If lambda_1

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(D) The normal vectors to the planes $P_1$ and $P_2$ are $\vec{n}_1 = 3 \hat{i} - 5 \hat{j} + \hat{k}$ and $\vec{n}_2 = \lambda \hat{i} + \hat{j} - 3

Vedclass · Accepted Answer

$315$

Vedclass · Answer

(D) The normal vectors to the planes $P_1$ and $P_2$ are $\vec{n}_1 = 3 \hat{i} - 5 \hat{j} + \hat{k}$ and $\vec{n}_2 = \lambda \hat{i} + \hat{j} - 3 \hat{k}$ respectively.Given the angle $	heta = \sin^{-1}\left(\frac{2 \sqrt{6}}{5}ight)$,we have $\sin 	heta = \frac{2 \sqrt{6}}{5}$.Thus,$\cos 	heta = \sqrt{1 - \sin^2 	heta} = \sqrt{1 - \frac{24}{25}} = \frac{1}{5}$.The angle between two planes is given by $\cos 	heta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|}$.$|\vec{n}_1| = \sqrt{3^2 + (-5)^2 + 1^2} = \sqrt{9 + 25 + 1} = \sqrt{35}$.$|\vec{n}_2| = \sqrt{\lambda^2 + 1^2 + (-3)^2} = \sqrt{\lambda^2 + 10}$.$\frac{1}{5} = \frac{|3\lambda - 5 - 3|}{\sqrt{35} \sqrt{\lambda^2 + 10}} = \frac{|3\lambda - 8|}{\sqrt{35} \sqrt{\lambda^2 + 10}}$.Squaring both sides: $\frac{1}{25} = \frac{(3\lambda - 8)^2}{35(\lambda^2 + 10)} \Rightarrow 35(\lambda^2 + 10) = 25(9\lambda^2 - 48\lambda + 64)$.$7(\lambda^2 + 10) = 5(9\lambda^2 - 48\lambda + 64) \Rightarrow 7\lambda^2 + 70 = 45\lambda^2 - 240\lambda + 320$.$38\lambda^2 - 240\lambda + 250 = 0 \Rightarrow 19\lambda^2 - 120\lambda + 125 = 0$.$(19\lambda - 25)(\lambda - 5) = 0$,so $\lambda_1 = \frac{25}{19}$ and $\lambda_2 = 5$.The point is $(38 	imes \frac{25}{19}, 10 	imes 5, 2) = (50, 50, 2)$.The perpendicular distance from $(x_0, y_0, z_0)$ to $ax + by + cz - d = 0$ is $d = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}$.$d = \frac{|3(50) - 5(50) + 1(2) - 7|}{\sqrt{35}} = \frac{|150 - 250 + 2 - 7|}{\sqrt{35}} = \frac{|-105|}{\sqrt{35}} = \frac{105}{\sqrt{35}}$.The square of the distance is $\left(\frac{105}{\sqrt{35}}ight)^2 = \frac{11025}{35} = 315$.

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