If `[(0,2beta,gamma),(alpha,beta,-gamma),(alpha,-beta,gamma)]` is orthogonal, then find the value of `2alpha^(2)+6beta^(2)+3gamma^(2).`

Explanation

Let `A = [(0, 2beta,gamma),(alpha,beta,-gamma),(alpha,-beta,gamma)],then A' [(0,alpha,alpha),(2beta,beta,-beta),(gamma,gamma,gamma)]`
since A is orthogonal.
`therefore" " A A'=I`
`rArr [(o,2beta,gamma),(alpha,beta,-gamma),(alpha,-beta, gamma)][(0,gamma,gamma),(2beta,beta,-beta),(gamma,-gamma,gamma)]=[(1,0,0),(0,1,0),(0,0,1)]`
Equating the corresponding elements, we get
`4beta^(2)+gamma^(2)=1`
`2beta^(2)-gamma^(2)=0`
and `alpha^(2)+beta^(2)+gamma^(2)=1`
From Eqs. (i) and (ii), we get ltbegt `beta^(2)=(1)/(2) and gamma^(2)=(1)/(3)`
From Eq. (iii)
`alpha^(2)=1-beta^(2)-gamma^(2)=-(1)/(6)-(1)/(3)=(1)/(2)`
Hence, `2alpha^(2)+6beta^(2)+3gamma^(2)=2xx(1)/(2)+6xx(1)/(6)+3xx(1)/(3)=3`
Aliter
the rows of matrix A are unit orthogonal vectors
`vecR_(1).vecR_(2)=0 rArr 2beta^(2)-gamma^(2)=0 rArr 2beta^(2)=gamma^(2)`
`vecR_(2).vecR_(3)=0 rArr alpha^(2)-beta^(2)-gamma^(2)=0 rArr beta^(2) + gamma^(2) = alpha^(2)` and `vecR_(1).vecR_(3)=1 rArr alpha^(2)+beta^(2)+gamma^(2)=1`
from Eqs. (i),(ii) and (iii), we get `alpha^(2)=(1)/(2),beta^(2)=(1)/(6) and gamma^(2)=(1)/(3)`
`therefore 2alpha^(2)+6beta^(2)+3gamma^(2)=3`