Find the angle between two vectors given other relations

Let $A,B,C \in \mathbb{R}^3$ be three vectors such that

$\lVert A \rVert = \lVert C \rVert = 5, \qquad \lVert B \rVert = 1, \qquad \lVert A - B + C \rVert = \lVert A + B + C \rVert.$

Given that the angle between $A$ and $B$ is equal to $\frac{\pi}{8}$ , determine the angle between $C$ and $C$ .

First, we have

$\begin{align*} && \lVert A - B + C \rVert &= \lVert A + B + C \rVert \\[9pt] \implies && \left ( (A - B + C ) \cdot (A - B + C) \right)^{\frac{1}{2}} &= \left( (A+B+C) \cdot (A+B+C) \right)^{\frac{1}{2}} \\[9pt] \implies && A^2 + B^2 + C^2 - 2A \cdot B - 2B \cdot C + 2A \cdot C &= A^2 + B^2 + C^2 + 2A\cdot B + 2 B \cdot C + 2 A \cdot C \\[9pt] \implies && 4A \cdot B + 4B \cdot C &= 0 \\[9pt] \implies && A \cdot B &= -B \cdot C. \end{align*}$

Then, since the angle between $A$ and $B$ is $\frac{\pi}{8}$ , we have

$\cos \frac{\pi}{8} = \frac{A \cdot B}{5}.$

This implies

$\cos \theta = -\frac{A \cdot B}{5} = - \cos \frac{\pi}{8} = \cos \frac{7\pi}{8} \quad \implies \quad \theta = \frac{7 \pi}{8}.$

Thus, the angle between $B$ and $C$ is $\frac{7 \pi}{8}$ .

Stumbling Robot

Find the angle between two vectors given other relations

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