JEE Main2019MathQ.Let α=3i+j^ and B=2i^−j^+3k^ If β=β1−β2, where β1 is parallel to α and β2 is perpendicular to α then β1×β2 is equal toA21(3i^−9j^+5k^)B21(−3i^+9j^+5k^)C−3i^+9j^+5k^D3i^−9j^−5k^SolutionView solution← Back to full paperPractice more MathConcept-wise practice with instant solutions on Fundamenthol.Start practicing →