If `veca,vecb,vecc` be unit vectors such that `veca.vecb=veca.vecc=0` and the angle between `vecb` and `vecc` is `pi//6`. Prove that `veca=pm2(vecbxxvecc)`

Explanation

`veca.vecb=veca. Vecc=0`
is perpendicular to vectors `vecb and vecc`. Thus
`veca=lambda(vecbxxvecc)`
`|veca|=|lambda(vecbxxvecc)=|lambdasqrt3/2|=1`
` |[veca vecbvecc]|=|veca.(vecbxxvecc)|`
`=|lambda||(vecbxxvecc)|^(2)`
`=2/sqrt3|vecb|^(2)|vecc|^(2)sin^(2)(pi/3)=2/sqrt3xx(sqrt3/2)^(2)=sqrt3/2`