If $\vec a,\vec b,\vec c$ be three vectors of magnitude $\sqrt{3},1,2$ such that $\vec a × (\vec a×\vec c)+3\vec b= 0$, if θ is the angle between $\vec a$ and $\vec c$, then $\cos^2θ$ is equal to
Answer & explanation
Correct answer: option 1
We have,
$\vec a × (\vec a×\vec c)+3\vec b= 0$
$⇒(\vec a. \vec c) \vec a- (\vec a. \vec a) \vec c +3\vec b=\vec 0$
$⇒(2\sqrt{3}\cos θ)\vec a-3\vec c+3\vec b=\vec 0$
$⇒(2\cos θ)\vec a-\sqrt{3}\vec c+\sqrt{3}\vec b=\vec 0$
$⇒|2\cos θ\vec a-\sqrt{3}\vec c|^2=|-\sqrt{3}\vec b|^2$
$⇒4\cos^2 θ|\vec a|^2+3|\vec c|^2-4\sqrt{3}\cos θ(\vec a.\vec c)=3|\vec b|^2$
$⇒12\cos^2 θ+12-4\sqrt{3}\cos θ×\sqrt{3}×2\cos θ=3$
$⇒12\cos^2 θ+9-24\cos^2 θ=0$
$⇒12\cos^2 θ=9⇒\cos^2 θ=\frac{9}{12}=\frac{3}{4}$