Find the cosine of the angle between two vectors given other relations

Let $A,B \in \mathbb{R}^3$ be two non-parallel vectors such that

$A \cdot B = 2, \qquad \lVert A \rVert = 1, \qquad \lVert B \rVert = 4.$

Define the vector $C = 2 (A \times B) - 3B$ . Compute $A \cdot (B+C), \ \lVert C \rVert$ , and the cosine of the angle between $B$ and $C$ .

First, we have

$\begin{align*} A \cdot (B+C) &= A \cdot B + A \cdot C \\ &= 2 + A \cdot (2 (A \times B) - 3 B) \\ &= 2 - 3(A \cdot B) \\ &= -4. \end{align*}$

Next, for the norm of $C$ we have

$\begin{align*} \lVert C \rVert &= \big( (2 (A \times B) - 3B) \cdot (2 (A \times B) - 3B) \big)^{\frac{1}{2}} \\ &= \big( 4(A \times B) \cdot (A\times B) + 9 B \cdot B \big)^{\frac{1}{2}} \\ &= \big( 4 (\lVert A \rVert^2 \lVert B \rVert^2 - (A \cdot B)^2) + 4 \cdot 36 \big)^{\frac{1}{2}} \\ &= \sqrt{192} \\ &= 8 \sqrt{3}. \end{align*}$

Finally, to compute the cosine of the angle between $B$ and $C$ , we first compute

$\begin{align*} \lVert B \times C \rVert^2 = \lVert B \rVert^2 \lVert C \rVert^2 - (B \cdot C)^2 \\ &= (16)(192) - (48)^2 \\ &= 768 \end{align*}$

Therefore $\lVert B \times C \rVert = \sqrt{768} = 16 \sqrt{3}$ . And so,

$\begin{align*} && \lVert B \times C \rVert &= \lVert B \rVert \lVert C \rVert \sin \theta \\ &&&= 32 \sqrt{3} \sin \theta \\ \implies && \sin \theta &= \frac{1}{2} \\ \implies && \cos \theta &= -\frac{\sqrt{3}}{2}. \end{align*}$

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Find the cosine of the angle between two vectors given other relations

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