If `a=hati+2hatj-3hatk` and `b=3hati+2hatk` then the angle between the vectors `a+b` and `a-b` is

Options

A. `60^(@)`
B. `90^(@)`
C. `45^(@)`
D. `55^(@)`

Explanation

(b) Given that `a=hati+2hatj-3hatk`
and `b=3hati-hatj+2hatk`
Now, `a+b=hati+2hatj-3hatk+3hati-hatj+2hatk=4hati+hatj-hatk`
and `a-b=(hati+2hatj-3hatk)=(3hati-hatj+2hatk)=-2hati+3hatj-5hatk`
Let `theta` be the angle between `a+b` and `a-b`.
`:. cos theta=((a+b).(a-b))/(|a+b||a-b|)=((4hati+hatj-hatk).(-2hati+3hatj-5hatk))/(|4hati+hatj-hatk|.|-2hati-3hatj-5hatk|)`
`=(-8+3+5)/(sqrt(16+1+1)sqrt(4+9+25))=0implies=90^(@)`