The vector `(veca +3vecb)` is perpendicular to `(7veca-5vecb)` and `(veca-4vecb)` is perpendicular to `(7veca-2vecb).` The angle between `veca` and `vecb` is:

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. None of these

Explanation

`(vec(a)+3vec(b)).(7vec(a)-5vec(b))=0`
`rArr 7a^(2)-15b^(2)+16 vec(a). vec(b)=0` …(i)
and, `(vec(a)-4vec(b)).(7vec(a)-2vec(b))=0`
`rArr 7a^(2)+8b^(2)-30 vec(a).vec(b)=0` …(ii)
By adding (i) and (ii)
`rArr -23b^(2)+46vec(a).vec(b)=0 rArr 2vec(a).vec(b)=b^(2)`
So `7a^(2)-15b^(2)+8b^(2)=0 rArr a^(2)=b^(2)`
`rArr 2ab cos theta=b^(2) rArr 2 cos theta=1`
`rArr theta=cos^(-1) (1//2)=60^(@)`