# Angle Between Non-Zero Vectors Always Defined

## Theorem

The angle between two non-zero vectors in $\R^n$ is always defined.

## Proof

### Case 1

Suppose that $\mathbf v$ and $\mathbf w$ are not scalar multiples of each other.

From Construction of Triangle from Given Lengths, it is sufficient to show that sum of the lengths of any two sides is greater than the third side.

Consider the side with length $\norm {\mathbf v}$.

From the triangle inequality for vectors:

\(\ds \norm {\mathbf v}\) | \(=\) | \(\ds \norm {\mathbf {w + v - w} }\) | ||||||||||||

\(\ds \) | \(<\) | \(\ds \norm {\mathbf w} + \norm {\mathbf {v - w} }\) |

Note that the equality is a strict inequality because the vectors are not scalar multiples of each other.

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Consider the side with length $\norm {\mathbf w}$.

\(\ds \norm {\mathbf w}\) | \(=\) | \(\ds \norm {\mathbf {v + w - v} }\) | ||||||||||||

\(\ds \) | \(<\) | \(\ds \norm {\mathbf v} + \norm {\mathbf {w - v} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \norm {\mathbf v} + \norm {\mathbf {v - w} }\) |

Lastly, Consider the side with length $\norm {\mathbf v - \mathbf w}$.

\(\ds \norm {\mathbf {v - w} }\) | \(=\) | \(\ds \norm {\mathbf {v + \paren {-w} } }\) | ||||||||||||

\(\ds \) | \(<\) | \(\ds \norm {\mathbf v} + \norm {\mathbf {-w} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \norm {\mathbf v} + \norm {\mathbf w}\) |

$\Box$

### Case 2

Suppose that $\mathbf v$ and $\mathbf w$ *are* scalar multiples of each other.

Then the existence of $\theta$ follows directly from the definition of the angle between vectors that are scalar multiples of each other.

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$\blacksquare$

## Sources

- For a video presentation of the contents of this page, visit the Khan Academy.