If` veca,vecb,vecc` are three mutually perpendicular vectors, then the vector which is equally inclined to these vectors is (A) `veca+vecb+vecc` (B) `veca/|veca|+vecb/|vecb|+vecc/|vecc|` (C) `veca/|veca|^2+vecb/|vecb|^2+vecc/|vecc|^2` (D) `|veca|veca-|vecb|vecb+|vecc|vecc`

Options

A. `veca+vecb+vecc`
B. `veca/|veca|+vecb/|vecb|+vecc/|vecc|`
C. `veca/|veca|^(2)+vecb/|vecb|^(2)+vecc/|vecc|^(2)`
D. `|veca|veca-|vecb|vecb+|vecc|vecc`

Answer

b

Explanation

Let `vecalpha= veca/|veca|+vecb/|vecb|+vecc/|vecc|`
since, `veca , vecb and vecc` ar mutually perpendicular vectors, if `vecalpha` makes angles `theta, phi,psi" with " veca,vecb and vecc,` respectively, then
`vecalpha.veca= (veca.veca)/|veca|`
`or |vecalpha|.|veca|costheta = |veca|`
`or cos theta= 1/|vecalpha|`
similarly, `cos theta = 1/|vecalpha| . cos psi = 1/|vecalpha|`
`theta= phi = psi`