If \(\lambda_1\) and \(\lambda_2\) are two values of \(\lambda\) such that the angle between the planes \(P_{1}: \vec{r}\cdot(3 \hat{i}-5 \hat{j}+\hat{k})=7\) and \(P_{2}: \vec{r} \cdot(\lambda \hat{i}+\hat{j}-3 \hat{k})=9\) is \(\sin ^{-1}\left(\frac{2 \sqrt{6}}{5}\right)\), then the square of the length of perpendicular from the point \((38 \lambda_{1}, 10 \lambda_{2}, 2)\) to the plane \(P_{1}\) is ______.

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Three Dimensional Geometry

λ_{1}

and

λ_{2}

are two values of

λ

such that the angle between the planes

P_{1} : r \cdot (3 \hat{i} - 5 \hat{j} + \hat{k}) = 7

and

P_{2} : r \cdot (λ \hat{i} + \hat{j} - 3 \hat{k}) = 9

sin^{- 1} (\frac{2 6}{5})

, then the square of the length of perpendicular from the point

(38 λ_{1}, 10 λ_{2}, 2)

to the plane

P_{1}

is ______.

Answer: 315

Since

sin θ = \frac{2 6}{5}

cos θ = \frac{1}{5}

, and with normals

n_{1} = (3, - 5, 1)

n_{2} = (λ, 1, - 3)

cos θ = \frac{∣3 λ - 8∣}{35 λ ^{2} + 10} = \frac{1}{5} .

Squaring gives

25 (3 λ - 8)^{2} = 35 (λ^{2} + 10)

, i.e.

19 λ^{2} - 120 λ + 125 = 0

, so

λ = 5

λ = \frac{25}{19}

. Taking

λ_{1} = \frac{25}{19}, λ_{2} = 5

, the point

(38 λ_{1}, 10 λ_{2}, 2) = (50, 50, 2)

. Its distance from

P_{1} : 3 x - 5 y + z = 7

\frac{∣150 - 250 + 2 - 7∣}{35} = \frac{105}{35}

, whose square is

\frac{10 5 ^{2}}{35} = 315.

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Integer answer