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 \(P\) ?

• A. \((2, -1, 1)\)
• B. \((0, 1, 1)\)
• C. \((1, 0, 1)\)
• D. \((-1, 1, 2)\)

Hint:

For any plane containing the intersection of two planes, the normal is a linear combination of the normals of the two planes. This "family of planes" approach is much faster than finding the line equation first.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The equation of a plane containing the intersection of two planes is given by \(P_1 + \lambda P_2 = 0\). We find \(\lambda\) by applying the condition that the normal to this plane is perpendicular to the normal of the given third plane.
Step 2: Key Formula or Approach:
1. Family of planes: \((x+2y+3z+1) + \lambda(x-y-z-6) = 0\). 2. Normal \(\vec{n_P} = (1+\lambda, 2-\lambda, 3-\lambda)\). 3. Perpendicularity: \(\vec{n_P} \cdot \vec{n_3} = 0\).
Step 3: Detailed Explanation:
Given \(\vec{n_3} = (-2, 1, 1)\). \[ (1+\lambda)(-2) + (2-\lambda)(1) + (3-\lambda)(1) = 0 \] \[ -2 - 2\lambda + 2 - \lambda + 3 - \lambda = 0 \implies 3 - 4\lambda = 0 \implies \lambda = 3/4 \] The equation of plane \(P\) is: \[ (x + 2y + 3z + 1) + \frac{3}{4}(x - y - z - 6) = 0 \] \[ 4x + 8y + 12z + 4 + 3x - 3y - 3z - 18 = 0 \implies 7x + 5y + 9z - 14 = 0 \] Now check the options: (A) \(7(2) + 5(-1) + 9(1) - 14 = 14 - 5 + 9 - 14 = 4 \neq 0\). (B) \(7(0) + 5(1) + 9(1) - 14 = 5 + 9 - 14 = 0\). (Lies on \(P\)) (C) \(7(1) + 0 + 9(1) - 14 = 2 \neq 0\). (D) \(7(-1) + 5(1) + 9(2) - 14 = -7 + 5 + 18 - 14 = 2 \neq 0\).
Step 4: Final Answer:
The point \((0, 1, 1)\) lies on the plane.