If equation of the plane that contains the point (-2,3,5) and is perpendicular to each of the planes 2 x+4 y+5 z=8 and 3 x-2 y+3 z=5 is α x+β y+γ z+97=0 then α+β+γ=

Solution

Maths sir · 4 steps

Perpendicular to both planes means its normal is perpendicular to both normals, so take their cross product.

1. Find the normal. Two planes are perpendicular when their normals are perpendicular. The normals of the given planes are (2, 4, 5) and (3, -2, 3), and the cross product is the vector perpendicular to both.

   n⃗ = |î, ĵ, k̂; 2, 4, 5; 3, -2, 3| = (12+10)î - (6-15)ĵ + (-4-12)k̂ = 22î + 9ĵ - 16k̂

2. Use the point. The plane through (-2, 3, 5) is 22(x+2) + 9(y-3) - 16(z-5) = 0. Expanding gives 22x + 9y - 16z + 44 - 27 + 80 = 0, that is 22x + 9y - 16z + 97 = 0.
3. Match the form. The constant is already 97, so no rescaling is needed: α = 22, β = 9, γ = -16.
4. So. α + β + γ = 22 + 9 - 16 = 15, which is option (a).

The idea to keep. A plane perpendicular to two given planes has its normal along the cross product of their normals. Then one point fixes the plane completely.

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