If the lines r⃗ = ( î - ĵ + k̂ ) + λ ( 3ĵ - k̂ ) and r⃗ = ( α î - ĵ ) + μ ( 2î - 3k̂ ) are co-planar, then the distance of the plane containing these two lines from the point (α, 0, 0) is :

Solution

Maths sir · 4 steps

Coplanarity fixes α, the cross product gives the normal, and the distance formula finishes it.

1. Normal to the plane. The plane contains both directions b⃗₁ = (0, 3, -1) and b⃗₂ = (2, 0, -3). So its normal is b⃗₁ × b⃗₂ = (-9, -2, -6), or more neatly n⃗ = (9, 2, 6), with |n⃗| = √(81 + 4 + 36) = 11.
2. Find alpha. The lines pass through A(1, -1, 1) and B(α, -1, 0). For coplanar lines, AB⃗ = (α - 1, 0, -1) must also lie in the plane, so AB⃗ · n⃗ = 0. This gives 9(α - 1) - 6 = 0, so α = 5/3.
3. Equation of plane. Through A with normal (9, 2, 6): 9(x - 1) + 2(y + 1) + 6(z - 1) = 0, that is 9x + 2y + 6z - 13 = 0. As a check, B(5/3, -1, 0) gives 15 - 2 - 13 = 0.

   d = (|9 · 5/3 + 2 · 0 + 6 · 0 - 13|)/√(9² + 2² + 6²) = (|15 - 13|)/11 = 2/11

4. So. The distance of the plane from (5/3, 0, 0) is 2/11, which is option (b).

The idea to keep. Two coplanar lines: normal is b⃗₁ × b⃗₂, and the join of their points must be perpendicular to it. That one dot product gives the unknown.

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