Question

If ${\lambda }_1<{\lambda }_2$ are two values of $\lambda$ such that the angle between the planes $P_1:\vec{r}(3\hat{i}-5\hat{j}+\hat{k})=7$ and $P_2:\vec{r}\cdot (\lambda \hat{i}+\hat{j}-3\hat{k})=9$ is ${\sin }^{-1}\left(\frac{2\sqrt{6}}{5}\right)$, then the square of the length of perpendicular from the point $\left(38{\lambda }_1,10{\lambda }_2,2\right)$ to the plane $P_1$ is ______

Answer 1

Answer: 315

$P_1=\vec{r}\cdot (3\hat{i}-5\hat{j}+\hat{k})=7$
$P_2=\vec{r}\cdot (\lambda \hat{i}+\hat{j}-3\hat{k})=9$
$\theta ={\sin }^{-1}\left(\frac{2\sqrt{6}}{5}\right)$
$\Rightarrow \sin \theta =\frac{2\sqrt{6}}{5}$
$\therefore \cos \theta =\frac{1}{5}.$
$\cos \theta =\frac{\overrightarrow{r}\cdot \overrightarrow{r}}{|\overrightarrow{r}||\overrightarrow{r_2}|}$
$=\frac{(3i-5j+K)(\lambda i+j-3K)}{\sqrt{35}\cdot \sqrt{{\lambda }^2+10}}$
$\frac{1}{5}=\left|\frac{3\lambda -8}{\sqrt{35}\cdot \sqrt{{\lambda }^2+10}}\right|$
$\text{ Square }\Rightarrow \frac{1}{25}=\frac{9{\lambda }^2+64-48\lambda }{35\left({\lambda }^2+10\right)}$
$\Rightarrow 19{\lambda }^2-120\lambda +125=0$
$\Rightarrow 19{\lambda }^2-95\lambda -25\lambda +125=0$
$\Rightarrow x=5,\frac{25}{19}$

Perpendicular distance of point
$\left(38{\lambda }_1,10{\lambda }_2,2\right)=(50,50,2)$ from plane $P_1$
$=\frac{|3\times 50-5\times 50+2-7|}{\sqrt{35}}=\frac{105}{\sqrt{35}}$
Square $=\frac{105\times 105}{35}=315$