उत्तर: 315

समाधान:

सूत्र: दो समतलों के बीच कोण

$ \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}(\dfrac{2 \sqrt{6}}{5}) \\ & \Rightarrow \sin \theta=\dfrac{2 \sqrt{6}}{5} \\ & \therefore \cos \theta=\dfrac{1}{5} \\ & \cos \theta=\dfrac{\overrightarrow{{}n_1} \cdot \overrightarrow{{}n_2}}{|\overrightarrow{{}n_1}||\overrightarrow{{}n_2}|} \\ &= \dfrac{(3 \hat{i}-5 \hat{j}+\hat{k})(\lambda \hat{i}+\hat{j}-3 \hat{k})}{\sqrt{35} \cdot \sqrt{\lambda^{2}+10}} \\ &\Rightarrow \dfrac{1}{5}=\dfrac{3 \lambda-8}{\sqrt{35} \cdot \sqrt{\lambda^{2}+10} } \end{aligned} $

$ \text{ वर्ग } \Rightarrow \dfrac{1}{25}=\dfrac{9 \lambda^{2}+64-48 \lambda}{35(\lambda^{2}+10)} $

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

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

$\lambda =5, \dfrac{25}{19}$

बिंदु $(38 \lambda_1 , 10 \lambda_2, 2) = (50,50,2)$

$P_1 : 3x-5y+z=7$

$d^2 = \left(\dfrac{3 \times 50 - 5 \times 50 +1 \times 2-7}{\sqrt{9+25+1}}\right)^2$

$d^2 = \left(\dfrac{-105}{\sqrt{35}}\right)^2$

$d^2 = 315$