Question

If AB = 0 where $A=\left[\begin{array}{cc}{cos}^2\theta & cos\theta sin\theta \\ cos\theta sin\theta & {sin}^2\theta \end{array}\right]$ and $B=\left[\begin{array}{cc}{cos}^2\varphi & cos\varphi sin\varphi \\ cos\varphi sin\varphi & {sin}^2\varphi \end{array}\right]$ then $|\theta -\varphi |$ is equal to

• A. $0$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{4}$
• D. $\pi$

Answer 1

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

$AB=\left[\begin{array}{cc}{\cos }^2\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta \end{array}\right]\left[\begin{array}{cc}{\cos }^2\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta \end{array}\right]$

= $\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$\therefore {\cos }^2\theta {\cos }^2\varphi +\cos \varphi \sin \theta \cos \theta {\sin }^2\theta =0$

$\cos \theta \cos \varphi (\cos \theta \cos \varphi +\sin \theta \sin \varphi )=0$

$\left(\cos \theta \cos \varphi \cos \right(\theta -\varphi )=0)$....(i)

$\cos \theta \cos \varphi \sin \varphi +\cos \theta \sin \theta {\sin }^2\varphi =0$

$\therefore \cos \theta \sin \varphi (\cos \theta \cos \varphi +\sin \theta \sin \varphi )=0$

$\left(\cos \theta \sin \varphi \cos \right(\theta -\varphi )=0)$...(ii)

$\cos \theta \sin \theta {\cos }^2\theta +{\sin }^2\theta \cos \varphi \sin \varphi =0$

$\sin \theta \cos \varphi (\cos \theta \cos \varphi +\sin \theta \sin \varphi )=0$

$\left(\sin \theta \cos \varphi \right(\cos (\theta -\varphi ))=0)$...(iii)

$\cos \theta \sin \theta \cos \varphi \sin \varphi +{\sin }^2\theta {\sin }^2\theta =0$

$\sin \theta \sin \varphi (\cos \theta \cos \varphi +\sin \theta \sin \varphi )=0$

$\left(\sin \theta \sin \varphi \cos \right(\theta -\varphi )=0)$...(iv)

(i), (ii), (iii) and (iv) hold together

$\therefore \cos (\theta -\varphi )=0$

$\therefore \theta -\varphi (2n+1)\frac{\pi }{2}$

here it can be $\frac{\pi }{2}$