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

$ 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} . $
$\cos \theta=\frac{\overrightarrow{ r } \cdot \overrightarrow{ r }}{|\overrightarrow{ r }|\left|\overrightarrow{ r _2}\right|}$
$ =\frac{(3 i-5 j+K)(\lambda i+j-3 K)}{\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 $=\frac{105 \times 105}{35}=315$