Calc3.2/Cross Product Proof

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

${\vec {u}}\times {\vec {w}}={\begin{pmatrix}u_{1}\\u_{2}\\u_{3}\end{pmatrix}}\times {\begin{pmatrix}w_{1}\\w_{2}\\w_{3}\end{pmatrix}}={\begin{pmatrix}u_{2}w_{3}-u_{3}w_{2}\\u_{3}w_{1}-u_{1}w_{3}\\u_{1}w_{2}-u_{2}w_{1}\end{pmatrix}}$

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 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}}\$.