`Let veca,vecb,and vecc` be three non-coplanar ubit vectors such the angle between every pair of them is `(pi)/(3). if vecaxxvecb+vecbxxvecc=pveca+qvecb+rvecc,` where p,q and r are scalars , then the value of `(p^(2)+2q^(2)+r^(2))/(q^(2))` is

Explanation

Given that `veca ,vecb and vecc` are three unit vectors inclined at an angle `theta` with each other.
Also `veca , vecb and vecc` are non-coplanar, therefore,
`[veca vecb vecc] ne 0`
Also given that `veca xx vecb + vecb xx vecc d= pvec + qvecb + rvecc`
taking dot product on both sides with `veca`. we get
`p + q costheta + r cos theta= [veca vecbvecc]`
Similarly m taking dot product on both sides with `vecb and vecc` we get , respectively.
`p costheta+q+r cos theta=0`
`and p cos theta + q cos theta+r[veca vecb vecc]`
Adding (i), (ii) and (iii) we get
`p+q+r= (2[veca vecbvecc])/(2costheta+1)`
Multiplying (iv) by `cos theta` and subtraching (i) from it, we get
`p(cos theta-1)= (2[veca vecbvecc]costheta)/(2costheta+1)-[veca vecb vecc]`
`or (-[veca vecb vecc])/(2costheta+1)`
`or P= ([veca vecb vecc])/((1-cos theta) (1+2 cos theta))`
Similarly, (iv), `xx cos theta - ` (ii) gives
`r(cos theta-1)=(2[veca vecb vecc]costheta)/(2costheta+1)-[veca vecb vecc]`
`or r= (-[veca vecbvecc])/((2costheta+1)(costheta-1))`
But we have to find p,q and r in terms of `theta` only, so let us find the value of `[veca vecb vecc]`
we know that
`[veca vecb vecc]^(2)=|{:(veca.veca,veca.vecb,veca.vecc),(vecb.veca,vecb.vecb,vecb.vec c),(vecc.veca,vecc.vecb,vecc.vecc):}|= |{:(1,costheta,costheta),(costheta,1,cos theta),(cos theta,cos theta,1):}|`
On operating `C_(1) to C_(1) + C_(2)+C_(3)`, we get
`|{:(1+2costheta,costheta,costheta),(1+2costheta,1,costheta),(1+2costheta,costheta,1):}|=(1+2costheta)|{:(1,costheta,cos theta),(1,1,costheta),(1,cos theta,1):||`
Operating `R_(1) to R_(1)-R_(2)and R_(2) to R_(2)- R_(3)`, we get
`(1+2costheta)|{:(0,costheta-1,0),(0,1-costheta,costheta),(1,cos theta,1):}|`
Expanding along `C_(1)`
`(1 + 2cos theta) (1-cos theta)^(2)`
`[veca vecb vecc] = (1-cos theta) sqrt(1+2costheta)`
thus , we get
`P= 1/(sqrt(1+2costheta)),q= (-2costheta)/(sqrt(1+2costheta)),`
`r=1/(sqrt(1+2cos theta))`