Let u=j^+4k^,v=i^−3k^ and w=cosθi^+sinθj^ Now, u×v=i^01j^10k^4−3 =i^(−3+j^−4(+k^)−1() =−3i^+4j^−k^ Now, (u×v)⋅w=(−3i^+4j^−k^)⋅(cosθi^+sinθj^)=−3cosθ+4sinθ Now, maximum possible value of ∣−3cosθ+4sinθ∣=(−3)2+(4)2=25=5
JEE Main 2012 — Mathematics Vectors & 3D Geometry
If u=j^+4k^,v=i^+3k^ and w=cosθi^+sinθj^ are vectors in 3-dimensional space, then the maximum possible value of ∣u×v⋅w∣ is
Held on 12 May 2012 · Verified 6 Jul 2026.
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