Question

Let the unit vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ be perpendicular to each other and the unit vector $\overrightarrow{c}$ be inclined at an angle $\theta$ to both $\overrightarrow{a}$ and $\overrightarrow{b}$. If $\overrightarrow{c}=x\overrightarrow{a}+y\overrightarrow{b}+z(\overrightarrow{a}\times \overrightarrow{b}),$ then

• A. $x=\cos \theta ,y=\sin \theta ,z=\cos 2\theta$
• B. $x=\sin \theta ,y=\cos \theta ,z=-\cos 2\theta$
• C. $x=y=\cos \theta ,z^2=\cos 2\theta$
• D. $x=y=\cos \theta ,z^2=-\cos 2\theta$

Answer 1

The correct option is D $x=y=\cos \theta ,z^2=-\cos 2\theta$

We have $c=ax+by+z(a\times b)$

$\Rightarrow c.a=x$

and $c.b=y$ $\Rightarrow x=y=\cos \theta$

Now, $c.c$$={\left|c\right|}^2$

$\Rightarrow (ax+by+z(a\times b)).(ax+by+z(a\times b))={\left|c\right|}^2$

$\Rightarrow 2x^2+z^2\{{\left|a\right|}^2{\left|b\right|}^2-{(a.b)}^2\}=1$

$\Rightarrow 2x^2+z^2\{1-0\}=1$ since $a$ and $b$ are perpendicular to each other.

$\Rightarrow 2x^2+z^2=1\Rightarrow z^2=1-2x^2=1-2{\cos }^2\theta =-\cos 2\theta$