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Q.

If a×b×c  is perpendicular to a×b×c, then we may have
 

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a

b.c=0

b

a.b=0

c

a.c=0

d

a.b=1

answer is A.

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Detailed Solution

a×(b×c)  a×b×c a·cb-(a·b)c·a·cb-(b·c)a=0 (a·c)2 b2-(a·b) (b·c) (a·c)                    -(a·b) (a·c) (b·c)+(a·b) (b·c) (c·a)=0 let  a·b=x   b·c=y  c·a=z z2|b|2-2xyz+xyz=0 z2 |b|2=xyz  z=0                              a·c =0 

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