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JEE Main 2019 · 12 April · Shift I

JEE Main 2019 Mathematics question on Three Dimensional Geometry

If the line x−23=y+12=z−1−1\frac{x-2}{3} = \frac{y+1}{2} = \frac{z-1}{-1} intersects the plane 2x+3y−z+13=02x + 3y - z + 13 = 0 at a point PP and the plane 3x+y+4z=163x + y + 4z = 16 at a point QQ, then PQPQ is equal to
  1. A.
    14
  2. B.
    14\sqrt{14}
  3. C.
    272\sqrt{7}
  4. D.
    2142\sqrt{14}
Show answer and solution

Correct option: D

Solution

Parametrize the line as x = 2+3t, y = -1+2t, z = 1-t.

For point P (intersection with 2x+3y-z+13=0):
2(2+3t) + 3(-1+2t) - (1-t) + 13 = 0
4+6t - 3+6t - 1+t + 13 = 0
13t + 13 = 0 → t = -1
P = (2-3, -1-2, 1+1) = (-1, -3, 2)

For point Q (intersection with 3x+y+4z=16):
3(2+3t) + (-1+2t) + 4(1-t) = 16
6+9t - 1+2t + 4-4t = 16
7t + 9 = 16 → t = 1
Q = (5, 1, 0)

PQ = √((5-(-1))² + (1-(-3))² + (0-2)²) = √(36+16+4) = √56 = 2√14

[JEE Main 2019 · 12 April · Shift I]

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