If the foot of the perpendicular drawn from (1, 9, 7) to the line passing through the point (3, 2, 1) and parallel to the planes x+2y+z=0 and 3y-z=3 is (α,β,γ), then α+β+γ is equal to :

Solution

Maths sir · 4 steps

Parallel to both planes means the line runs along the cross product of their normals. Find that first.

1. Direction of line. A line parallel to a plane is perpendicular to that plane's normal. The normals here are (1, 2, 1) and (0, 3, -1), so the direction is their cross product.

   d⃗ = |î, ĵ, k̂; 1, 2, 1; 0, 3, -1| = (-2-3)î - (-1-0)ĵ + (3-0)k̂ = (-5, 1, 3)

2. General point. Any point on the line is (3-5t, 2+t, 1+3t). The vector from (1, 9, 7) to it is (2-5t, -7+t, -6+3t).
3. Perpendicular condition. At the foot, this vector is perpendicular to d⃗: -5(2-5t) + 1(-7+t) + 3(-6+3t) = 0, which gives 35t - 35 = 0, so t = 1.
4. So. The foot is (α, β, γ) = (-2, 3, 4), and α + β + γ = -2 + 3 + 4 = 5. The answer is (d).

The idea to keep. A line parallel to two planes runs along n⃗₁ × n⃗₂. For the foot of a perpendicular, take a general point with parameter t and set its dot product with the direction to zero.

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