If the vectors `3 vec p+ vec q ;5p-3 vec qa n d2 vec p+ vec q ;3 vec p-2 vec q` are pairs of mutually perpendicular vectors, then find the angle between vectors ` vec pa n d vec qdot`

Explanation

`3 vecp + vecq and 5vecp-3vecq` are perpendicular. thereforem,
`(3vecP +vecq). (5vecp -3vecq)=0`
`15vecp^(2)-3vecq^(2)=4vecp.vecq`
`2vecp +vecq and 4vecp -2vecq` are perpendicular , therefore,
`(2vecp +vecq).(4vecp - 2vecq)=0`
`8vecp^(2)=2vecq^(2)`
`vecq^(2)=4vecp^(2)`
Now, `cos theta= (vecp.vecq)/(|vecp||vecq|)`
Substituting `vecq^(2)=4vecp^(2) "in" (i), 3vecp^(2)=4vecp.vecq`
`cos theta = 3/4.vecp^(2)/(|vecp|2\|vecp|)=3/8`
or ` theta=cos^(-1)""3/8`