A space vector makes the angles 150^{circ} an | vedclass

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(A) We know that for a space vector making angles $\alpha, \beta$,and $\gamma$ with the positive direction of the $x, y$,and $z$-axes respectively,the

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$90$

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(A) We know that for a space vector making angles $\alpha, \beta$,and $\gamma$ with the positive direction of the $x, y$,and $z$-axes respectively,the direction cosines satisfy the relation: $\cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1$ Given that $\alpha = 150^{\circ}$ and $\beta = 60^{\circ}$. Substituting these values into the equation: $\cos^{2} 150^{\circ} + \cos^{2} 60^{\circ} + \cos^{2} \gamma = 1$ Since $\cos 150^{\circ} = -\frac{\sqrt{3}}{2}$ and $\cos 60^{\circ} = \frac{1}{2}$,we have: $(-\frac{\sqrt{3}}{2})^{2} + (\frac{1}{2})^{2} + \cos^{2} \gamma = 1$ $\frac{3}{4} + \frac{1}{4} + \cos^{2} \gamma = 1$ $1 + \cos^{2} \gamma = 1$ $\cos^{2} \gamma = 0$ $\cos \gamma = 0$ Therefore,$\gamma = 90^{\circ}$.

$A$ space vector makes the angles $150^{\circ}$ and $60^{\circ}$ with the positive direction of $x$- and $y$-axes. The angle made by the vector with the positive direction of the $z$-axis is (in $^{\circ}$)

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