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

The correct answer is 315.

$P_{1}=\vec{r}.(3\hat{i}-5\hat{j}+\hat{k})=7$

$P_{2}=\vec{r}.(\lambda \hat{i}+\hat{j}-3\hat{k})=9$

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

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

$\therefore cos\theta =\frac{1}{5}.$

$cos\theta =\frac{\vec{r}.\vec{r}}{|\vec{r}||\vec{r_{2}}|}$

$=\frac{(3i-5j+k)(\lambda {i}+{j}-3{k})}{\sqrt{35}.\sqrt{\lambda ^{2}+10}}$

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

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

Perpendicular distance of point

$(38\lambda _{1},10\lambda _{2},2)=(50,50,2)$ from plane P1

$=\frac{|30\times50-5\times50+2-7}{\sqrt{35}}$

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