Question

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

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

Answer 1

The correct option is D $\frac{\pi }{2}$

$\begin{array}{c}a=\cos \theta i+\sin \theta j\\ b=-\sin \theta i+\cos \theta j\\ c=\hat{k}\\ a.b=\left(\cos \theta i+\sin \theta j\right)\left(-\sin \theta i+\cos \theta j\right)\\ =-\cos \theta .\sin \theta +\sin \theta .\cos \theta \\ a.b=\left|a\right|\left|b\right|\cos \alpha \left[\begin{array}{c}0=1\times 1\cos \alpha \\ 0=\cos \alpha \\ \alpha =\frac{\pi }{2}\end{array}\right]\\ 0=\sqrt{{\cos }^2\theta +{\sin }^2\theta }.\sqrt{{\sin }^2\theta +{\cos }^2\theta }\cos \alpha \end{array}$