Let the plane \(\mathrm{P}: 4 \mathrm{x}-\mathrm{y}+\mathrm{z}=10\) be rotated by an angle \(\frac{\pi}{2}\) about its line of intersection with the plane \(\mathrm{x}+\mathrm{y}-\mathrm{z}=4\). If \(\alpha\) is the distance of the point \((2,3,-4)\) from the new position of the plane P, then \(35 \alpha\) is

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Three Dimensional Geometry

Let the plane

P : 4 x - y + z = 10

be rotated by an angle

\frac{π}{2}

about its line of intersection with the plane

x + y - z = 4

. If

α

is the distance of the point

(2, 3, - 4)

from the new position of the plane P, then

35 α

Answer: (D)

The family of planes through the line of intersection is

(4 x - y + z - 10) + λ (x + y - z - 4) = 0

. After rotation by

\frac{π}{2}

, the normal vectors are perpendicular:

(4, - 1, 1) \cdot (4 + λ, - 1 + λ, 1 - λ) = 0

. Expanding:

16 + 4 λ + 1 - λ + 1 - λ = 0

, so

18 + 2 λ = 0

, giving

λ = - 9

. The new plane is

- 5 x - 10 y + 10 z + 26 = 0

. The distance from

(2, 3, - 4)

α = \frac{∣ - 10 - 30 - 40 + 26∣}{25 + 100 + 100} = \frac{54}{15} = \frac{18}{5} .

Thus

35 α = 35 \cdot \frac{18}{5} = 126

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Single correct option