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Question

Unit vector perpendicular to the plane of the triangle ABC with position vectors of the vertices A,B,C, is ( where Δ is the area of the triangle ABC ) .

A
(a×b+b×c+c×a)Δ
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B
(a×b+¯¯b×c+c×a)2Δ
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C
(a×b+b×c+c×a)3Δ
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D
(a×b+b×c+c×a)4Δ
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Solution

The correct option is B (a×b+¯¯b×c+c×a)2Δ
VectorperpendiculartotheplaneofΔABCisAB×AC=(ba)×(ca)=b×c+a×b+c×aAreaoftriangle=12AB×BC2Δ=AB×ACUnitvector=(b×c+a×b+c×a)b×c+a×b+c×a=(b×c+a×b+c×a)2ΔHence,theoptionBisthecorrectanswer.

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