−−→OA=→a,−−→OC=→b,−−→OB=10→a+2→b where O, A, C are non-collinear points. Let p denote the area of quadrilateral OABC & q denote the area of parallelogram with OA and OC as adjacent sides. If p=λq, then λ is
A
3
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B
5
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C
6
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D
8
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Solution
The correct option is C 6 q=|→a×→b| p=Ar(ΔOAB)+Ar(ΔOBC) =12{|→a×(10→a+2→b)|+|→b×(10→a+2→b)|} =12{|2(→a×→b)|+|10(→b×→a)|} p=6|→a×→b| ∴p=6q⇒λ=6