Question

Let the unit vector $a$ and $b$ be perpendicular to each other and the unit vector $c$ be inclined at an angle $\theta$ to both $a$ and $b$. If $c=xa+yb+z(a\times b)$, 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 $c.c={\left|c\right|}^2$

$\Rightarrow [xa+yb+z(a\times b)][xa+yb+z(a\times b)]={\left|c\right|}^2$

$\Rightarrow 2x^2+z^2{|a\times b|}^2=1$

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

$\Rightarrow 2x^2+z^2[1-0]=1[\because a\perp b\therefore a.b=0]$

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