Cross Product of Antiparallel Vectors is Zero

Theorem

Let $\mathbf u$ and $\mathbf v$ be antiparallel vectors.

Then:

$\mathbf u \times \mathbf v = 0$

where $\times$ denotes vector cross product.

We have by hypothesis that $\mathbf u$ and $\mathbf v$ are antiparallel.

Hence we can define $\mathbf u$ and $\mathbf v$ as:

$\ds \mathbf u$

$=$

$\ds x \mathbf i + y \mathbf j + z \mathbf k$

$\ds \mathbf v$

$=$

$\ds \paren {-x} \mathbf i + \paren {-y} \mathbf j + \paren {-z} \mathbf k$

where $x$, $y$ and $z$ are arbitrary real numbers.

$\mathbf u \times \mathbf v = \begin {vmatrix} \mathbf i & \mathbf j & \mathbf k \\ x & y & z \\ -x & -y & -z \end {vmatrix}$

It is seen that the $3$rd row is a multiple of the $2$nd row.

$\blacksquare$