If →a1,→b1,→c1 is the reciprocal system of vector triad of →a,→b,→c, then →a×→a1+→b×→b1+→c×→c1 is:
A
→0
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B
→a
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C
→b
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D
→c
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Solution
The correct option is A→0 We know, for the reciprocal system, →a1=→b×→c[→a→b→c]
Similarly, →b1=→c×→a[→a→b→c]
and →c1=→a×→b[→a→b→c]
Hence →a×→a1+→b×→b1+→c×→c1=Σ→a×(→b×→c)[→a→b→c]=[(→a⋅→c)→b−(→a⋅→b)→c]+[(→b⋅→a)→c−(→b⋅→c)→a]+[(→c⋅→b)→a−(→c⋅→a)→b][→a→b→c]=0