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Question

Let a=^i^j, b=^j^k, c=^k^i. If d is a unit vector such that ad=0=[bcd], then d is equal to:

A
±(^i+^j2^k6)
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B
±(^i+^j^k3)
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C
±(^i+^j+^k3)
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D
±^k
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Solution

The correct option is A ±(^i+^j2^k6)
d is perpendicular to a and coplanar with b and c.
Hence, d is a vector collinear with a×(b×c)
d=±a×(b×c)|a×(b×c)|
Now
a×(b×c)=(ac)b(ab)c
=1(^j^k)+1(^k^i)
=^i^j+2^k
Hence, d=±(^i+^j2^k6)

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