Let the line l: x=(1-y)/-2=(z-3)/λ, λ ∈ ℝ meet the plane P: x+2 y+3 z=4 at the point (α, β, γ). If the angle between the line l and the plane P is cos ⁻¹(√(5/14)), then α+2 β+6 γ is equal to ___________.

Solution

Maths sir · 5 steps

The angle with a plane uses sine, not cosine, so turn the given cosine into a sine first.

1. Read the line. Rewrite (1-y)/-2 as (y-1)/2. The line passes through (0, 1, 3) with direction d⃗ = (1, 2, λ). The plane's normal is n⃗ = (1, 2, 3).
2. Use sine. The angle θ between a line and a plane is 90° minus the angle between the line and the normal, so sin θ = (|n⃗ · d⃗|)/|n⃗||d⃗|. Here cos θ = √(5/14), so sin θ = √(9/14) = 3/√14.

   (|1 + 4 + 3λ|)/(√14 √(5+λ²)) = 3/√14 ⇒ (5+3λ)² = 9(5+λ²)

3. Find lambda. Expanding gives 25 + 30λ + 9λ² = 45 + 9λ². The λ² terms cancel, so 30λ = 20 and λ = 2/3.
4. Meet the plane. A general point on the line is (t, 1+2t, 3+2t/3). Putting it into the plane: t + 2 + 4t + 9 + 2t = 4, so 7t = -7 and t = -1. The point is (-1, -1, 7/3).
5. So. α + 2β + 6γ = -1 - 2 + 14 = 11.

The idea to keep. Line with plane: use sin θ = (|n⃗ · d⃗|)/|n⃗||d⃗|. Line with line uses cosine. Rewrite each fraction as (y-y₁)/b before reading the direction.

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