The shortest distance between the lines (x-4)/4=(y+2)/5=(z+3)/3 and (x-1)/3=(y-3)/4=(z-4)/2 is :

Solution

Maths sir · 4 steps

Take the normal common to both lines from the cross product, then project the joining vector onto it.

1. Read off the lines. Line 1 passes through A(4, -2, -3) with direction b⃗₁ = (4, 5, 3). Line 2 passes through B(1, 3, 4) with direction b⃗₂ = (3, 4, 2).
2. Common perpendicular. The shortest distance is measured along a direction perpendicular to both lines, which is b⃗₁ × b⃗₂ = (15 - 12)î... worked out fully as (5 × 2 - 3 × 4, 3 × 3 - 4 × 2, 4 × 4 - 5 × 3) = (-2, 1, 1). Its length is √(4 + 1 + 1) = √6.

   d = (| AB⃗ · (b⃗₁ × b⃗₂) |)/(| b⃗₁ × b⃗₂ |)

3. Put in numbers. AB⃗ = (1 - 4, 3 + 2, 4 + 3) = (-3, 5, 7). The dot product with (-2, 1, 1) is 6 + 5 + 7 = 18, so d = 18/√6 = 18√6/6.
4. So. The shortest distance is 3√6, which is option (a).

The idea to keep. For two skew lines, the cross product of the directions gives the common perpendicular. Project the vector joining any two points on the lines onto it, and take the absolute value.

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