JEE Main 2023 — Three Dimensional Geometry Question with Solution

Given:

$P_1:\overrightarrow{r}·\left(3\hat{i}-5\hat{j}+\hat{k}\right)=7$

So, normal vector is ${\overrightarrow{n}}_1=3\hat{i}-5\hat{j}+\hat{k}$

$P_2:\overrightarrow{r}·\left(\lambda \hat{i}+\hat{j}-3\hat{k}\right)=9$

So, normal vector is ${\overrightarrow{n}}_2=\lambda \hat{i}+\hat{j}-3\hat{k}$

Now, angle between planes $P_1$ and $P_2$ is

$\theta ={\sin }^{-1}\left(\frac{2\sqrt{6}}{5}\right)$

$\Rightarrow \sin \theta =\frac{2\sqrt{6}}{5}$

$\Rightarrow \cos \theta =\frac{1}{5}$

Also, 

$\cos \theta =\frac{{\overrightarrow{n}}_1·{\overrightarrow{n}}_2}{\left|{\overrightarrow{n}}_1\right|\left|{\overrightarrow{n}}_2\right|}$

$\Rightarrow \frac{1}{5}=\frac{\left(3\hat{i}-5\hat{j}+\hat{k}\right)·\left(\lambda \hat{i}+\hat{j}-3\hat{k}\right)}{\sqrt{35}\sqrt{{\lambda }^2+10}}$

$\Rightarrow \frac{1}{5}=\left|\frac{3\lambda -8}{\sqrt{35}·\sqrt{{\lambda }^2+10}}\right|$

Squaring both sides, we get

 $\Rightarrow \frac{1}{25}=\frac{9{\lambda }^2+64-48\lambda }{35\left({\lambda }^2+10\right)}$

$\Rightarrow 7\left({\lambda }^2+10\right)=45{\lambda }^2+320-240\lambda$

$\Rightarrow 38{\lambda }^2+250-310\lambda =0$

$\Rightarrow 19{\lambda }^2-120\lambda +125=0$

$\Rightarrow 19{\lambda }^2-95\lambda -25\lambda +125=0$

$\Rightarrow x=5, \frac{25}{19}$

Since, ${\lambda }_2>{\lambda }_1$, so ${\lambda }_2=5, {\lambda }_1=\frac{25}{19}$

Perpendicular distance of point

$\left(38{\lambda }_1,10{\lambda }_2,2\right)\equiv (50,50,2)$ from plane $P_1$ is

$=\left|\frac{3\times 50-5\times 50+2-7}{\sqrt{35}}\right|=\frac{105}{\sqrt{35}}$

Square of the distance is

$=\frac{105\times 105}{35}=315$