The shortest distance, between lines L₁ and L₂, where L₁: (x-1)/2=(y+1)/-3=(z+4)/2 and L₂ is the line, passing through the points A(-4,4,3), B(-1,6,3) and perpendicular to the line (x-3)/-2=y/3=(z-1)/1, is

Solution

Maths sir · 4 steps

Take L₂ along AB⃗, then use the skew-line distance formula.

1. Read off both lines. L₁ passes through P(1,-1,-4) with direction d₁⃗ = (2,-3,2). L₂ passes through A(-4,4,3) and B(-1,6,3), so its direction is d₂⃗ = AB⃗ = (3,2,0). As a check, (3,2,0)·(-2,3,1) = -6+6+0 = 0, so L₂ is indeed perpendicular to the given line.
2. Common perpendicular. d₁⃗×d₂⃗ = (2,-3,2)×(3,2,0) = (-4, 6, 13). Its length is √(16+36+169) = √221.

   d = (|PA⃗·(d₁⃗×d₂⃗)|)/(|d₁⃗×d₂⃗|)

3. Put in numbers. PA⃗ = (-4-1, 4+1, 3+4) = (-5,5,7). Then PA⃗·(-4,6,13) = 20+30+91 = 141.
4. So. The shortest distance is 141/√221, which is option (a).

The idea to keep. For skew lines, project the vector joining one point on each line onto d₁⃗×d₂⃗. When a line is given by two points, their difference is its direction.

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