Question

Let $\text{P}(x, y, z)$ be a point in the first octant, whose projection in the xy-plane is the point $\text{Q}$. Let $\text{OP} = \gamma$; the angle between $\text{OQ}$ and the positive x-axis be $\theta$; and the angle between $\text{OP}$ and the positive z-axis be $\phi$, where $\text{O}$ is the origin. Then the distance of $\text{P}$ from the x-axis is:

• A. $\gamma \sqrt{1 - \sin^2 \phi \cos^2 \theta}$
• B. $\gamma \sqrt{1 + \cos^2 \theta \sin^2 \phi}$
• C. $\gamma \sqrt{1 - \sin^2 \theta \cos^2 \phi}$
• D. $\gamma \sqrt{1 + \cos^2 \phi \sin^2 \theta}$

Answer 1

The Correct Option is A

$P(x, y, z), Q(x, y, 0); \quad x^2 + y^2 + z^2 = \gamma^2$

$OQ = xi + yj$ 

$\cos \theta = \frac{x}{\sqrt{x^2 + y^2}}$ 

$\cos \phi = \frac{x}{\sqrt{x^2 + y^2 + z^2}}$ 

$\implies \sin^2 \phi = \frac{x^2 + y^2}{x^2 + y^2 + z^2}$ 

$\textbf{Distance of } P \textbf{ from x-axis} = \sqrt{y^2 + z^2}$ 

$\implies \sqrt{\gamma^2 - x^2} \implies \gamma \sqrt{1 - \frac{x^2}{\gamma^2}}$ 

$= \gamma \sqrt{1 - \cos^2 \theta \sin^2 \phi}$