Calc3.2.CrossProductProof

Theorem

In Cartesian coordinates, the cross product, defined as having length equal to the area of the parallelogram bordered on two sides by <math>\vec{u} \ </math> and <math>\vec{w} \ </math>, and direction perpendicular to either vector, with a right handed orientation, is given by

<math>\vec{u}\times\vec{w}= \begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix} \times\begin{pmatrix}w_1\\w_2\\w_3\end{pmatrix} =\begin{pmatrix}u_2 w_3 - u_3 w_2\\u_3w_1-u_1w_3\\u_1w_2-u_2w_1\end{pmatrix}</math>

Proof

To prove this theorem, we will need several identities about the cross product:

<math>(c\vec{u})\times\vec{w}=c(\vec{u}\times\vec{w})</math>

<math>\vec{u} \times \vec{w} = -(\vec{w} \times \vec{u}) \ </math>

<math>\vec{p}\times(\vec{q}+\vec{r})=(\vec{p}\times\vec{q})+(\vec{p}\times\vec{r})</math>

Let's set out to prove these three.

The area of a a parallelogram with sides <math>x,y \ </math> and interior acute angle <math>\theta \ </math> is <math>xy\sin(\theta) \ </math>. Therefore, by definition, we have

<math>\left\|{\vec{u} \times \vec{w}}\right\| = \left\|{\vec{u}}\right\| \left\|{\vec{w}}\right\| \sin(\theta)</math>

where <math>\theta \ </math> is the angle between the two vectors.

First Identity

This definitional equation for the magnitude of the cross product immediately gives us an identity - for scalar (ie, non-vector) value c, we have:

<math>(c\vec{u})\times\vec{w}=c(\vec{u}\times\vec{w})</math>

Second Identity

Since our definitional equation for the magnitude of the cross product also has the property that <math>\left\|{\vec{u} \times \vec{w}}\right\| = \left\|{\vec{w} \times \vec{u}}\right\|</math>, we can combine this fact with the "right-handed orientation" part of the definition to get another identity:

<math>\vec{u} \times \vec{w} = -(\vec{w} \times \vec{u}) \ </math>

Third Identity

The final identity we will require takes a little bit more finesse. Suppose we have three vectors, <math>\vec{p},\vec{q},\vec{r} \ </math>. Suppose <math>\vec{p}\neq\vec{0} \ </math>. Let <math>P \ </math> be the plane through the origin which is perpendicular to <math>\vec{p} \ </math> (ie, the plane <math>p_1x+p_2y+p_3z=0</math>). If <math>\vec{q} \ </math> is any vector, let <math>\vec{q}_*</math> be the projection of <math>\vec{q}</math> into <math>P \ </math>. Clearly, <math>\left\|{ \vec{q}_* }\right\|=\left\|{ \vec{q} \ }\right\|\sin(\theta) \ </math>, and so <math>\vec{p}\times\vec{q}=\vec{p}\times\vec{q}_* \ </math>. The reader may also assure themselves if they wish that <math>(\vec{q}+\vec{r})_* = \vec{q}_*+\vec{r}_* \ </math>.

Hence, to prove the third identity, it would suffice to prove instead <math>\vec{p}\times(\vec{q}_*+\vec{r}_*) = \vec{p}\times\vec{q}_* + \vec{p}\times\vec{r}_* \ </math>. The advantage is that the vectors <math>\vec{q}_*, \vec{r}_*, \ </math> and <math>\vec{q}_*+\vec{r}_* \ </math> are all either <math>\vec{0}</math> or perpendicular to <math>\vec{p} \ </math>.

If either <math>\vec{q}_*, \vec{r}_*=\vec{0}</math>, then the identity we seek is certainly true, and if they are collinear, then one is a multiple of the other and the identity we seek follows from application of our first two identities.

So, let us assume that <math>\vec{q}_*, \vec{r}_*</math> both have non-zero magnitude and are not collinear. If we imagine <math>\vec{p}</math> sticking out of our computer monitors, pointing towards the reader, then the vectors in questions all look something like the picture on the right.