Representation of a Plane and Linear Independence 
I want to prove that if a set of linearly independent vectors $\{u,v,w,\}$ 
   in $\mathbb{R}^3$ gets combined as follows
  $x= u + \lambda_1 (v-u) + \mu_1 (w-u)$ then $x$ is a representation of a plane in $\mathbb{R}^3$ (not necessarily a linear subspace, but an affine linear subspace). 

This would be defined as follows $x= u + \lambda_2 y + \mu_2 z $, where $u$, $y$ and $z$ are linearly independent.
If we just let $y=v-u$ and $z=w-u$
We get $x= u + \lambda_1 y + \mu_1 z $. Since $\lambda_1$ and $\mu_1$ were arbitrarily chosen, this is a representation as desired.
I just need to prove that $u$,  $y$ and $z$ are linearly independent. Is there some handy theorem I can use for combining a linearly independent set or do I need to prove it by writing out $a_1 u + a_2 y + a_3 z=0$ given that $b_1 u +b_2 v + b_3w =0$ if and only if the $b_i$s are $0$?
 A: Note that $$a_1 u + a_2 y + a_3 z = (a_1 - \lambda_1 a_2 -\mu_1 a_3)u + a_2 v + a_3 w = 0$$
implies $a_3 = 0$, $a_2=0$, and $a_1 - \lambda_1 a_2 -\mu_1 a_3 =a_1 =0$.
A: Suppose $y$ and $z$ are linearly dependent; then there are $\sigma, \tau \in \Bbb R$, not both zero, such that
$\sigma y + \tau z = 0; \tag 1$
we note in passing that in fact neither $\sigma$ or $\tau$ vanishes; if, say, $\sigma = 0$, then (1) becomes
$\tau z = 0, \tag 2$
which since 
$z \ne 0, \tag 3$
forces 
$\tau = 0; \tag 4$
(3) binds since
$z = w - u, \tag 5$
with $u$ and $w$ linearly independent. Now (4) is impossible since then
$\sigma = \tau = 0, \tag 6$
contradicting our hypothesis that $\sigma$, $\tau$ are not both zero.  Thus we have
$\sigma \ne 0 \ne \tau; \tag 7$ 
similar remarks apply to the case $\tau = 0$.  
In any event, with (1) in place, we find that
$\sigma v + \tau w - (\sigma + \tau) u = \sigma(v - u) + \tau(w - u) = \sigma y + \tau z = 0 \tag 8$
which, by virtue of (7), contradicts the hypothesized linear independence of $u, v, w$; thus $y = v - u$ and $z = v - w$ are linearly independent.
