Let `vecp and vecq` any two othogonal vectors of equal magnitude 4 each. Let `veca ,vecb and vecc` be any three vectors of lengths `7sqrt15 and 2sqrt33`, mutually perpendicular to each other. Then find the distance of the vector `(veca.vecp)vecp+(veca.vecq)vecq+(veca.(vecpxxvecq))(vecpxxvecq)+(vecb.vecp)vecp+(vecb.vecp)vecq+ (vecb.(vecb.vecq))(vecpxxvecq)+(vecc.vecp)vecp+(vecc.vecq)vecq+(vecc.(vecpxxvecq))(vecpxxvecq)` from the origin.

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`