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|+vec/|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

`veca.vecb.=vecb,vecc=vecc.veca=0`
`"Let " vecr=(veca)/(|veca|)+(vecb)/(|vecb|)+(vecc)/(|vecc|)`
`" cos "0 =(vecr.veca)/(|vecr||vecb|)=(1)/(|vecr|)`
`" cos "phi = (vecr.vecb)/(|vecr||vecc|) =(1)/(|vecr|)`
`" cos "Psi = (vecr.vecc)/(|vecr||vecc|) =(1)/(|vecr|)`
`:. 0 = phi = Psi`