A line `O P` through origin `O` is inclined at `30^0a n d45^0toO Xa n dO Y ,` respectivley. Then find the angle at which it is inclined to `O Zdot`

Explanation

Let `l, m and n` be the direction coisines of the given vector. Then `l^(2)+m^(2)+n^(2)=1`.
If `l=cos30^(@)=sqrt(3)//2, m=cos45^(@)=1//sqrt(2),"then "(3)/(4)+(1)/(2)+n^(2)=1`.
`rArr" "n^(2)=-1//4`, which is not possible. So, such a line cannot exist.