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(C) We know that the sum of the squares of the direction cosines of a vector is $1$. $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$ Given that the

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$45^{\circ}$ or $135^{\circ}$

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(C) We know that the sum of the squares of the direction cosines of a vector is $1$. $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$ Given that the vector makes equal angles $\alpha$ with $x$ and $y$ axes,so $\alpha = \beta$. Also,it makes $90^{\circ}$ with the $z$-axis,so $\gamma = 90^{\circ}$. Substituting these values into the identity: $\cos^2 \alpha + \cos^2 \alpha + \cos^2 90^{\circ} = 1$ $2 \cos^2 \alpha + 0 = 1$ $\cos^2 \alpha = \frac{1}{2}$ $\cos \alpha = \pm \frac{1}{\sqrt{2}}$ Therefore,$\alpha = 45^{\circ}$ or $135^{\circ}$.

$A$ vector makes equal angles $\alpha$ with $x$ and $y$ axes and $90^{\circ}$ with $z$-axis. Then $\alpha=$

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