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(C) Let the direction angles of the line be $\alpha, \beta, \gamma$. Since the line makes equal angles with the coordinate axes,we have $\alpha = \bet

Vedclass · Accepted Answer

$\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$

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(C) Let the direction angles of the line be $\alpha, \beta, \gamma$. Since the line makes equal angles with the coordinate axes,we have $\alpha = \beta = \gamma$. The direction cosines are given by $l = \cos \alpha, m = \cos \beta, n = \cos \gamma$. We know the identity $l^2 + m^2 + n^2 = 1$,which implies $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$. Substituting $\alpha = \beta = \gamma$,we get $3 \cos^2 \alpha = 1$. This gives $\cos^2 \alpha = \frac{1}{3}$,so $\cos \alpha = \pm \frac{1}{\sqrt{3}}$. Since the angles are given to be acute,$\cos \alpha, \cos \beta, \cos \gamma$ must be positive. Therefore,the direction cosines are $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$.

The direction cosines of a line which makes equal acute angles with the coordinate axes are

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