JEE MAIN - Mathematics (2024 - 8th April Morning Shift - No. 12)
Let $$P(x, y, z)$$ be a point in the first octant, whose projection in the $$x y$$-plane is the point $$Q$$. Let $$O P=\gamma$$; the angle between $$O Q$$ and the positive $$x$$-axis be $$\theta$$; and the angle between $$O P$$ and the positive $$z$$-axis be $$\phi$$, where $$O$$ is the origin. Then the distance of $$P$$ from the $$x$$-axis is
$$\gamma \sqrt{1-\sin ^2 \phi \cos ^2 \theta}$$
$$\gamma \sqrt{1+\cos ^2 \theta \sin ^2 \phi}$$
$$\gamma \sqrt{1+\cos ^2 \phi \sin ^2 \theta}$$
$$\gamma \sqrt{1-\sin ^2 \theta \cos ^2 \phi}$$
Explanation
$$\begin{aligned}
& \overrightarrow{O P}=x \hat{i}+y \hat{j}+z \hat{k} \\
& \overrightarrow{O Q}=x \hat{i}+y \hat{j} \\
& |O P|=\gamma=\sqrt{x^2+y^2+z^2} \\
& \cos \theta=\frac{x}{\sqrt{x^2+y^2}} \Rightarrow \cos ^2 \theta=\frac{x^2}{\gamma^2-z^2}=\frac{x^2}{\gamma^2-\gamma^2 \cos ^2 \phi} \\
& \cos \phi=\frac{z}{\sqrt{x^2+y^2+z^2}}=\frac{z}{\gamma} \\
& \text { Distance of } P \text { from } x \text {-axis }=\sqrt{y^2+z^2} \\
& d=\sqrt{\gamma^2-x^2} \\
& \Rightarrow x^2=\gamma^2 \sin ^2 \phi \cos ^2 \theta \\
& \Rightarrow d=\sqrt{\gamma^2-\gamma^2 \sin ^2 \phi \cos ^2 \theta} \\
& =\gamma \sqrt{1-\sin ^2 \phi \cos ^2 \theta}
\end{aligned}$$
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