Q.
p→=b→×c→[a→b→c→],q→=c→×a→[a→b→c→] and r→=a→×b→[a→b→c→] where a→,b→ and c→ are thee non-coplanar vectors, then the value of the expression (a→+b→+c→)⋅(p→+q→+r→) is
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a
3
b
2
c
1
d
0
answer is A.
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Detailed Solution
We have,a→⋅p→=a→⋅(b→×c→)[a→b→c→]=a→⋅(b→×c→)[a→b→c→]=[a→b→c→][a→b→c→]=1a→⋅q→=a→⋅c→×a→[a→b→c→]=[a→c→a→][a→b→c→]=0 Similarly, a→⋅r→=0,b→⋅p→=0,b→⋅q→=1,b→⋅r→=0c→⋅p→=0,c→⋅q→=0 and c→⋅r→=1∴ (a→+b→+c→)⋅(p→+q→+r→)=a→⋅p→+a→⋅q→+a→⋅r→+b→⋅p→+b→⋅q→+b→⋅r→+c→⋅p→+c→⋅q→+c→⋅r→=1+1+1=3
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