Here is the task that I need to solve:

"Let a,b,c are non-coplanar vectors and p,q,r are three vectors. Show that p,q,r are non-coplanar if and only if there exist numbers $\displaystyle x_i,y_i,z_i (i=1,2,3)$ where $\displaystyle a=x_1p + x_2q+x_3r, b=y_1p+y_2q+y_3r, c=z_1p+z_2q+z_3r$"

I know that I need to show it in both ways that is:

If p,q,r are non-coplanar then there exist numbers $\displaystyle x_i,y_i,z_i (i=1,2,3)$ where $\displaystyle a=x_1p + x_2q+x_3r, b=y_1p+y_2q+y_3r, c=z_1p+z_2q+z_3r$

and

If exist numbers $\displaystyle x_i,y_i,z_i (i=1,2,3)$ where $\displaystyle a=x_1p + x_2q+x_3r, b=y_1p+y_2q+y_3r, c=z_1p+z_2q+z_3r$ then p,q,r are non-coplanar.

I will begin with the second case.

I need to show that u*p+v*q+w*r=0 leads to u=v=w=0.

I know the fact that m*a+n*b+l*c=0 leads to m=n=l=0

So that $\displaystyle m*(x_1p + x_2q+x_3r)+n*(y_1p+y_2q+y_3r)$+$\displaystyle l*(z_1p+z_2q+z_3r)=0$

and
$\displaystyle
p*(x_1*m+x_2*m+x_3*m)+q*(y_1*n+y_2*n+y_3*n)$+$\displaystyle r*(z_1*l+z_2*l+z_3*l)=0
$

Only I do not know if its valid to take:

$\displaystyle u=(x_1*m+x_2*m+x_3*m)$
$\displaystyle v=(y_1*n+y_2*n+y_3*n)$
$\displaystyle r=(z_1*l+z_2*l+z_3*l)$

In this case because of m=n=l=0 then u=v=r=0.


Please help.