A) b . b
B) \[{{a}^{2}}b\]
C) 0
D) \[{{a}^{2}}+ab\]
Correct Answer: C
Solution :
Since \[\mathbf{a}\times \mathbf{b}\] is perpendicular to \[\mathbf{a}\] and \[\mathbf{b}\] both, therefore \[\mathbf{a}\,.\,(\mathbf{a}\times \mathbf{b})=0.\] \[(\because \]Scalar product of two perpendicular vector is zero)
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