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
3
2
1
0
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→=0
c→⋅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