The shortest distance between the line 4x−3=−11y+7=5z−1 and 3x−5=−6y−9=1z+2 is :
(A)563187✓ Correct
(B)563178
(C)563185
(D)563179
Answer: (A)
Solution
Direction of the common perpendicular: n=(4,−11,5)×(3,−6,1)=(19,11,9), ∣n∣=563. With points A(3,−7,1) and B(5,9,−2), AB=(2,16,−3), so d=∣n∣∣AB⋅n∣=563∣38+176−27∣=563187 — option (1).
JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Single correct option