Prove that the angle between two given vectors is double the angle between two other vectors

by RoRi

April 28, 2016

Let

$A = (1,2,1), \quad B = (2,1,-1), \quad C = (1,4,1), \quad D = (2,5,5).$

Prove that the angle between $A$ and $B$ is twice the angle between $C$ and $D$ .

Proof. Let $\theta$ denote the angle between $A$ and $B$ , and let $\phi$ denote the angle between $C$ and $D$ . Then, we compute

$\begin{align*} \cos \theta &= \frac{A \cdot B}{\lVert A \rVert \lVert B \rVert} = \frac{3}{\sqrt{6} \sqrt{6}} = \frac{1}{2} & \implies && \theta &= \frac{\pi}{3}. \\[9pt] \cos \phi &= \frac{C \cdot D}{\lVert C \rVert \lVert D \rVert} = \frac{27}{\sqrt{18} \sqrt{54}} = \frac{3}{\sqrt{12}} = \frac{\sqrt{3}}{2} & \implies && \phi &= \frac{\pi}{6}. \end{align*}$

Thus, $\theta = 2 \phi. \qquad \blacksquare$

Stumbling Robot

Prove that the angle between two given vectors is double the angle between two other vectors

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