Let a plane P contain two lines
r⃗ = î + λ ( î + ĵ ), λ ∈ R and
r⃗ = - ĵ + μ ( ĵ - k̂ ), μ ∈ R
If Q(α, β, γ) is the foot of the perpendicular drawn from the point M(1, 0, 1) to P, then 3(α + β + γ) equals _______.

Solution

Maths sir · 4 steps

Cross the two direction vectors for the normal, then slide from M along that normal until you hit the plane.

1. Normal to P. Both lines lie in P, so the normal is perpendicular to both directions (1, 1, 0) and (0, 1, -1). Their cross product is (-1, 1, 1), so take n = (1, -1, -1).
2. Equation of P. P passes through (1, 0, 0) on the first line, so x - y - z = 1. The point (0, -1, 0) on the second line also satisfies it: 0 + 1 - 0 = 1.

   Q = M + t n, t = -((1 - 0 - 1) - 1)/(1² + 1² + 1²) = 1/3

3. Find Q. Q = (1, 0, 1) + 1/3(1, -1, -1) = (4/3, -1/3, 2/3). Check: 4/3 + 1/3 - 2/3 = 1, so Q lies on P.
4. So. α + β + γ = 4/3 - 1/3 + 2/3 = 5/3, and 3(α + β + γ) = 5.

The common slip: 7. Dropping the minus sign in t gives t = -1/3 and Q = (2/3, 1/3, 4/3), which leads to 7. That point is not on P: 2/3 - 1/3 - 4/3 = -1, not 1.

The idea to keep. A plane holding two lines has normal = cross product of their directions. For the foot of a perpendicular, move from the point along the normal by t = -(ax₀ + by₀ + cz₀ + d)/(a² + b² + c²).

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