Let the shortest distance between the lines (x-3)/3=(y-α)/-1=(z-3)/1 and (x+3)/-3=(y+7)/2=(z-β)/4 be 3 √30. Then the positive value of 5 α+β is

Solution

Maths sir · 4 steps

Project the joining vector onto b⃗₁ × b⃗₂, set it equal to 3√30, and solve.

1. Read off the lines. Line 1 passes through A(3, α, 3) with direction b⃗₁ = (3, -1, 1). Line 2 passes through B(-3, -7, β) with direction b⃗₂ = (-3, 2, 4).
2. Common perpendicular. b⃗₁ × b⃗₂ = ((-1)(4) - (1)(2), (1)(-3) - (3)(4), (3)(2) - (-1)(-3)) = (-6, -15, 3). Its length is √(36 + 225 + 9) = √270 = 3√30.

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

3. Put in numbers. AB⃗ = (-6, -7-α, β - 3), so the dot product is 36 + 105 + 15α + 3β - 9 = 132 + 15α + 3β. Setting d = 3√30 gives |132 + 15α + 3β| = 3√30 × 3√30 = 270.
4. So. Dividing by 3, |44 + 5α + β| = 90, so 5α + β = 46 or -134. The positive value is 46, option (d).

The idea to keep. Shortest distance between skew lines is the length of the projection of the joining vector on b⃗₁ × b⃗₂. The modulus gives two cases, so check both before picking one.

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