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The component of vecA along vecB is sqrt...

The component of `vecA` along `vecB` is `sqrt(3)` times that of the component of `vecB` along `vecA`. Then A:B is

A

`1:sqrt(3)`

B

`sqrt(3):1`

C

`2:sqrt(3)`

D

`sqrt(3):2`

Text Solution

Verified by Experts

The correct Answer is:
B
`A cos theta=(vecA.vecB)/(|vecB|)` and `B cos theta=(vecA.vecB)/(|vecA|)`,
`A cos theta=sqrt(3) B cos theta`
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