If →a and →b are perpendicular unit vectors and vector →c is such that →c=→a+→b, then (→a×→b).(→b×→c)+(→b×→c).(→c×→a)+(→c×→a).(→a×→b) is
A
0
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B
1
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C
-1
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D
→b.→c+→c.→a
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Solution
The correct option is C -1 (→a×→b).(→b×→c)+(→b×→c).(→c×→a)+(→c×→a).(→a×→b) =(→a.→b)(→b.→c)−(→a.→c)(→b.→b)+(→b.→c)(→c.→a)−(→b.→a)(→c.→c)+(→c.→a)(→a.→b)−(→c.→a)(→a.→b)−(→a.→a) =0−(→a.→c)+(→b.→c)(→c.→a)−0+0−(→c.→b)(∵→a.→b=0) =(→b.→c)(→c.→a)−(→c.→a)−(→b.→c)+1−1 =((→b.→c)−1)((→c.→a)−1)−1 =(1−1)(1−1)−1=−1(∵→c=→a+→b)