Practicing Success
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 |
$\frac{π}{3}$ $\frac{π}{4}$ $\frac{π}{6}$ $\frac{π}{2}$ |
$\frac{π}{6}$ |
Let θ be the angle between $\vec a$ and $\vec c$. Then, $\vec a × (\vec a×\vec c) + \vec b = \vec 0$ $⇒(\vec a.\vec c)\vec a-(\vec a.\vec a)\vec c+\vec b= \vec 0$ $⇒(2\cos θ)\vec a-\vec c+\vec b= \vec 0$ $[∵\vec a.\vec c=\cos θ\,and\,|\vec a|=|\vec b|=1,|\vec c|=2]$ $⇒(2\cos θ)\vec a-\vec c=-\vec b$ $⇒|(2\cos θ)\vec a-\vec c|^2=|-\vec b|^2$ $⇒4\cos^2θ|\vec a|^2+|\vec c|^2-4\cos θ(\vec a.\vec c)=|\vec b|^2$ $⇒4\cos^2θ+4-4\cos θ(2\cos θ)=1$ $⇒4\cos^2θ=3⇒\cos θ=\frac{\sqrt{3}}{2}⇒θ=\frac{π}{6}$ |