Question

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

Answer 1

Answer: 3

$\overrightarrow{c}=x\overrightarrow{a}+y\overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{b}$
$\overrightarrow{c}.\overrightarrow{a}=xandx=2cos\alpha$
$\overrightarrow{c}.\overrightarrow{b}=yandy=2cos\alpha$
Also, $\left|\overrightarrow{a}\times \overrightarrow{b}\right|=1$
$\therefore \overrightarrow{c}=2cos\alpha {\left(\left(\overrightarrow{a}+\overrightarrow{b}\right)+\overrightarrow{a}\right)}^{}\times \overrightarrow{b}$
${\left(\overrightarrow{c}\right)}^2=4{\left(cos\right)}^2\alpha {\left(\overrightarrow{a}+\overrightarrow{b}\right)}^2+{\left(\overrightarrow{a}\times \overrightarrow{b}\right)}^2+4cos\alpha \left(\overrightarrow{a}+\overrightarrow{b}\right).\left(\overrightarrow{a}\times \overrightarrow{b}\right)$
$4=8co{s}^2\alpha +1$
$8co{s}^2\alpha =3$