Question

Let $\overset{ \rightarrow }{a}$ and $\overset{ \rightarrow }{b}$ be two unit vectors such that $\overset{ \rightarrow }{a}.\overset{ \rightarrow }{b}=0.$ For some $x,y\in R,$ let $\overset{ \rightarrow }{c}=x\overset{ \rightarrow }{a}+y\overset{ \rightarrow }{b}+\left(\overset{ \rightarrow }{a} \times \overset{ \rightarrow }{b}\right).$ If $\left|\overset{ \rightarrow }{c}\right|=2$ and the vector $\overset{ \rightarrow }{c}$ is inclined at the same angle $\alpha $ to both $\overset{ \rightarrow }{a}$ and $\overset{ \rightarrow }{b},$ then the value of $8 cos^{2} \alpha $ is

Answer 1

$\overset{ \rightarrow }{c}=x\overset{ \rightarrow }{a}+y\overset{ \rightarrow }{b}+\overset{ \rightarrow }{a}\times \overset{ \rightarrow }{b}$
$\overset{ \rightarrow }{c}.\overset{ \rightarrow }{a}=x \, and \, x=2cos \alpha $
$\overset{ \rightarrow }{c}.\overset{ \rightarrow }{b}=y \, and \, y=2cos \alpha $
Also, $\left|\overset{ \rightarrow }{a} \times \overset{ \rightarrow }{b}\right|=1$
$\therefore \, \, \overset{ \rightarrow }{c}=2cos \alpha \left(\left(\overset{ \rightarrow }{a} + \overset{ \rightarrow }{b}\right) + \overset{ \rightarrow }{a}\right)^{ \, }\times \overset{ \rightarrow }{b}$
$\left(\overset{ \rightarrow }{c}\right)^{2}=4\left(cos\right)^{2} \alpha \left(\overset{ \rightarrow }{a} + \overset{ \rightarrow }{b}\right)^{2}+\left(\overset{ \rightarrow }{a} \times \overset{ \rightarrow }{b}\right)^{2}+4cos⁡\alpha \left(\overset{ \rightarrow }{a} + \overset{ \rightarrow }{b}\right).\left(\overset{ \rightarrow }{a} \times \overset{ \rightarrow }{b}\right)$
$4=8cos^{2} \alpha +1$
$8cos^{2} \alpha =3$