JEE Main 2020 · 6 September · evening · Single correct · +4 / −1

A plane P meets the coordinate axes at A, B and C respectively. The centroid of ΔABC is given to be (1, 1, 2). Then the equation of the line through this centroid and perpendicular to the plane P is :

  1. (a) (x - 1)/1 = (y - 1)/1 = (z - 2)/2
  2. (b) (x - 1)/2 = (y - 1)/1 = (z - 2)/1
  3. (c) (x - 1)/2 = (y - 1)/2 = (z - 2)/1
  4. (d) (x - 1)/1 = (y - 1)/2 = (z - 2)/2
Answer and solution

Answer (c) (x - 1)/2 = (y - 1)/2 = (z - 2)/1

Solution

Maths sir · 3 steps

Get the intercepts from the centroid, then the normal comes from their reciprocals.

  1. Intercepts from centroid. Let the plane cut the axes at A(a,0,0), B(0,b,0) and C(0,0,c). The centroid is (a/3,b/3,c/3) = (1,1,2), so a = 3, b = 3, c = 6.

    x/3+y/3+z/6=1 ⇒ 2x+2y+z=6

  2. Read the normal. The coefficients of x, y, z give the normal to the plane: (2, 2, 1). A line perpendicular to the plane runs along this normal.
  3. So. The line through (1,1,2) with direction ratios 2, 2, 1 is (x-1)/2=(y-1)/2=(z-2)/1, which is option (c).

The idea to keep. In intercept form x/a+y/b+z/c=1, the normal is (1/a,1/b,1/c). Use the reciprocals of the intercepts, not the intercepts themselves.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 300 times in JEE Main since 2020. This question is from 2020.