If alpha, 2alpha, 3alpha are angles made by a | vedclass

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(D) The direction cosines of a ray making angles $\alpha, \beta, \gamma$ with the coordinate axes satisfy the relation $\cos^2 \alpha + \cos^2 \beta +

Vedclass · Accepted Answer

$\frac{\pi}{6}, \frac{\pi}{4}$

Vedclass · Answer

(D) The direction cosines of a ray making angles $\alpha, \beta, \gamma$ with the coordinate axes satisfy the relation $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$. Given $\beta = 2\alpha$ and $\gamma = 3\alpha$,we have $\cos^2 \alpha + \cos^2 2\alpha + \cos^2 3\alpha = 1$. Using the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$,the equation becomes: $\frac{1 + \cos 2\alpha}{2} + \frac{1 + \cos 4\alpha}{2} + \frac{1 + \cos 6\alpha}{2} = 1$ $3 + \cos 2\alpha + \cos 4\alpha + \cos 6\alpha = 2$ $\cos 2\alpha + \cos 4\alpha + \cos 6\alpha = -1$ $(\cos 6\alpha + \cos 2\alpha) + \cos 4\alpha = -1$ $2 \cos 4\alpha \cos 2\alpha + \cos 4\alpha = -1$ $\cos 4\alpha (2 \cos 2\alpha + 1) = -1$. Testing the options: For $\alpha = \frac{\pi}{6}$: $\cos^2 \frac{\pi}{6} + \cos^2 \frac{\pi}{3} + \cos^2 \frac{\pi}{2} = \frac{3}{4} + \frac{1}{4} + 0 = 1$. For $\alpha = \frac{\pi}{4}$: $\cos^2 \frac{\pi}{4} + \cos^2 \frac{\pi}{2} + \cos^2 \frac{3\pi}{4} = \frac{1}{2} + 0 + \frac{1}{2} = 1$. Thus,the possible values are $\frac{\pi}{6}$ and $\frac{\pi}{4}$.

If $\alpha, 2\alpha, 3\alpha$ are angles made by a ray with $OX, OY, OZ$ axes respectively,then all the possible values of $\alpha$ are:

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