
b1=2(b⋅a)a^
=2(3j^+4k^)⋅(i^+j^)(2i^+j^)
=2×23(i^+j^)=23(i^+j^)
b1+b2=b
b2=b−b1
=(3j^+4k^)−23(i^+j^)
b2=−23i^+23j^+4k^
b1×b2=∣i^23−23j^2323k^04∣
⇒i^(6)−j^(6)+k^(49+49)=6i^−6j^+29k^
JEE Main 2017 — Mathematics Vectors & 3D Geometry
If the vector b=3j^+4k^ is written as the sum of a vector b1, parallel to a=i^+j^ and a vector b2, perpendicular to a, then b1×b2 is equal to :
Held on 9 Apr 2017 · Verified 6 Jul 2026.
6i^−6j^+29k^
−3i^+3j^−9k^
−6i^+6j^−29k^
3i^−3j^+9k^
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