Calc3.2.CrossProductProof

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Theorem

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

\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}

Proof

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

(c\vec{u})\times\vec{w}=c(\vec{u}\times\vec{w})

\vec{u} \times \vec{w} = -(\vec{w} \times \vec{u}) \

\vec{p}\times(\vec{q}+\vec{r})=(\vec{p}\times\vec{q})+(\vec{p}\times\vec{r})

Let's set out to prove these three.

The area of a a parallelogram with sides x,y \ and interior acute[1] angle \theta \ is xy\sin(\theta) \ . Therefore, by definition, we have

\left\|{\vec{u} \times \vec{w}}\right\| = \left\|{\vec{u}}\right\| \left\|{\vec{w}}\right\| \sin(\theta)

where \theta \ 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:

(c\vec{u})\times\vec{w}=c(\vec{u}\times\vec{w})

Second Identity

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

\vec{u} \times \vec{w} = -(\vec{w} \times \vec{u}) \

Third Identity

The final identity we will require takes a little bit more finesse. Suppose we have three vectors, \vec{p},\vec{q},\vec{r} \ . Suppose \vec{p}\neq\vec{0} \ . Let P \ be the plane through the origin which is perpendicular to \vec{p} \ (ie, the plane p1x + p2y + p3z = 0). If \vec{q} \ is any vector, let \vec{q}_* be the projection of \vec{q} into P \ . Clearly, \left\|{  \vec{q}_*   }\right\|=\left\|{  \vec{q}   \ }\right\|\sin(\theta) \ , and so \vec{p}\times\vec{q}=\vec{p}\times\vec{q}_* \ . The reader may also assure themselves if they wish that (\vec{q}+\vec{r})_* = \vec{q}_*+\vec{r}_* \ .

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

If either \vec{q}_*, \vec{r}_*=\vec{0}, 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.

We imagine \vec{p} sticking out of the screen towards the reader. You may need to enlarge this picture.

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

If \vec{s} is any vector in the plane of your screen (ie, p1x + p2y + p3z = 0), then \vec{p}\times\vec{s} is also in the plane of your screen. It will make a 90^\circ=\pi/2 angle with \vec{s} and have magnitude \left\|{  \vec{p}  }\right\| \left\|{  \vec{s}   }\right\|, since \sin(90^\circ)=1. This gives rise to the picture at right. Letting \vec{s} be each of \vec{q}_*, \vec{r}_*, \vec{q}_*+\vec{r}_*, we see that the vectors \vec{p}\times\vec{q}_*, \vec{p}\times\vec{r}_*, \vec{p}\times(\vec{q}_*+\vec{r}_*) have the same configuration as \vec{q}_*, \vec{r}_*, \vec{q}_*+\vec{r}_*, just rotated 90^\circ and multiplied by \left\|{  \vec{p}   \ }\right\|. Convince yourself that this picture demonstrates \vec{p}\times(\vec{q}_*+\vec{r}_*) = \vec{p}\times\vec{q}_* + \vec{p}\times\vec{r}_* \ .

Putting it all together

Since the vectors \vec{i},\vec{j},\vec{k} are all perpendicular to each other, their cross products are very easy to compute:

\vec{i}\times\vec{i}=\vec{0} \ \ \ \ \ \ \ \ \ \ \vec{i}\times\vec{j}=\vec{k} \ \ \ \ \ \ \ \vec{i}\times\vec{k}=-\vec{j}

\vec{j}\times\vec{i}=-\vec{k} \ \ \ \ \ \ \vec{j}\times\vec{j}=\vec{0} \ \ \ \ \ \ \ \vec{j}\times\vec{k}=\vec{i}

\vec{k}\times\vec{i}=\vec{j} \ \ \ \ \ \ \ \ \ \vec{k}\times\vec{j}=-\vec{i} \ \ \ \ \ \vec{k}\times\vec{k}=\vec{0}

Using these formulas, and our identities, we can break take two vectors given by their components, and obtain the formula for the cross product. We leave this simple task as an exercise.

Notes

  1. The acute descriptor is unnecessary, since the obtuse angle will then be \pi-\theta \ , and hence the sine will be equal.
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