Answer:
Option B
Explanation:
We have
$\overrightarrow{c}=x\overrightarrow{a}+y\overrightarrow{b}+\overrightarrow{a}\times\overrightarrow{b}$ and $\overrightarrow{a}.\overrightarrow{b}=0$
$\mid\overrightarrow{a}\mid=\mid\overrightarrow{b}\mid=1$ and $\mid\overrightarrow{c}\mid=2$
Also , given $\overrightarrow{c}$ is inclined on $\overrightarrow{a}$ and $\overrightarrow{b}$ with same angle $\alpha$
$\therefore$ $\overrightarrow{a}$ . $\overrightarrow{c}$ = $x\mid\overrightarrow{a}\mid^{2}+y(\overleftarrow{a}.\overrightarrow{b})+\overrightarrow{a}.(\overrightarrow{a}\times\overrightarrow{b})$
$\mid\overrightarrow{a}\mid^{}.\mid\overrightarrow{c}\mid^{}\cos\alpha=x+0+0$
x= $2\cos\alpha$
Similarly,
$\mid\overrightarrow{b}\mid^{}.\mid\overrightarrow{c}\mid^{}\cos\alpha=0+y+0$
y = $2\cos\alpha$
$\mid\overrightarrow{c}\mid^{2}=x^{2}+y^{2}+\mid\overrightarrow{a}\times\overrightarrow{b}\mid^{2}$
$4= 8\cos^{2}\alpha+\mid a\mid^{2}\mid b\mid^{2}\sin^{2}90^{0}$
$4= 8\cos ^{2}\alpha+1$
$\Rightarrow$ $8\cos^{2}\alpha=3$