`vecp, vecq and vecr` are three mutually prependicular vectors of the same magnitude . If vector `vecx` satisfies the equation `vecp xx((vecx-vecq) xxvecp)+ vecq xx ((vecx -vecr)xxvecq)+vecrxx((vecx-vecp)xxvecr)=vec0 " then " vecx` is given by

Options

A. A. `1/2(vecp + vecq-2vecr)`
B. B. `1/2(vecp + vecq+vecr)`
C. C. `1/3(vecp + vecq+vecr)`
D. D. `1/3(2vecp + vecq-vecr)`

Answer

b

Explanation

As `vecp, vecq and vecr` are three mutually pependicular vectors of same magnitude , so let us consider.
`vecp=ahati,vecq=ahatj,vecr=ahatk`
Also let ` vecx = x_(1)hati + y_(1)hatj + z_(1)hatk`
Given that `vecx` satisfies the equation
` vecP xx [(vecx-vecq)xxvecp]=vecpxx[(vecx-vecr)xx vecq]`
`+vecr xx[(vecx-vecp)xxvecr]=0`
Now `vecpxx[(vecx-vecq)xx vecp] = vecp xx [vecx xx vecp-vecqxx vecp]`
`=vecpxx (vecx xx vecp) -vecp xx(vecq xx vecp)`
`(vecp.vecp)vecx.(vecp.vecx) vecp-(vecp.vecp) vecq + (vecp.vecq) vecp `
`a^(2) vecx-a^(2)xhati - a_(3)hatj+0`
Similarly
`vecqxx[(vecx-vecr)xx vecq] =a^(2)vecx-a^(2)x-a^(2)y_(1)hatj -a^(3)hatk`
`and vecrxx[(vecx- vecp) xx vecr] =a^(2)vecx-a^(2)z_(1)hatk -a_(3)hati`
Substituting these values in the equations, we get
`3a^(2)vecx-a^(2)(x_(1)hati +y_(1)hatj+z_(1)hatk)`
`-a^(2)(ahati +ahatj +ahatk)=0`
`or 3a^(2)vecx -a^(2)vecx-a^(2)(p+q+r) =vec0`
`or 2a^(2)vecx = (vecp +vecq +vecr)a^(2)`
`or vecx= 1/2(vecp +vecq +vecr) a^(2)`