a⋅b=0
1+15+αβ=0
αβ=−16...(1)
Also,
∣b×c∣2=75
(10+β2)⋅14−(5−3β)2=75
5β2+30β+40=0
β=−4,−2
α=4,8
∣a∣max2=(26+α2)max=90
JEE Main 2021 — Mathematics Vectors & 3D Geometry
Let a=i^+5j^+αk^,b=i^+3j^+βk^ and c=−i^+2j^−3k^ be three vectors such that, ∣b×c∣=53 and a is perpendicular to b. Then the greatest amongst the values of ∣a∣2 is ________.
Held on 27 Aug 2021 · Verified 6 Jul 2026.
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