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JEE Main 2023 · 24 Jan · Shift 1 · Paper-1(B.E./B. Tech.)

JEE Main 2023 Mathematics question on Three Dimensional Geometry

The distance of the point (−1,9,−16)(-1, 9, -16) from the plane 2x+3y−z=52x + 3y - z = 5 measured parallel to the line x+43=2−y4=z−312\dfrac{x+4}{3} = \dfrac{2-y}{4} = \dfrac{z-3}{12} is
  1. A.
    20220\sqrt{2}
  2. B.
    3131
  3. C.
    2626
  4. D.
    13213\sqrt{2}
Show answer and solution

Correct option: C

Solution

The line through (-1, 9, -16) parallel to the given direction (3, -4, 12) is parametrized as:
x = -1 + 3t, y = 9 - 4t, z = -16 + 12t.

Substitute into 2x + 3y - z = 5:
2(-1 + 3t) + 3(9 - 4t) - (-16 + 12t) = 5
-2 + 6t + 27 - 12t + 16 - 12t = 5
41 - 18t = 5
18t = 36, so t = 2.

The point of intersection is (-1+6, 9-8, -16+24) = (5, 1, 8).

Distance = √[(5-(-1))² + (1-9)² + (8-(-16))²] = √[36 + 64 + 576] = √676 = 26.

[JEE Main 2023 · 24 Jan · Shift 1 · Paper-1(B.E./B. Tech.)]

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