Let the plane passing through the point (-1, 0, -2) and perpendicular to each of the planes 2x + y - z = 2 and x - y - z = 3 be ax + by + cz + 8 = 0. Then the value of a + b + c is equal to :

Solution

Maths sir · 4 steps

Perpendicular to two planes means your normal is the cross product of their two normals.

1. Why cross product. If two planes are perpendicular, their normals are perpendicular. So the required normal is perpendicular to both n⃗₁ = (2, 1, -1) and n⃗₂ = (1, -1, -1), which makes it parallel to n⃗₁ × n⃗₂.

   n⃗₁ × n⃗₂ = |î, ĵ, k̂; 2, 1, -1; 1, -1, -1| = -2î + ĵ - 3k̂

2. Write the plane. Take the normal as (2, -1, 3). Through (-1, 0, -2) the plane is 2(x+1) - y + 3(z+2) = 0, that is, 2x - y + 3z + 8 = 0.
3. Match the constant. The constant term is already +8, as the question requires, so a = 2, b = -1, c = 3. Check: 2(-1) - 0 + 3(-2) + 8 = 0.
4. So. a + b + c = 2 - 1 + 3 = 4, which is option (d).

The idea to keep. A plane perpendicular to two given planes has its normal along the cross product of their normals. Then scale it so the constant term matches the given form.

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