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

Question

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

Accepted Answer

$$\begin{aligned}
& \overrightarrow{O P}=x \hat{i}+y \hat{j}+z \hat{k} \
& \overrightarrow{O Q}=x \hat{i}+y \hat{j} \
& |O P|=\gamma=\sqrt{x^2+y^2+z^2} \
& \cos \theta=\frac{x}{\sqrt{x^2+y^2}} \Rightarrow \cos ^2 \theta=\frac{x^2}{\gamma^2-z^2}=\frac{x^2}{\gamma^2-\gamma^2 \cos ^2 \phi} \
& \cos \phi=\frac{z}{\sqrt{x^2+y^2+z^2}}=\frac{z}{\gamma} \
& \text { Distance of } P \text { from } x \text {-axis }=\sqrt{y^2+z^2} \
& d=\sqrt{\gamma^2-x^2} \
& \Rightarrow x^2=\gamma^2 \sin ^2 \phi \cos ^2 \theta \
& \Rightarrow d=\sqrt{\gamma^2-\gamma^2 \sin ^2 \phi \cos ^2 \theta} \
& =\gamma \sqrt{1-\sin ^2 \phi \cos ^2 \theta}
\end{aligned}$$

Solutions · 1