If the vectors →a,→b,→c from the sides BC,CA,AB respectively of triangle ABC, then
A
→a.→b+→b.→c+→c.→a=0
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B
→a×→b=→b×→c=→c×→a
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C
→a.→b=→b.→c=→c.→a
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D
→a×→b+→b×→c+→c×→a=0
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Solution
The correct option is C→a×→b=→b×→c=→c×→a Given: →a+→b+→c=0 →a×(→a+→b+→c) =→a×→a+→a×→b+→a×→c=0 But →a×→a=0 ⇒→a×→b+→a×→c=0 ⇒→a×→b=−→a×→c ⇒→a×→b=→c×→a →b×(→a+→b+→c) ⇒→b×→a+→b×→b+→b×→c=0 But →b×→b=0⇒→b×→c=→a×→b ∴→a×→b=→c×→a ∴→a×→b=→b×→c=→c×→a