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Q.

If a→ and b→ are perpendicular unit vectors and vector c→ is  such that c→=a→+b→, then the value of (a→×b→)⋅(b→×c→)+(b→×c→)⋅(c→×a→)+(c→×a→)⋅(a→×b→) is

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answer is -1.

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Detailed Solution

(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→⋅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→)
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