Three Dimensional Geometry Question 21
Question 21 - 30 January - Shift 1
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 $\sin ^{-1}(\frac{2 \sqrt{6}}{5})$, then the square of the length of perpendicular from the point $(38 \lambda_1, 10 \lambda_2, 2)$ to the plane $P_1$ is ___________
Show Answer
Answer: 315
Solution:
Formula: Angle between two planes
$ \begin{aligned} & P_1=\overrightarrow{{}r} \cdot(3 \hat{i}-5 \hat{j}+\hat{k})=7 \\ & P_2=\overrightarrow{{}r} \cdot(\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} \cdot \\ & \cos \theta=\frac{\overrightarrow{{}r} \cdot \overrightarrow{{}r}}{|\overrightarrow{{}r}||\overrightarrow{{}r_2}|} \\ & \frac{(3 i-5 j+K)(\lambda i+j-3 K)}{\sqrt{35} \cdot \sqrt{\lambda^{2}+10}} \\ & .\frac{1}{5}=\frac{3 \lambda-8}{\sqrt{35} \cdot \sqrt{\lambda^{2}+10} \mid} \rvert, m \end{aligned} $
$ \text{ Square } \Rightarrow \frac{1}{25}=\frac{9 \lambda^{2}+64-48 \lambda}{35(\lambda^{2}+10)} $
$\Rightarrow 19 \lambda^{2}-120 \lambda+125=0$
$\Rightarrow 19 \lambda^{2}-95 \lambda-25 \lambda+125=0$
