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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 $z$-axis is
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Verified Answer
The correct answer is:
$90^{\circ}$
We know that, the condition when a space vector makes the angles $\alpha, \beta$ and $\gamma$ with the positive direction of $\mathrm{x}, \mathrm{y}$ and $\mathrm{z}$-axes respectively is
$$
\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1
$$
Given that, $\alpha=150^{\circ}, \beta=60^{\circ}, \gamma=$ ?
From Eq (i), $\cos ^{2} 150^{\circ}+\cos ^{2} 60^{\circ}+\cos ^{2} \gamma=1$
$\left(\sin ^{2} 60^{\circ}+\cos ^{2} 60^{\circ}\right)+\cos ^{2} \gamma=1$
$1+\cos ^{2} \gamma=1$
$\Rightarrow \quad \cos ^{2} \gamma=0$
$\Rightarrow \quad \cos \gamma=0=\cos 90^{\circ}$
$\Rightarrow \quad \gamma=90^{\circ}$
$$
\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1
$$
Given that, $\alpha=150^{\circ}, \beta=60^{\circ}, \gamma=$ ?
From Eq (i), $\cos ^{2} 150^{\circ}+\cos ^{2} 60^{\circ}+\cos ^{2} \gamma=1$
$\left(\sin ^{2} 60^{\circ}+\cos ^{2} 60^{\circ}\right)+\cos ^{2} \gamma=1$
$1+\cos ^{2} \gamma=1$
$\Rightarrow \quad \cos ^{2} \gamma=0$
$\Rightarrow \quad \cos \gamma=0=\cos 90^{\circ}$
$\Rightarrow \quad \gamma=90^{\circ}$
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