Question

A unit vector $d$ is equally inclined at an angle $\alpha$ with the vectors $a=\cos \theta .\hat{i}+\sin \theta .\hat{j},b=-\sin \theta .\hat{i}$ $+\cos \theta .\hat{j}$ and $c=\hat{k}$. Then, $\alpha$ is equal to

• A. ${\cos }^{-1}\frac{1}{\sqrt{2}}$
• B. ${\cos }^{-1}\frac{1}{\sqrt{3}}$
• C. ${\cos }^{-1}\frac{1}{3}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (B)

Let $d\cdot a=d\cdot b=d\cdot c=\cos \alpha$
$\Rightarrow d\cdot (a-\hat{k})=0$ and $d\cdot (b-\hat{k})=0$
$\Rightarrow d$ is perpendicular to $(a-\hat{k})$ and $(b-\hat{k})$.
Hence, dis parallel to $(a-\hat{k})\times (b-\hat{k})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \cos \theta & \sin \theta & -1\\ -\sin \theta & \cos \theta & -1\end{array}\right|$
$\therefore d=\frac{(\cos \theta -\sin \theta )\hat{i}+(\cos \theta +\sin \theta )\hat{j}+\hat{k}}{\sqrt{3}}$
$\cos \alpha =d\cdot \hat{k}=\frac{1}{\sqrt{3}}\Rightarrow \alpha ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$