Solve for `vec x,vecx xx vec b=veca`, where `veca,vecb` are two given vectors such that `veca` is perpendicular to `vecb`.

Explanation

`vec x xxvec b=veca` where `veca` bot `vecb` i.e., `veca.vecb=0` since `veca,vecb,vecaxxvecb`are three non-coplanar vectors, we can express `vec x` as a linear combination of these vectors.
`vecx=lambdaveca+muvecb+v(vecaxxvecb)`
`:. {lamdaveca+muvecb+v(vecaxxvecb)}xxvecb=veca`
`rArr lambda(vecaxxvecb)+mu(vecbxxvecb)+v(vecaxxvecb)xxvecb=veca`
`rArr lambda(vecaxxvecb)+v{(veca.xxvec b)vecb-(vecb.vecb)veca}=veca`
`rArr lambda(vecaxxvecb)-v|vecb|^(2)veca=veca+0(vecaxxvecb)`
Since `veca` and `vecaxxvecb` are non-collinear vectors equating cor=efficient of `vecaxxvecb` and `veca`on both sides we get,
`lambda=0" "-v|vecb|^(2)=1`
`rArr lambda=0" " v-(1)/(b^(2))," " {|vecb|^(2)=b," say"}`
With these values, our expression becomes
`x=-(1)/(b^(2))(vecaxxvecb)+muvecb`,where `mu` is an arbitrary scalar.
Comaring the above equation with `vecr=veca+tvecb`, we can say, geometrically that the points whose position vector `-(1)/(b^(2))(vecaxxvecb)` and is parallel to the vector `vecb`.