If the vectors →a,→b and →c form the sides BC, CA and AB respectively of a ΔABC, 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 B→a×→b=→b×→c=→c×→a By triangle law, →a+→b+→c=→0
Taking cross product by →a,→b,→c respectively, →a×(→a+→b+→c)=→a×→0=→0⇒→a×→a+→a×→b+→a×→c=0⇒→a×→b=→c×→a[∵→a×→a=→0]Similarly,→a×→b=→b×→c∴→a×→b=→b×→c=→c×→a