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If vecA = vecB + vecC and the magnitude...

If `vecA = vecB + vecC` and the magnitudes of `vecA, vecB and vecC` are 5,4 and 3 units respecetively, the angle between `vecA` and `vecC` is :

A

`cos^(-1) ((3)/(5))`

B

`cos^(-1)((4)/(5))`

C

`(pi)/(2)`

D

`sin^(-1) ((3)/(4))`

Text Solution

Verified by Experts

The correct Answer is:
A
Given : `A=|vecA|=5 , B=|vecB|=4`
`C= |vecC|=3`
As `5^(2)=4^(2)+3^(2)`
`A^(2)=B^(2)+C^(2)`
`:.` Angle between `vecB` and `vecC` is `90^(@)`.
If `theta` is the angle between `vecA` and `vecC` then
`cos theta= (C )/(A) = (3)/(5)`
`theta= cos^(-1) ((3)/(5))`
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