The line l₁ passes through the point (2, 6, 2) and is perpendicular to the plane 2x+y-2z=10. Then the shortest distance between the line l₁ and the line (x+1)/2=(y+4)/-3=z/2 is :

Solution

Maths sir · 4 steps

The normal of the plane is the direction of l₁. After that, it is the standard skew-lines formula.

1. Set up both lines. A line perpendicular to a plane runs along its normal, so l₁ passes through A(2, 6, 2) with direction d₁⃗ = (2, 1, -2). The second line passes through B(-1, -4, 0) with direction d₂⃗ = (2, -3, 2).

   d = (| AB⃗ · (d₁⃗ × d₂⃗) |)/(| d₁⃗ × d₂⃗ |)

2. Common perpendicular. d₁⃗ × d₂⃗ = (1 × 2 - (-2)(-3), (-2)(2) - 2 × 2, 2(-3) - 1 × 2) = (-4, -8, -8). This is -4(1, 2, 2), so use n⃗ = (1, 2, 2) with |n⃗| = 3.
3. Project AB. AB⃗ = (-1-2, -4-6, 0-2) = (-3, -10, -2). Then AB⃗ · n⃗ = -3 - 20 - 4 = -27.
4. So. d = (|-27|)/3 = 9, which is option (a).

The idea to keep. For the shortest distance between skew lines, project the vector joining any two points onto d₁⃗ × d₂⃗. Pull common factors out of the cross product first, because it keeps the arithmetic small.

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