Let $\vec a,\vec b$ and $\vec c$ be three vectors having magnitudes 1, 1 and 2 respectively such that $\vec a × (\vec a×\vec c) + \vec b = \vec 0$, then the acute angle between $\vec a$ and $\vec c$ is
Answer & explanation
Correct answer: option 3
Let θ be the angle between $\vec a$ and $\vec c$. Then,
$\vec a × (\vec a×\vec c) + \vec b = \vec 0$
$⇒(\vec a.\vec c)\vec a-(\vec a.\vec a)\vec c+\vec b= \vec 0$
$⇒(2\cos θ)\vec a-\vec c+\vec b= \vec 0$ $[∵\vec a.\vec c=\cos θ\,and\,|\vec a|=|\vec b|=1,|\vec c|=2]$
$⇒(2\cos θ)\vec a-\vec c=-\vec b$
$⇒|(2\cos θ)\vec a-\vec c|^2=|-\vec b|^2$
$⇒4\cos^2θ|\vec a|^2+|\vec c|^2-4\cos θ(\vec a.\vec c)=|\vec b|^2$
$⇒4\cos^2θ+4-4\cos θ(2\cos θ)=1$
$⇒4\cos^2θ=3⇒\cos θ=\frac{\sqrt{3}}{2}⇒θ=\frac{π}{6}$