Dot Product of Antiparallel Vectors is Negative

Theorem

Let $\mathbf u$ and $\mathbf v$ be non-zero antiparallel vectors.

Then:

$\mathbf u \cdot \mathbf v < 0$

where $\cdot$ denotes dot 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 such that at least one of $x$, $y$ and $z$ is not zero.

By definition of dot product:

$\ds \mathbf u \cdot \mathbf v$

$=$

$\ds x \paren {-x} + y \paren {-y} + z \paren {-z}$

Definition of Dot Product

$\ds $

$=$

$\ds -\paren {x^2 + y^2 + z^2}$

$\ds $

$<$

$\ds 0$

as $x^2 + y^2 + z^2 > 0$

$\blacksquare$