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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 (7,−3,−4)(7, -3, -4) from the plane passing through the points (2,−3,1)(2, -3, 1), (−1,1,−2)(-1, 1, -2) and (3,−4,2)(3, -4, 2) is:
  1. A.
    424\sqrt{2}
  2. B.
    55
  3. C.
    525\sqrt{2}
  4. D.
    44
Show answer and solution

Correct option: C

Solution

Find two vectors in the plane:
v1 = (-1-2, 1-(-3), -2-1) = (-3, 4, -3)
v2 = (3-2, -4-(-3), 2-1) = (1, -1, 1)

Normal = v1 × v2:
i: (4·1 - (-3)·(-1)) = 4-3 = 1
j: -((-3)·1 - (-3)·1) = -(-3+3) = 0
k: ((-3)·(-1) - 4·1) = 3-4 = -1

Normal = (1, 0, -1), so plane equation: 1(x-2) + 0(y+3) - 1(z-1) = 0 → x - z - 1 = 0.

Distance from (7, -3, -4): |7 - (-4) - 1| / √(1²+0²+(-1)²) = |7+4-1| / √2 = 10/√2 = 5√2.

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

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