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Question

The unit vector along A×B is

A
A×B|A||B|
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B
A×B|A||B|sinθ
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C
A×B|A||B|cosθ
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D
AB|A||B|sinθ
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Solution

The correct option is B A×B|A||B|sinθ
We know,
|A×B|=|A||B|sinθ
A vector divided by it's own magnitude gives unit vector along itself.
A×B|A||B|sinθ will be unit vector along A×B.

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