MCQ 11 Mark
Let $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ be a point in the first octant, whose projection in the xy-plane is the point $\mathrm{Q}$. Let $\mathrm{OP}=\gamma$; the angle between $OQ$ and the positive $\mathrm{x}$-axis be $\theta$; and the angle between $\mathrm{OP}$ and the positive $\mathrm{z}$-axis be $\phi$, where $\mathrm{O}$ is the origin. Then the distance of $\mathrm{P}$ from the $\mathrm{x}$-axis is :
- ✓$\gamma \sqrt{1-\sin ^2 \phi \cos ^2 \theta}$
- 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
View full question & answer→Correct option: A.
$\gamma \sqrt{1-\sin ^2 \phi \cos ^2 \theta}$
a
$ \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 $
$ \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{xi}+\mathrm{y} $
$ \cos \theta=\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} $
$ \text { distance of } \mathrm{P} \text { from } \mathrm{x} \text {-axis } \sqrt{\mathrm{y}^2+\mathrm{z}^2} $
$ \Rightarrow \sqrt{\gamma^2-\mathrm{x}^2} \Rightarrow \gamma \sqrt{1-\frac{\mathrm{x}^2}{\gamma^2}} $
$ =\gamma \sqrt{1-\cos ^2 \theta \sin ^2 \phi}$