Let the line of the shortest distance between the lines
L₁: r⃗=(î+2 ĵ+3 k̂)+λ(î-ĵ+k̂) and ; ; L₂: r⃗=(4 î+5 ĵ+6 k̂)+μ(î+ĵ-k̂)
intersect L₁ and L₂ at P and Q respectively. If (α, β, γ) is the mid point of the line segment PQ, then 2(α+β+γ) is equal to ____________.

Solution

Maths sir · 4 steps

Write P and Q as general points, then make PQ perpendicular to both direction vectors.

1. General points. Take P = (1+λ, 2-λ, 3+λ) on L₁ and Q = (4+μ, 5+μ, 6-μ) on L₂. Then PQ⃗ = (3+μ-λ, 3+μ+λ, 3-μ-λ).
2. Perpendicular to both. The shortest-distance segment is perpendicular to both lines. A dot product with (1,-1,1) gives 3-μ-3λ = 0. A dot product with (1,1,-1) gives 3+3μ+λ = 0.

   μ + 3λ = 3, 3μ + λ = -3 ⇒ λ = 3/2, μ = -3/2

3. Find P and Q. P = (5/2, 1/2, 9/2) and Q = (5/2, 7/2, 15/2). Check: PQ⃗ = (0, 3, 3), and its dot product with each direction vector is 0.
4. So. The midpoint is (5/2, 2, 6), so α+β+γ = 21/2 and 2(α+β+γ) = 21.

The common slip: 10.5. Stopping at α+β+γ gives 10.5. The question asks for twice that sum.

The idea to keep. For the feet of the common perpendicular, write general points on both lines and set the joining vector perpendicular to each direction. Two dot products give two equations in λ and μ.

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