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Angle between vec(a) and vec(b) is 60^(@...

Angle between `vec(a)` and `vec(b)` is `60^(@)` then

A

The compound of `vec(a)-vec(b)` will be `(a^(2)-b^(2))/sqrt(a^(2)+b^(2)+ab)`

B

`vec(a)xx vec(b)` is perpendicular to resultant of `(vec(a)+2vec(b))` and `(vec(a)- vec(b))`

C

The component of `vec(a)-vec(b)` along `vec(a)+vec(b)` will be `(a^(2)-b^(2))/sqrt(a^(2)+b^(2)+2ab)`

D

The component of `vec(a)+vec(b)` along `vec(a)-vec(b)` will be `(a^(2)-b^(2))/sqrt(a^(2)+b^(2)+sqrt(3)ab)`

Text Solution

Verified by Experts

The correct Answer is:
A, B
For (A) : Required component `=((vec(a)-vec(b)).(vec(a)+vec(b)))/(|vec(a)+vec(b)|)=(a^(2)-b^(2))/sqrt(a^(2)+b^(2)+2ab cos 60^(@))=(a^(2)-b^(2))/sqrt(a^(2)+b^(2)+ab)`
For (B) : `vec(a)+2vec(b)+vec(a)-vec(b)=2vec(a)+vec(b)` which lies in the plane of `vec(a)` and `vec(b)`
`rArr` resultant is perpendicular to `vec(a)xxvec(b)`
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