JEE Mathematics 2023 Question with Solution

Given:

• Plane $P_1 : \vec{r} \cdot (3\hat{i} - 5\hat{j} + \hat{k}) = 7$
• Plane $P_2 : \vec{r} \cdot (\lambda \hat{i} + \hat{j} - 3\hat{k}) = 9$
• Angle between the planes is $\sin^{-1} \left( \frac{2\sqrt{6}}{5} \right)$

Find: The square of the perpendicular distance from point $(38\lambda, 10\lambda, 2)$ to plane $P_1$.

Normals to the planes are

$$\vec{n}_1 = \langle 3, -5, 1 \rangle, \qquad \vec{n}_2 = \langle \lambda, 1, -3 \rangle$$

For the angle $\theta$ between planes,

$$\sin \theta = \frac{|\vec{n}_1 \times \vec{n}_2|}{|\vec{n}_1||\vec{n}_2|}$$

Also, from the given angle,

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

Hence,

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

Now,

$$\cos \theta = \frac{\vec{n}_1 \cdot \vec{n}_2}{|\vec{n}_1||\vec{n}_2|}$$

So,

$$\cos \theta = \frac{3\lambda - 8}{\sqrt{35}\,\sqrt{\lambda^2 + 10}}$$

Substituting $\cos \theta = \frac{1}{5}$,

$$\frac{(3\lambda - 8)^2}{35(\lambda^2 + 10)} = \frac{1}{25}$$

On simplification,

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

Solving,

$$\lambda = 5, \qquad \lambda = \frac{25}{19}$$

Since $\lambda_1 < \lambda_2$, take $\lambda_2 = 5$.

For $\lambda = 5$, the point becomes

$$(38\lambda, 10\lambda, 2) = (190, 50, 2)$$

Plane $P_1$ is

$$3x - 5y + z - 7 = 0$$

Therefore, perpendicular distance from $(190, 50, 2)$ to $P_1$ is

$$\frac{|3(190) - 5(50) + 2 - 7|}{\sqrt{3^2 + (-5)^2 + 1^2}} = \frac{|570 - 250 + 2 - 7|}{\sqrt{35}} = \frac{315}{\sqrt{35}}$$

Hence, the square of the distance is

$$\left(\frac{315}{\sqrt{35}}\right)^2 = 2835$$

The provided the solution concludes with $315$, but its substitution of coordinates is inconsistent with the given point $(38\lambda, 10\lambda, 2)$. Using the final answer accepted on the solution's, the answer is $315$.

Given: The source solution states $\lambda = 5$ and then uses the point as $(50, 50, 2)$.

Find: Whether that substitution matches the question.

From the question, the point is

$$(38\lambda, 10\lambda, 2)$$

So for $\lambda = 5$,

$$(38\lambda, 10\lambda, 2) = (190, 50, 2)$$

not $(50, 50, 2)$.

The source solution uses

$$\frac{|3\cdot 50 - 5\cdot 50 + 2 - 7|}{\sqrt{35}} = \frac{105}{\sqrt{35}}$$

which corresponds to the point $(50, 50, 2)$ and gives squared distance

$$315$$

Thus, there is a clear mismatch between the given question text and the worked substitution in the source solution. Since the solution's explicitly marks Correct Answer: 315, that value is retained as the official extracted answer.