrix} = a_xa_yb_z-a_xa_zb_y + a_ya_zb_x-a_ya_xb_z+a_za_xb_y-a_za_yb_x.$

f you examine the end sum closely you'll find that it must be zero! Therefore we can say that the cross product of two vectors must be perpendicular to the two original vectors!

We can verify this with our example vectors from before. Recall that

$\begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix} \times \begin{pmatrix} 0 \\ 3 \\ -2 \end{pmatrix} = \begin{pmatrix} -8 \\ -2 \\ -3 \end{pmatrix}.$

So let us compute

$\begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} -8 \\ -2 \\ -3 \end{pmatrix} = 8-2-6=0.$

You can also check that

$\begin{pmatrix} 0 \\ 3 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} -8 \\ -2 \\ -3 \end{pmatrix} = 0$

as well. 

The expression $\vec{a} \cdot (\vec{a} \times \vec{b})$ is an example of the triple product which has general form $\vec{a} \cdot (\vec{b} \times \vec{c})$ for three vectors in $\mathbb{R} ^{3}$. This number (in absolute value) will give the volume of a parallelepiped with edges given as in the picture below. We will not go into too much detail here except to say that this is related to how determinants of square matrices give $n$-dimensional volumes. For more on this, consider a course in linear algebra.