Let `hata` be a unit vector and `hatb` a non zero vector non parallel to `veca`. Find the angles of the triangle tow sides of which are represented by the vectors. `sqrt(3)(hatxxvecb)and vecb-(hata.vecb)hata`

Options

A. `pi//4, pi//4, pi//2`
B. `pi//4, pi//3, pi//12`
C. `pi//6, pi//3, pi//2`
D. none of these

Answer

C

Explanation

We have
`hatax(vecbxxhata)=(hata.hata)vecb-(hata.vecb)hata`
`implieshataxx(vecbxxhata)=vecb-(hata.hatb)hata [ :' hata.hata=|hata|^(2)=1]`
Thus two sides of the triangles are represented by the vectors
`sqrt(3)(hataxxvecb)` and `hataxx(vecbxxhata)`
Siince `hataxx(vecbxxhata)` is perpendicular to `hata` and `vecbxxhata`. Therefore, angle between the sides represented by the given vectors is a right angle.
Now
`|vecb-(hata.vecb)hata|=|hataxx(vecbxxhata)|`
`implies|vecb-(hata.vecb)hata|=|hata||vecbxxhata|sinpi//2 [ :' hata_|_(vecbxxhata)]`
`implies|vecb-(hata.vecb)hata|=|vecbxxhata|=|hataxxvecb|`
Thus the lengths of the sides containing the right angle of the triangle are in the ratio `1:sqrt(3)` So, the other two angles are `pi//6` and `pi//3`.