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
Solution
<p>$$\begin{aligned}
& \overrightarrow{O P}=x \hat{i}+y \hat{j}+z \hat{k} \\
& \overrightarrow{O Q}=x \hat{i}+y \hat{j} \\
& |O P|=\gamma=\sqrt{x^2+y^2+z^2} \\
& \cos \theta=\frac{x}{\sqrt{x^2+y^2}} \Rightarrow \cos ^2 \theta=\frac{x^2}{\gamma^2-z^2}=\frac{x^2}{\gamma^2-\gamma^2 \cos ^2 \phi} \\
& \cos \phi=\frac{z}{\sqrt{x^2+y^2+z^2}}=\frac{z}{\gamma} \\
& \text { Distance of } P \text { from } x \text {-axis }=\sqrt{y^2+z^2} \\
& d=\sqrt{\gamma^2-x^2} \\
& \Rightarrow x^2=\gamma^2 \sin ^2 \phi \cos ^2 \theta \\
& \Rightarrow d=\sqrt{\gamma^2-\gamma^2 \sin ^2 \phi \cos ^2 \theta} \\
& =\gamma \sqrt{1-\sin ^2 \phi \cos ^2 \theta}
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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