Question

Consider the planes
$ P_1: x=y \sin \alpha+z \sin \beta $
$ P_2: y=z \sin \theta+x \sin \alpha $
$ P_3: z=x \sin \beta+y \sin \theta$
If $\mathrm{P}_1, \mathrm{P}_2, \mathrm{P}_3$ intersect in a line $\frac{\mathrm{x}}{\cos \theta}=\frac{\mathrm{y}}{\cos \beta}=\frac{\mathrm{z}}{\cos \alpha}$ then

• A. $\sin ^2 \alpha+\sin ^2 \beta+2 \sin \theta \sin \alpha \sin \beta=\cos ^2 \theta$
• B. $\sin ^3 \alpha+\sin ^3 \beta+\sin ^3 \theta=3 \sin \alpha \sin \beta \sin \theta$
• C. $\theta+\alpha+\beta=\frac{\pi}{2}$
• D. $\theta+\alpha+\beta=\pi$

Answer 1

Answer: (A), (C)

Three planes intersect in a line
$\therefore \mathrm{D}=\mathrm{D}_1=\mathrm{D}_2=\mathrm{D}_3=0$
Here $\mathrm{D}_1=\mathrm{D}_2=\mathrm{D}_3=0, \therefore \mathrm{D}=0$
$\Delta=0$
$\left|\begin{array}{ccc}-1 & \sin \alpha & \sin \beta \\ \sin \alpha & -1 & \sin \theta \\ \sin \beta & \sin \theta & -1\end{array}\right|=0$
$-1\left(1-\sin ^2 \theta\right)-\sin \alpha(-\sin \alpha-\sin \theta \sin \beta)$
$ +\sin \beta(\sin \alpha \sin \theta+\sin \beta)=0$
$ \sin ^2 \theta+\sin ^2 \alpha+\sin ^2 \beta+2 \sin \alpha \sin \beta \sin \theta=1$
Now D.R's of line
$\left|\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ -1 & \sin \alpha & \sin \beta \\ \sin \alpha & -1 & \sin \theta\end{array}\right|$
$=i(\sin \alpha \sin \theta+\sin \beta)+j(\sin \alpha \sin \beta+\sin \theta)+k\left(1-\sin ^2 \alpha\right)$
This vector is parallel to line.
$\therefore \frac{\sin \alpha \sin \theta+\sin \beta}{\cos \theta}=\frac{\sin \alpha \sin \beta+\sin \theta}{\cos \beta}=\frac{1-\sin ^2 \theta}{\cos \alpha} $
$ \sin \alpha \sin \theta+\sin \beta=\cos \theta \cos \alpha$
$ \cos (\theta+\alpha)=\cos \left(\frac{\pi}{2}-\beta\right) $
$ \therefore \alpha+\beta+\theta=2 \mathrm{n} \pi+\frac{\pi}{2}$