Cosine Formula for Dot Product/Proof 2

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Theorem

Let $\mathbf v, \mathbf w$ be two non-zero vectors in $\R^n$.

The dot product of $\mathbf v$ and $\mathbf w$ can be calculated by:

$\mathbf v \cdot \mathbf w = \norm {\mathbf v} \norm {\mathbf w} \cos \theta$

where:

$\norm {\, \cdot \,}$ denotes vector length and
$\theta$ is the angle between $\mathbf v$ and $\mathbf w$.


Proof

Let $\mathbf v$ and $\mathbf w$ be considered to be embedded in a Cartesian plane $\CC$.

By Dot Product is Invariant under Coordinate Rotation, we may rotate $\CC$ arbitrarily, and $\mathbf v \cdot \mathbf w$ will not change.

So, let us rotate $\CC$ to $\CC'$ such that the $x$-axis is parallel to $\mathbf v$.

Hence $\mathbf v$ can be expressed as:

\(\ds \mathbf v\) \(=\) \(\ds \norm {\mathbf v} \mathbf i + 0 \mathbf j + 0 \mathbf k\)
\(\ds \mathbf w\) \(=\) \(\ds \norm {\mathbf w} \cos \theta \mathbf i + \norm {\mathbf w} \sin \theta \mathbf j + 0 \mathbf k\)

Hence: by definition of dot product:

\(\ds \mathbf v \cdot \mathbf w\) \(=\) \(\ds \norm {\mathbf v} \times \norm {\mathbf w} \cos \theta + 0 \times \norm {\mathbf w} \sin \theta + 0 \times 0\) Definition of Dot Product
\(\ds \) \(=\) \(\ds \norm {\mathbf v} \norm {\mathbf w} \cos \theta\)

Hence the result.

$\blacksquare$


Sources

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