If the square of the shortest distance between the lines (x-2)/1=(y-1)/2=(z+3)/-3 and (x+1)/2=(y+3)/4=(z+5)/-5 is m/n, where m, n are coprime numbers, then m+n is equal to :

Solution

Maths sir · 4 steps

Project the joining vector onto b⃗₁ × b⃗₂; that projection is the shortest distance.

1. Read off the lines. Line 1 passes through A(2, 1, -3) with direction b⃗₁ = (1, 2, -3). Line 2 passes through B(-1, -3, -5) with direction b⃗₂ = (2, 4, -5).

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

2. Common perpendicular. b⃗₁ × b⃗₂ = ((2)(-5) - (-3)(4), (-3)(2) - (1)(-5), (1)(4) - (2)(2)) = (2, -1, 0). Its length is √(4 + 1 + 0) = √5.
3. Project AB. AB⃗ = (-1-2, -3-1, -5+3) = (-3, -4, -2). Its dot product with (2, -1, 0) is -6 + 4 + 0 = -2, so d = 2/√5.
4. So. d² = 4/5, so m = 4 and n = 5, which are coprime. Then m + n = 9, option (d).

The idea to keep. For skew lines, cross the two direction vectors to get the common perpendicular, then project any vector joining the lines onto it. Take the modulus at the end, because distance cannot be negative.

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