Let a, b be unit vectors, angle between a and c is 12, and b= c+2( c a).

Question

Let $\hat a,\hat b$ be unit vectors, angle between $\hat a$ and $\vec c$ is $\dfrac\pi{12}$, and $\hat b=\vec c+2(\vec c	imes\hat a)$. Then $6|\vec c|^2$ is:

Accepted Answer

$|\hat b|^2 = |\vec c|^2+4|\vec c\times\hat a|^2$ (the cross term $\vec c\cdot(\vec c\times\hat a)=0$).

$|\vec c\times\hat a|^2=|\vec c|^2\sin^2\left(\dfrac\pi{12}\right)$, and $\sin^2 15^\circ = \dfrac{2-\sqrt3}4$:

$$|\vec c|^2\left(1+4\cdot\frac{2-\sqrt3}4\right)=1 \implies |\vec c|^2(3-\sqrt3)=1 \implies |\vec c|^2=\frac1{3-\sqrt3}=\frac{3+\sqrt3}6$$

$$6|\vec c|^2 = 3+\sqrt3$$

*(Independent derivation gives $3+\sqrt3$, matching option B numerically; the source's answer key states option C, $6(3+\sqrt3)$, which is 6 times larger --- likely reflecting a difference in how the "6" in the question was OCR'd (e.g. possibly $6|\vec c|^{-2}$ or a different coefficient in the original). The source's declared answer is retained below.)*

Answer: C

Solutions · 1