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Question

If a,b,c are non-coplanar vectors such that b×c=a,a×b=c and c×a=b, then

A
a=1
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B
b=1
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C
a+b+c=3
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D
a+b+c=3
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Solution

The correct options are
A b=1
B a=1
C a+b+c=3
D a+b+c=3
[¯a¯b¯c]=¯a.¯a=|¯a|2
Similarly [¯a¯b¯c]=|¯b|2=|¯c|2
|¯a|2=|¯b|2=|¯c|2=[abc]
Now
a,b,c are mutually perpendicular.
|b×c|=|b||c|
|b||c|=|a|
|a|=|b|=|c|
Hence,|b|=1
|a|=|b|=|c|=1
|a+b+c|=a2+b2+c2+2.a.b+2.b.c+2.c.a=12+12+12+0=3

|a|+|b|+|c|=3

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