Question

Let the unit vectors $a$ and $b$ be perpendicular to each other and the unit vector $c$ be inclined at an angle $\theta$ to both $a$ and $b$. If $c=xa+yb+x(x\times b)$, 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 $c=xa+yb+z(a\times b)$
$\Rightarrow c.a=x$ and $c.b=y\Rightarrow x=y=\cos \theta$
Now, $c.c={|c|}^2$
$\Rightarrow [xa+yb+z(a\times b\left)\right].[xa+yb+z(a\times b\left)\right]={|c|}^2$
$\Rightarrow 2x^2+z^2{|a\times b|}^2=1\Rightarrow x^2+z^2[{|a|}^2{|b|}^2-{(a.b)}^2]=1$
$\Rightarrow 2x^2+z^2[1-0]=1$ $[\because a\perp b\therefore a.b=0]$
$\Rightarrow 2x^2+z^2=1\Rightarrow z^2=1-2x^2=1-2{\cos }^2\theta =-\cos 2\theta$