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Question

Let a,b and c be three non-coplanar vectors, and let p,q and r be the vectors defined by the relation
p=b×c[abc], q=c×a[abc], and r=a×b[abc],
Then the value of the expression (a+b)p+(b+c)q+(c+a)r is equal to

A
0
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B
1
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C
2
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D
3
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Solution

The correct option is D 3
ap=a(b×c)[abc]=[abc][abc]=1
bp=b(b×c)[abc]=0[abc]=0
bq=b(c×a)[abc]=(b×c)a[abc]a(b×c)[abc][abc][abc]=1
cq=c(c×a)[abc]=0
cr=(a×b)c[abc]=1 and ar=0
Therefore, the given expression is equal to 1+0+1+0+1+0=3.

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