Show that $\vec u+\vec v$, $\vec u-3\vec w$ and $2\vec v-3\vec w$ are linearly independent 
Question:
  Let $\vec u,\vec v$ and $\vec w$ be vectors in $\Bbb R^n$. Suppose that $\vec u,\vec v, \vec w$  are linearly independent. Show that $\vec u+\vec v$, $\vec u-3\vec w$  and $2\vec v-3\vec w$ are linearly independent. 

I understand that if a vectors are linear independent if the equation 
$$a_1\vec v_1 + a_2\vec v_2 + \cdots + a_k\vec v_k = 0$$
so when 
$$a_1 = a_2 = \cdots= a_k = 0.$$
I just don't understand how to explain, any help would be appreciated, thanks guys!
 A: Ok, here's the idea. A (finite) collection of vectors is linearly independent if whenever you have a linear combination of them resulting in the zero vector, you have that the coefficients have to be zero. So if you want to check that the set $$\{u+2v, u-3w,2v-3w\}$$is linearly independent, you must write $$a(u+2v)+b(u-3w)+c(2v-3w)=0$$and somehow conclude that $a=b=c=0$. Now, the only information you have is that $\{u,v,w\}$ is linearly independent. So reorganize the above expression as $$(a+b)u + (2a+2c)v+(-3b-3c)w=0.$$Since $\{u,v,w\}$ is linearly independent, you may safely conclude that $$\begin{cases} a+b=0 \\ 2a+2c=0 \\ -3b-3c=0\end{cases}$$Now solve this system to conclude that $a=b=c=0$, as you wanted.
A: you need to show that given scalars $a,b,c$ then if
$$ a (u + 2v) + b (u - 3w ) + c (2v - 3w) = {\bf 0 } $$
then $a=b=c=0$, but notice that after rearranging we obtain
$$ (a+b) u + (2a+2c) v + (-3b-3c) w = {\bf 0 } $$
Since $\{ u,v,w \}$ is LI, then 
$$ a+b = a+c=b+c = 0 $$
And we get what we want
