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Given vecA=2hati+3hatj and vecB=hati+hat...

Given `vecA=2hati+3hatj and vecB=hati+hatj` . The component of vector `vecA` along vector `vecB` is

A

`(1)/(sqrt(2))`

B

`(3)/(sqrt(2))`

C

`(5)/(sqrt(2))`

D

`(7)/(sqrt(2))`

Text Solution

Verified by Experts

The correct Answer is:
C
Here `vecA=2hati+3hatj, vecB= hati+hatj`
`:. |vecA|= sqrt((2)^(2)+(3)^(2))=sqrt(13)`
`|vecB|= sqrt((1)^(2)+(1)^(2))=sqrt(2)`
Let `theta` be the angle between `vecA` and `vecB` .
`:. costheta= (vecA.vecB)/(|vecA||vecB|)`
The component of vector `vecA` along `vecB` is
`=|vecA|cos theta=(vecA.vecB)/(|vecB|)`
`=((2hati+3hatj).(hati+hatj))/(sqrt(2))=(5)/(sqrt(2))`
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