If the shortest distance between the lines
L₁: r⃗=(2+λ) î+(1-3 λ) ĵ+(3+4 λ) k̂,, λ ∈ ℝ; L₂: r⃗=2(1+μ) î+3(1+μ) ĵ+(5+μ) k̂,, μ ∈ ℝ
is m/√n, where gcd(m, n)=1, then the value of m+n equals

Solution

Maths sir · 4 steps

Read a point and direction off each line, then project AB onto b⃗₁ × b⃗₂.

1. Point and direction. L₁ passes through A(2, 1, 3) with direction b⃗₁ = (1, -3, 4). Expand L₂ as (2 + 2μ)î + (3 + 3μ)ĵ + (5 + μ)k̂, so it passes through B(2, 3, 5) with direction b⃗₂ = (2, 3, 1).
2. Common normal. b⃗₁ × b⃗₂ = ((-3)(1) - (4)(3), (4)(2) - (1)(1), (1)(3) - (-3)(2)) = (-15, 7, 9). Its length is √(225 + 49 + 81) = √355.

   d = (|AB⃗ · (b⃗₁ × b⃗₂)|)/(|b⃗₁ × b⃗₂|) = (|(0, 2, 2) · (-15, 7, 9)|)/√355 = 32/√355

3. Check lowest terms. AB⃗ = (0, 2, 2), and the dot product is 0 + 14 + 18 = 32. Since 32 = 2⁵ and 355 = 5 × 71, gcd(32, 355) = 1. So m = 32 and n = 355.
4. So. m + n = 32 + 355 = 387, which is option (b).

The idea to keep. Shortest distance between skew lines is |AB⃗ · (b⃗₁ × b⃗₂)| over |b⃗₁ × b⃗₂|. Expand forms like 2(1+μ) before reading the point.

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