Let the equation of the plane passing through the line x - 2y - z - 5 = 0 = x + y + 3z - 5 and parallel to the line x + y + 2z - 7 = 0 = 2x + 3y + z - 2 be ax + by + cz = 65. Then the distance of the point (a, b, c) from the plane 2x + 2y - z + 16 = 0 is ____________.

Solution

Maths sir · 5 steps

Take the family of planes through the line, then fix λ by making its normal perpendicular to the second line.

1. Family of planes. Every plane through the first line has the form (x - 2y - z - 5) + λ(x + y + 3z - 5) = 0. Its normal is (1+λ, -2+λ, -1+3λ).
2. Direction of second line. The second line lies in both planes, so it is perpendicular to both normals. Its direction is (1, 1, 2) × (2, 3, 1) = (1 - 6, 4 - 1, 3 - 2) = (-5, 3, 1).
3. Parallel condition. The plane is parallel to the line, so its normal is perpendicular to (-5, 3, 1). That gives -5(1+λ) + 3(-2+λ) + (-1+3λ) = 0, that is, λ - 12 = 0, so λ = 12.
4. The plane. Putting in λ = 12 gives 13x + 10y + 35z - 65 = 0, that is, 13x + 10y + 35z = 65. So (a, b, c) = (13, 10, 35).

   d = (|2(13) + 2(10) - 35 + 16|)/√(2² + 2² + 1²) = 27/3

5. So. The distance of (13, 10, 35) from the plane is 9.

The common slip: 3. Dividing by 2² + 2² + 1² = 9 instead of its square root gives 3, not 9.

The idea to keep. A plane through the line of intersection of two planes is P₁ + λ P₂ = 0. A plane parallel to a line has its normal perpendicular to that line, and this one dot product fixes λ.

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