Angle Between Non-Zero Vectors Always Defined

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Theorem

Let $\mathbf v$ and $\mathbf w$ be non-zero vectors in real Euclidean space $\R^n$.

Then the angle between $\mathbf v$ and $\mathbf w$ is always defined.


Proof

Case 1

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

AngleBetweenTwoVectors.png

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

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