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 $ heta$; and the angle between $O P$ and the positive $z$-axis be $\phi$, where $O$ is the origin. Then the distance of $P$…

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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 $	heta$; 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

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z}), \mathrm{Q}(\mathrm{x}, \mathrm{y}, \mathrm{O}) ; \mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2=\gamma^2 \ & \overline{\mathrm{OQ}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}} \ & \cos 	heta=\frac{\mathrm{x}}{\sqrt{\mathrm{x}^2+\mathrm{y}^2}} \ & \cos \phi=\frac{\mathrm{z}}{\sqrt{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}} \ & \Rightarrow \sin ^2 \phi=\frac{\mathrm{x}^2+\mathrm{y}^2}{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}\end{aligned}$ distance of $\mathrm{P}$ from $\mathrm{x}$-axis $\sqrt{\mathrm{y}^2+\mathrm{z}^2}$ $\begin{aligned} & \Rightarrow \sqrt{\gamma^2-x^2} \Rightarrow \gamma \sqrt{1-\frac{x^2}{\gamma^2}} \ & =\gamma \sqrt{1-\cos ^2 	heta \sin ^2 \phi}\end{aligned}$

Let $P(x, y, z)$ be a point in the first octant, whose projection in the $x y$-plane is the point $Q$. Let…

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