Use vector methods to prove the law of cosines

by RoRi

April 28, 2016

Let $\theta$ denote the angle between two vectors $A$ and $B$ both of which are nonzero. Prove that

$\lVert A - B \rVert^2 = \lVert A \rVert^2 + \lVert B \rVert^2 - 2 \lVert A \rVert \lVert B \rVert \cos \theta.$

Proof. We compute

$\begin{align*} \lVert A - B \rVert^2 &= (A-B) \cdot (A-B) \\ &= A \cdot A - 2A \cdot B + B \cdot B \\ &= \lVert A \rVert^2 + \lVert B \rVert^2 - 2 \left( \lVert A \rVert \lVert B \rVert \cos \theta). \qquad \blacksquare \end{align*}$

Stumbling Robot

Use vector methods to prove the law of cosines

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