Question

$P$ is a point and $PM,PN$ are perpendiculars from $P$ to the $ZX$ and $XY$ planes respectively. If $OP$ makes angles $\theta ,\alpha ,\beta$ $\gamma$ with the plane $OMN$ and the $XY,YZ,ZX$ plane respectively then ${\sin }^2\theta \left(cose{c}^2\alpha +cose{c}^2\beta +cose{c}^2\gamma \right)$ is equal to

• A. 0
• B. 1
• C. -1
• D. None of these

Answer 1

Answer: (B)

Let $P$ be $\left(x_1,y_1,z_1\right).$ Point $M$ is $\left(x_1,0,z_1\right)$ and $\dot{N}$ is $\left(x_1,y_1,0\right)$
So normal to plane $OMN$ is $\overrightarrow{OM}\times \overrightarrow{ON}=\vec{x}($ say $)$
i.e. $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ x_1& 0& z_1\\ x_1& y_1& 0\end{array}\right|$
$=\hat{i}\left(-y_1{z}_1\right)-\hat{j}\left(-x_1{z}_1\right)+\hat{k}\left(x_1{y}_1\right)$
therefore, $\sin \theta =\frac{-x_1{y}_1{z}_1+x_1{y}_1{z}_1+x_1{y}_{1}z}{\sqrt{x_1^2+y_1^2+z_1^2}\sqrt{\sum x_1^2{y}_1^2}}$
or $\left(\because \sin \theta =\frac{\vec{n}\times \overrightarrow{OP}}{|n\Vert \overline{OP}|}\right)$
$\Rightarrow cose{c}^2\theta =\frac{\sum x_1^2\sum x_1^2{y}_1^2}{{\left(x_1{y}_1{z}_1\right)}^2}$
$=\frac{\sum x_1^2}{x_1^2}+\frac{\sum x_1^2}{y_1^2}+\frac{\sum x_1^2}{z_1^2}$
Now, $\sin \alpha =\frac{\overrightarrow{OP}\cdot \hat{k}}{|\overrightarrow{OP}|}=\frac{z_1}{\sqrt{\sum x_1^2}}$
$\sin \beta =\frac{x_1}{\sqrt{\sum x_1^2}}$ and $\sin \gamma =\frac{y_1}{\sqrt{\sum x_1^2}}$
Now, $cose{c}^2\alpha +cose{c}^2\beta +cose{c}^2\gamma$
$=\frac{x_1^2+y_1^2+z_1^2}{x_1^2}+\frac{\sum x_1^2}{y_1^2}+\frac{\sum x_1^2}{z_1^2}=cose{c}^2\theta$