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Questions  

p=b×c[abc],q=c×a[abc] and r=a×b[abc] where a,b and c  are thee non-coplanar vectors, then  the value of the expression (a+b+c)(p+q+r) is

a
3
b
2
c
1
d
0

detailed solution

Correct option is A

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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Similar Questions

[(a×b)×(b×c) (b×c)×(c×a) (c×a)×(a×b)]  is equal to (where a,b and c are non-zero non-coplanar vectors)


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