Question

A unit vector $d$ is equally inclined at an angle $\alpha$ with the vectors $a =\cos \theta . \hat{ i }+\sin \theta . \hat{ j }, \quad b =-\sin \theta . \hat{ i }$ $+\cos \theta . \hat{ j }$ 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 $(a-\hat{k}) \times(b-\hat{k})=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ \cos \theta & \sin \theta & -1 \\ -\sin \theta & \cos \theta & -1\end{vmatrix}$
$\therefore d=\frac{(\cos \theta-\sin \theta) \hat{i}+(\cos \theta+\sin \theta) \hat{j}+\hat{k}}{\sqrt{3}}$
$\cos \alpha=d \cdot \hat{k}=\frac{1}{\sqrt{3}} \Rightarrow \alpha=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)$