Question

A plane P contains the line x + 2y + 3z + 1 = 0 = x – y – z – 6, and is perpendicular to the plane –2x + y + z + 8 = 0. Then which of the following points lies on P?

Options

• (1, 0, 1)

• (2, –1, 1)

• (–1, 1, 2)

• (0, 1, 1)

Answer 1

(0, 1, 1)

Explanation:

A plane which contains the line x + 2y + 3z + 1 = 0 = x – y – z – 6 is (x + 2y + 3z + 1) + k(x – y – z – 6) = 0

⇒ (1 + k)x + (2 – k)y + (3 – k)z + 1 – 6k = 0

This plane is perpendicular to the plane –2x + y + z + 8 = 0 then –2(1 + k) + 1(2 – k) + 1(3 – k) = 0 ⇒ k = `3/4`

Equation of plane is `7/4x + 5/4y + 9/4z = 14/4`

⇒ 7x + 5y + 9z = 14  ...(i)

For point (1, 0, 1)

7(1) + 5(0) + 9 (1) = 16 ≠ 14

For point (2, –1, 1)

7(2) + 5(–1) + 9(1) = 18 ≠ 14

For point (1, 1, 2)

7(–1) + 5(1) + 9(2) = 16 ≠ 14

For point (0, 1, 1)

7(0) + 5(1) + 9(1) = 14

So, (0, 1, 1) lies on the plane.