The line L₁ is parallel to the vector a⃗=-3 î+2 ĵ+4 k̂ and passes through the point (7,6,2) and the line L₂ is parallel to the vector b⃗=2 î+ĵ+3 k̂ and passes through the point (5,3,4). The shortest distance between the lines L₁ and L₂ is :

Solution

Maths sir · 4 steps

Join the two given points, then project that join onto a⃗ × b⃗. That projection is the gap.

1. The formula. For skew lines r⃗ = p⃗₁ + λ a⃗ and r⃗ = p⃗₂ + μ b⃗, the common perpendicular points along a⃗ × b⃗. The shortest distance is the part of p⃗₂ - p⃗₁ that lies along it.

   d = (| (p⃗₂ - p⃗₁) · (a⃗ × b⃗) |)/(| a⃗ × b⃗ |)

2. Cross product. With a⃗ = (-3, 2, 4) and b⃗ = (2, 1, 3), a⃗ × b⃗ = (2 × 3 - 4 × 1)î - (-3 × 3 - 4 × 2)ĵ + (-3 × 1 - 2 × 2)k̂ = 2î + 17ĵ - 7k̂. Its length is √(4 + 289 + 49) = √342 = 3√38.
3. Join the points. p⃗₂ - p⃗₁ = (5 - 7, 3 - 6, 4 - 2) = (-2, -3, 2). Its dot product with (2, 17, -7) is -4 - 51 - 14 = -69.
4. So. d = 69/3√38 = 23/√38, which is option (a).

The idea to keep. Shortest distance between skew lines: take the vector joining one point on each line, dot it with a⃗ × b⃗, and divide by |a⃗ × b⃗|. Take the absolute value at the end.

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