The distance of the point (-1,2,3) from the plane r⃗ ·(î-2 ĵ+3 k̂)=10 parallel to the line of the shortest distance between the lines r⃗=(î-ĵ)+λ(2 î+k̂) and r⃗=(2 î-ĵ)+μ(î-ĵ+k̂) is :

Solution

Maths sir · 4 steps

Measure along the shortest-distance line, whose direction is the cross product of the two directions.

1. Direction of SD line. The line of shortest distance is perpendicular to both direction vectors b⃗₁ = 2î+k̂ and b⃗₂ = î-ĵ+k̂. So its direction is b⃗₁ × b⃗₂.

   b⃗₁ × b⃗₂ = |î, ĵ, k̂; 2, 0, 1; 1, -1, 1| = î - ĵ - 2k̂

2. Line through the point. A general point on the line through (-1,2,3) in this direction is (-1+t, 2-t, 3-2t). The plane is x - 2y + 3z = 10.
3. Meet the plane. Substitute: (-1+t) - 2(2-t) + 3(3-2t) = 4 - 3t = 10, so t = -2. The line meets the plane at (-3, 4, 7).
4. So. The distance is |t| × |î-ĵ-2k̂| = 2√(1+1+4) = 2√6, which is option (c).

The idea to keep. Distance from a point to a plane along a given direction is not the perpendicular distance. Draw the line through the point in that direction, find where it meets the plane, then measure.

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