Question

Let the unit vectors $\overline{a}$ and $\overline{b}$ be perpendicular to each other and the unit vector $\overline{c}$ be inclined at an angle ${}^{\prime }{\theta }^{\prime }$ to both $\overline{a}$ and $\overline{b}$. If $\overline{c}=\alpha \overline{a}+\beta \overline{b}+\gamma (\overline{a}\times \overline{b})$, then

• A. $\alpha =\beta =\cos \theta ,{\gamma }^2=\cos 2\theta$
• B. $\alpha =\beta =\cos \theta ,{\gamma }^2=-\cos 2\theta$
• C. $\alpha =\cos \theta ,\beta =\sin \theta ,{\gamma }^2=\cos 2\theta$
• D. $\alpha =\sin \theta ,\beta =\cos \theta ,{\gamma }^2=-\cos 2\theta$

Answer 1

The correct option is B $\alpha =\beta =\cos \theta ,{\gamma }^2=-\cos 2\theta$

$\overline{a}\cdot \overline{b}=0,\overline{a}\cdot \overline{c}=\cos \theta ,\overline{b}\cdot \overline{c}=\cos \theta$ ($\because \overline{a},\overline{b},\overline{c}$ are unit vectors)
Also $\overline{c}=\alpha \overline{a}+\beta \overline{b}+\gamma (\overline{a}\times \overline{b})$
$\therefore \overline{c}\cdot \overline{a}=\alpha +\beta \overline{a}\cdot \overline{b}+\gamma (\overline{a}\times \overline{b})\cdot \overline{a}$
$\Rightarrow \cos \theta =\alpha$

Similarly, $\beta =\cos \theta$

Now, $\gamma (\overline{a}\times \overline{b})=\overline{c}-\alpha a-\beta \overline{b}$
${\left|\gamma (\overline{a}\times \overline{b})\right|}^2$
$={\left|\overline{c}\right|}^2+{\alpha }^2{\left|\overline{a}\right|}^2+{\beta }^2{\left|\overline{b}\right|}^2-2\alpha \overline{a}\cdot \overline{c}-2\beta \overline{b}\cdot c$
$\Rightarrow {\gamma }^2=1+{\alpha }^2+{\beta }^2-2\alpha \cos \theta -2\beta \cos \theta$
$\Rightarrow {\gamma }^2=1-2{\cos }^2\theta =-\cos 2\theta$