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Q.

 Let a,b and c be three non-coplanar vectors and p,q and r the vectors defined by the relations 

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 

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a

0

b

1

c

2

d

3

answer is D.

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Detailed Solution

detailed_solution_thumbnail

 Given that a,b,c are non-coplanar. Therefore, [abc]0

ALSO ,p=b×c[abc], q=c×a[abc],r=a×b[abc]

now,(a+b)p+(b+c)q+(c+a)r=    (a+b)b×c[abc]+(b+c)c×a[abc]+(c+a)a×b[abc]

=ab×c[abc]+bc×a[abc]+ca×b[abc]

[bb×c=cc×a=aa×b=0]=[abc][abc]+[abc][abc]+[abc][abc]=1+1+1=3

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