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JEE Main 2023 · 24 Jan · Shift 2

JEE Main 2023 Mathematics question on Three Dimensional Geometry

If the foot of the perpendicular drawn from (1,9,7)(1, 9, 7) to the line passing through the point (3,2,1)(3, 2, 1) and parallel to the planes x+2y+z=0x + 2y + z = 0 and 3y−z=33y - z = 3 is (α,β,γ)(\alpha, \beta, \gamma), then α+β+γ\alpha + \beta + \gamma is equal to
  1. A.
    5
  2. B.
    3
  3. C.
    1
  4. D.
    −1-1
Show answer and solution

Correct option: A

Solution

The line passes through (3,2,1) and is parallel to both planes x+2y+z=0 and 3y-z=3. The direction vector is the cross product of the normals (1,2,1) and (0,3,-1):

d = (1,2,1)×(0,3,-1) = (2·(-1)-1·3, 1·0-1·(-1), 1·3-2·0) = (-5, 1, 3)

Parametric line: (3-5t, 2+t, 1+3t)

For the foot of perpendicular from (1,9,7), the vector from (1,9,7) to the point on the line must be perpendicular to d=(-5,1,3):

(3-5t-1, 2+t-9, 1+3t-7)·(-5,1,3) = 0
(2-5t, t-7, 3t-6)·(-5,1,3) = 0
-10+25t + t-7 + 9t-18 = 0
35t = 35 → t = 1

Foot: (3-5, 2+1, 1+3) = (-2, 3, 4)

α+β+γ = -2+3+4 = 5

[JEE Main 2023 · 24 Jan · Shift 2]

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