Linear combination of vectors in $\mathbb{R}^3$ Show that any linear combination of $\pmatrix{1\\\frac{3}{2}\\0}$ and $\pmatrix{0\\3\\6}$ is also a linear combination of $\pmatrix{2\\3\\0}$ and $\pmatrix{0\\1\\2}$
I'm not sure how to do this. I have a proof sketch for showing that any 2-dimensional vector is a linear combination of any other two nonparallel 2d vectors, but as far as I understand, this is not the case for 3d vectors. So how should I go about showing this?
Thanks.
 A: Hint: Your third and fourth vectors are just scaled versions of the first two vectors.
A: A linear combination of $\pmatrix{1\\\frac{3}{2}\\0}$ and $\pmatrix{0\\3\\6}$ is a vector that can be expressed as
$$a\pmatrix{1\\\tfrac{3}{2}\\0}+b\pmatrix{0\\3\\6}$$
for some real numbers $a$ and $b$.
A linear combination of $\pmatrix{2\\3\\0}$ and $\pmatrix{0\\1\\2}$ is a vector that can be expressed as 
$$c\pmatrix{2\\3\\0}+d\pmatrix{0\\1\\2}$$
for some real numbers $c$ and $d$.
But
$$2\pmatrix{1\\\tfrac{3}{2}\\0}=\pmatrix{2\\3\\0}\qquad \tfrac{1}{3}\pmatrix{0\\3\\6}=\pmatrix{0\\1\\2}.$$
A: Suppose $x=\alpha u_1 + \beta u_2$.  Then $x=(\frac{\alpha}{2})u_1'+(3\beta)u_2'$, where $u_1, u_2$ are your first two vectors and $u_1', u_2'$ are your second two vectors.
A: every combination of $\pmatrix{1\\\frac{3}{2}\\0}$ and $\pmatrix{0\\3\\6}$ will be as $c_1\pmatrix{1\\\frac{3}{2}\\0}+c_2\pmatrix{0\\3\\6}=\frac{c_1}{2}\pmatrix{2\\3\\0}+3c_3\pmatrix{0\\1\\2} $ 
so :
any linear combination of$ \pmatrix{1\\\frac{3}{2}\\0}$ and $\pmatrix{0\\3\\6}$ is also a linear combination of $ \pmatrix{2\\3\\0}$and$\pmatrix{0\\1\\2}$
