If `alpha, beta, gamma` are direction angles of a line l, then prove that
`cos^(2)alpha+cos^(2)beta+cos^(2)gamma=1`.
Hence, deduce that `sin^(2) alpha + sin^(2) beta + sin^(2) gamma = 2`.

Explanation

Consider line L passing through origin and parallel to the line AB.
Let P (x, y, z )be a point on L and l (OP)=r.
Position vectro of point P is
`vec(OP)=xhati+yhatj+zhatk" "......(A)`
If `alpha,beta&gamma` are the direction angles and l, m, n, are direction cosines of line OP.
Then `l=cosalpha,m=cosbeta,n=cosx`
Consider `vec(OP).hati=(xhati+yhatj+zhatk.hati=x" ".......(i)`
Also `vec(OP)hati=|vec(OP)||hati|.cosalpha=rcosalpha" "......(ii)`
From equation (i) and (ii) , we get x=r .`cosalpha`
Similarly, we can have y=r .`cosbeta,z=r.cosgamma`
`:.vec(OP)=rcosalphahati+rcosbetahatj+rcosgamma.hatk`
Now, `vec(OP).vec(OP)=|vec(OP)|^(2)`
`:.(rcosalpha)^(2)+(rcosbeta)^(2)+(rcosgamma)^(2)=r^(2)`
`impliesr^(2)(cos^(2)alpha+cos^(2)beta+cos^(2)gamma)=r^(2)`
`impliescos^(2)alpha+cos^(2)beta+cos^(2)gamma=1`
Consider `sin^(2)alpha+sin^(2)beta+sin^(2)gamma`
`=(1-cos^(2)alpha)+(1-cos^(2)beta)+(1-cos^(2)gamma)`
`=3-(cos^(2)alpha+cos^(2)beta+cos^(2)gamma)`
`=3-1=2`
`:.sin^(2)alpha+sin^(2)beta+sin^(2)gamma=2`.