If the equation of a plane P, passing through the intersection of the planes,
x + 4y - z + 7 = 0 and 3x + y + 5z = 8 is ax + by + 6z = 15 for some a, b ∈ R, then the distance of the point (3, 2, -1) from the plane P is...........

Solution

Maths sir · 4 steps

Write the family of planes through the line, then match the z-coefficient and the constant to fix λ.

1. Family of planes. Every plane through the line of intersection has the form (x + 4y - z + 7) + λ(3x + y + 5z - 8) = 0. Its z-coefficient is 5λ - 1 and its constant term is 7 - 8λ.
2. Find lambda. Plane P is ax + by + 6z - 15 = 0. We know its z-coefficient and constant, so those two must be in the same ratio. This gives (5λ - 1)/6 = (7 - 8λ)/-15, so 15 - 75λ = 42 - 48λ and λ = -1.
3. The plane. With λ = -1 the plane is -2x + 3y - 6z + 15 = 0, that is 2x - 3y + 6z - 15 = 0. This matches 6z and 15, with a = 2 and b = -3.

   d = (|2(3) - 3(2) + 6(-1) - 15|)/√(2² + 3² + 6²) = (|-21|)/√49 = 21/7

4. So. The distance of (3, 2, -1) from P is 3.

The idea to keep. For a plane through the intersection of two planes, use P₁ + λ P₂ = 0. Fix λ from whichever coefficients the question already gives you, then use the point-to-plane distance formula.

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