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(A) The direction cosines of a line are given by $\cos \alpha, \cos \beta, \cos \gamma$,where $\alpha, \beta, \gamma$ are the angles made with the $x,

Vedclass · Accepted Answer

$\frac{\pi}{6}$

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(A) The direction cosines of a line are given by $\cos \alpha, \cos \beta, \cos \gamma$,where $\alpha, \beta, \gamma$ are the angles made with the $x, y, z$ axes respectively.Here,$\alpha = 90^{\circ}, \beta = 60^{\circ}, \gamma = 	heta$.The sum of the squares of the direction cosines is always $1$,so $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.Substituting the values: $\cos^2 90^{\circ} + \cos^2 60^{\circ} + \cos^2 	heta = 1$.$(0)^2 + (\frac{1}{2})^2 + \cos^2 	heta = 1$.$0 + \frac{1}{4} + \cos^2 	heta = 1$.$\cos^2 	heta = 1 - \frac{1}{4} = \frac{3}{4}$.Since $	heta$ is acute,$\cos 	heta = \frac{\sqrt{3}}{2}$.Therefore,$	heta = 30^{\circ}$ or $\frac{\pi}{6}$ radians.

If a line makes angles $90^{\circ}, 60^{\circ}$ and $\theta$ with $x, y$ and $z$ axes respectively,where $\theta$ is acute,then the value of $\theta$ is

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