Why does this worked problem in determining the equation of a plane in space multiply vectors by scalars?

Question

This i.imgur.com/b6BIiAK.png is given as a solution in Larson 6th ed for a problem to compute equation of a plane from 3 points. Can anyone please tell me why they multiply u and v by 1/2 and -1 respectively?

Answer 1

The unit vector perpendicular to $u$ and $v$ is precisely the unit vector perpendicular to $\frac{1}{2}u$ and $-v$. The only reason to make this change is in order to simplify computations.

If we make these computations directly on $u$ and $v$ we obtain; $$ u\times v = -8i+6j-8k = -2(4i - 3j + 4k) $$ So normalizing the above we obtain the same result as we would with the vectors $\frac{1}{2}u, -v$