L1:2x−1=3y−2=4z−3L1:1x=αy=1z−5x=i^21j^3αk^41=i^(3−4α)−j^(−2)+k^(2α−3) S.D. =∣n∣BA⋅n=∣n∣(i^+2j^−2k^)⋅n⇒6(13−8α)2=25((4α−3)2+(2α−3)2+16)6(64a2−280α+169)=25(20α2−36α+34)⇒116α2+348α−164=0α1+α2=116−348=−3
JEE Main 2025 — Mathematics Vectors & 3D Geometry
If the shortest distance between the lines 2x−1=3y−2=4z−3 and 1x=αy=1z−5 is 65, then the sum of all possible values of α is
Held on 7 Apr 2025 · Verified 6 Jul 2026.
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