Question

$a,b,c$ are three vectors of magnitudes $\sqrt{3},1,2$ such that $a\times (a\times c)+3b=0$. If $\theta$ is tha angle between $a$ and $c$, then ${\cos }^2\theta$ is equal to

Answer 1

We have, $a\times (a\times c)+3b=0$

$\Rightarrow (a.c)a-(a.a)c+3b=0$

$\Rightarrow \left(2\sqrt{3}\cos \theta \right)a-3c+3b=0$

$\Rightarrow \left(2\cos \theta \right)a-\sqrt{3}c+\sqrt{3}b=0$

$\Rightarrow {|\left(2\cos \theta \right)a-\sqrt{3}c|}^2={|-\sqrt{3}b|}^2$

$\Rightarrow 4{\cos }^2\theta {\left|a\right|}^2+3{\left|c\right|}^2-4\sqrt{3}\cos \theta (a.c)=3{\left|b\right|}^2$

$\Rightarrow 12{\cos }^2\theta +12-4\sqrt{3}\cos \theta \times \sqrt{3}\times 2\cos \theta =3$

$\Rightarrow 12{\cos }^2\theta +9-24{\cos }^2\theta =0$

$\Rightarrow 12{\cos }^2\theta =9$

$\Rightarrow {\cos }^2\theta =\frac{9}{12}=\frac{3}{4}.$