Solveeit Logo
Login

Question

Question: Let A = \(\begin{bmatrix} \cos^{2}\theta & \sin\theta\cos\theta \\ \cos\theta\sin\theta & \sin^{2}\t...

Let A = [cos⁡2θsin⁡θcos⁡θcos⁡θsin⁡θsin⁡2θ]\begin{bmatrix} \cos^{2}\theta & \sin\theta\cos\theta \\ \cos\theta\sin\theta & \sin^{2}\theta \end{bmatrix} and

B = [cos⁡2φsin⁡φcos⁡φcos⁡φsin⁡φsin⁡2φ]\begin{bmatrix} \cos^{2}\varphi & \sin\varphi\cos\varphi \\ \cos\varphi\sin\varphi & \sin^{2}\varphi \end{bmatrix}then AB = 0, if-

A

q = nf, n = 0, 1, 2, …….

B

q + f = np n = 0, 1, 2, ……

C

q = f + (2n +1) π2\frac{\pi}{2}, n = 0, 1, 2, ……

D

q = f + nπ2\frac{\pi}{2}, n = 0, 1, 2, …..

Answer

q = f + (2n +1) π2\frac{\pi}{2}, n = 0, 1, 2, ……

Explanation

Solution

AB = [cos⁡2θcos⁡2φ+sin⁡θcos⁡θcos⁡φsin⁡φcos⁡2θsin⁡φcos⁡φ+sin⁡θcos⁡θsin⁡2φcos⁡θsin⁡θcos⁡2φ+sin⁡2θcos⁡φsin⁡φcos⁡θsin⁡θsin⁡φcos⁡φ+sin⁡2θsin⁡2φ]\begin{bmatrix} \cos^{2}\theta\cos^{2}\varphi + \sin\theta\cos\theta\cos\varphi\sin\varphi & \cos^{2}\theta\sin\varphi\cos\varphi + \sin\theta\cos\theta\sin^{2}\varphi \\ \cos\theta\sin\theta\cos^{2}\varphi + \sin^{2}\theta\cos\varphi\sin\varphi & \cos\theta\sin\theta\sin\varphi\cos\varphi + \sin^{2}\theta\sin^{2}\varphi \end{bmatrix}

AB = [cos⁡θcos⁡φcos⁡(θ−φ)sin⁡φcos⁡θcos⁡(θ−φ)sin⁡θcos⁡φcos⁡(θ−φ)sin⁡θsin⁡φcos⁡(θ−φ)]\begin{bmatrix} \cos\theta\cos\varphi\cos(\theta - \varphi) & \sin\varphi\cos\theta\cos(\theta - \varphi) \\ \sin\theta\cos\varphi\cos(\theta - \varphi) & \sin\theta\sin\varphi\cos(\theta - \varphi) \end{bmatrix}

Q AB = 0 Ž cos (q – f) = 0

Ž cos (q – f) = cos (2n +1) π2\frac{\pi}{2} where n = 0, 1, 2, …

q = (2n +1) π2\frac{\pi}{2}+ fwhere n = 0, 1, 2, ……