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Question

# If the vectors →a and →b are perpendicular to each other, then a vector →v in terms of →a and →b satisfying the equations →v.→a=0,→v.→b=1 and [→v →a →b]=1 is

A

b|b|2+a×b|a×b|2

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B

b|b|+a×b|a×b|2

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C

b|b|2+a×b|a×b|

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D

None of these

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Solution

## The correct option is A →b|→b|2+→a×→b|→a×→b|2 Since a⊥b,∴ a, b, a×b are linerly independent ∴→v can be expressed uniquely in terms of →a,→b and (→a×→b) Let →v=x→a+y→b+z(→a×→b) Given: →a.→b=0, →v.→a=0, →v.→b=1, [→v →a →b]=1⇒→v.→a=x|→a|2[∵→a.→b=0, →a.→a×→b=0]⇒0=xa2⇒x=0→v.→b=yb2⇒1=yb2∴y=1b2→v.→a×→b=z|→a×→b|2⇒1=z|→a×→b|2⇒z=1|→a×→b|2Hence, →v=1|→b|2→b+1|→a×→b|2(→a×b)

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