` vec p , vec q ,a n d vec r` are three mutually perpendicular vectors of the same magnitude. If vector ` vec x` satisfies the equation ` vec pxx(( vec x- vec q)xx vec p)+ vec qxx(( vec x- vec r)xx vec q)+ vec rxx(( vec x- vec p)xx vec r)=0,t h e n vec x` is given by `1/2( vec p+ vec q-2 vec r)` b. `1/2( vec p+ vec q+ vec r)""` c. `1/3( vec p+ vec q+ vec r)` d. `1/3(2 vec p+ vec q- vec r)`

Answer

224

Explanation

`vecp,vecq and vecp xx vecq` are perpendicular to each other. We have,
`(veca.vecp) + ( veca.vecq) vecq`
`+(veca.(vecpxxvecq))(vecpxxvecq) =veca |vecp|^(2)`
`(vecb.vecp) vecp + (vecb.vecq)vecq`
`= (vecb.(vecpxxvecq))(vecpxxvecq) =vecb |vecp|^(2)`
`(vecc.vecp) + (vecc.vecq) vecq)`
`= (vecc.(vecpxxvecq))(vecpxxvecq) =vecc|vecp|^(2)`
Hence, the required distance is `|(veca +vecb +vecc)||vecp|^(2)`
`sqrt(|veca|^(2)+|vecb|^(2)+|vecc|^(2))xx |vecp|^(2)`
` 14 xx 4^(2)= 224`