a→ and b→ are two given vectors. On these vectors as adjacent sides, a parallelogram is constructed. The vector which is the altitude of the parallelogram and perpendicular to a→ is given by:

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a→ ⋅ b→|a→|2 a→ − b→

a→ ⋅ b→|b→|2 b→ − a→

b→ × (a→ × b→)|b→|2

a→ ⋅ b→|a→|2 a→ + b→

answer is A.

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Detailed Solution

The altitude required is BN→BN→=ON→−OB→= projection of OB→ on ON→−b→= (b→ ⋅ a∧) a∧ − b→= b→ ⋅ a→|a→| a→|a→| − b→ = (b→ ⋅a→) a→|a→|2 − b→

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