JEE Main 2020 · 2 September · morning · Single correct · +4 / −1

The plane passing through the points (1, 2, 1),
(2, 1, 2) and parallel to the line, 2x = 3y, z = 1
also passes through the point :

  1. (a) (0, 6, –2)
  2. (b) (–2, 0, 1)
  3. (c) (0, –6, 2)
  4. (d) (2, 0 –1)
Answer and solution

Answer (b) (–2, 0, 1)

Solution

Maths sir · 3 steps

The joining vector and the line's direction both lie along the plane, so cross them for the normal.

  1. Direction of the line. Write 2x = 3y as x/3 = y/2. Since z stays fixed at 1, the line's direction is d⃗ = (3, 2, 0).
  2. Normal to the plane. The vector AB⃗ = (2-1, 1-2, 2-1) = (1, -1, 1) lies in the plane. The plane is parallel to d⃗, so the normal is AB⃗ × d⃗ = (-2, 3, 5).

    -2(x-1) + 3(y-2) + 5(z-1) = 0 ⇒ 2x - 3y - 5z + 9 = 0

    Point2x - 3y - 5z + 9On the plane?
    (a) (0, 6, -2)0 - 18 + 10 + 9 = 1No
    (b) (-2, 0, 1)-4 - 0 - 5 + 9 = 0Yes
    (c) (0, -6, 2)0 + 18 - 10 + 9 = 17No
    (d) (2, 0, -1)4 - 0 + 5 + 9 = 18No
  3. So. Only (-2, 0, 1) satisfies 2x - 3y - 5z + 9 = 0, so the answer is (b).

The idea to keep. For a plane through two points and parallel to a line, cross the joining vector with the line's direction to get the normal. Turn 2x = 3y into x/3 = y/2 first.

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

Study Three Dimensional Geometry

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