JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

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 $P_2: \vec{r}.(\lambda\hat{i} + \hat{j} - 3\hat{k}) = 9$ is $\sin^{-1}\left(\frac{2\sqrt{6}}{5}\right)$ , then the square of the length of perpendicular from the point $(38\lambda_1, 10\lambda_2, 2)$ to the plane $P_1$ is:

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Explanation

Solution

Angle $\theta = \sin^{-1}\left(\frac{2\sqrt{6}}{5}\right) \implies \cos \theta = \sqrt{1 - \frac{24}{25}} = \frac{1}{5}$

$\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}||\vec{n_2}|} = \frac{|3\lambda - 5 - 3|}{\sqrt{35}\sqrt{\lambda^2 + 10}} = \frac{1}{5}$

$25(3\lambda - 8)^2 = 35(\lambda^2 + 10) \implies 5(9\lambda^2 - 48\lambda + 64) = 7\lambda^2 + 70$

$38\lambda^2 - 240\lambda + 250 = 0 \implies 19\lambda^2 - 120\lambda + 125 = 0$

$(19\lambda - 25)(\lambda - 5) = 0 \implies \lambda_1 = \frac{25}{19}, \lambda_2 = 5$

Point: $(38 \cdot \frac{25}{19}, 10 \cdot 5, 2) = (50, 50, 2)$

Distance $d = \frac{|3(50) - 5(50) + 1(2) - 7|}{\sqrt{3^2 + (-5)^2 + 1^2}} = \frac{|150 - 250 + 2 - 7|}{\sqrt{35}} = \frac{105}{\sqrt{35}}$

Square of distance $d^2 = \frac{105 \times 105}{35} = 3 \times 105 = 315$

Answer: 315

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