If the shortest distance between the lines (x-λ)/3=(y-2)/-1=(z-1)/1 and (x+2)/-3=(y+5)/2=(z-4)/4 is 44/√30, then the largest possible value of |λ| is equal to _________.

Solution

Maths sir · 4 steps

Use the skew-line formula, then keep both signs of the modulus: each sign gives its own λ.

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

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

2. Cross and dot. b⃗₁ × b⃗₂ = (-6, -15, 3), with magnitude √(36 + 225 + 9) = √270 = 3√30. With AB⃗ = (-2-λ, -7, 3), the dot product is 12 + 6λ + 105 + 9 = 6λ + 126.
3. Set equal. d = (|6λ + 126|)/3√30 = (|2λ + 42|)/√30. Setting this equal to 44/√30 gives |2λ + 42| = 44.
4. So. 2λ + 42 = 44 gives λ = 1, and 2λ + 42 = -44 gives λ = -43. The largest possible |λ| is 43.

The common slip: 1. Solving only 2λ + 42 = 44 gives λ = 1. The negative case 2λ + 42 = -44 gives λ = -43, which has the larger modulus.

The idea to keep. A distance condition always gives an equation with a modulus. Solve both the plus and the minus case before you pick the answer the question asks for.

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