But as (b) is a sum rather than a vector in the form (x,y,z) do I just multiply that whole long equation of x by v1,v2,v3? It just feels like I’m missing something when I tried to expand out x on its own in my workings here because it is such a long equation …
I think you are doing this the hard way. Let's consider a simpler case:
Consider a new basis (1, 3), (2, -1) and a vector (5.9) written in the usual [imath]( \hat{i}, \hat{j} )[/imath] basis. Then in order to change the vector from the old basis to the new we have the following transformation matrix:
[imath]T = \left ( \begin{matrix} 1 & 2 \\ 3 & -1 \end{matrix} \right )[/imath]
(Note the position of the numbers with respect to the new basis vectors.)
Then, in terms of the vector in the "old basis"
[imath]\left ( \begin{matrix} 5 \\ 9 \end{matrix} \right ) = T \left ( \begin{matrix} x_1 \\ x_2 \end{matrix} \right )[/imath]
and we want to find
[imath]\left ( \begin{matrix} x_1 \\ x_2 \end{matrix} \right ) = T^{-1} \left ( \begin{matrix} 5 \\ 9 \end{matrix} \right ) = \left ( \begin{matrix} 1 & 2 \\ 3 & -1 \end{matrix} \right ) ^{-1} \left ( \begin{matrix} 5 \\ 9 \end{matrix} \right ) = \dfrac{1}{7} \left ( \begin{matrix} 23 \\ 6 \end{matrix} \right )[/imath]
(My students always complained that my examples never came out with nice numbers!)
Now, this might seem a little backward as we have to invert the matrix we get but that's because we have to write the transformation matrix T in terms of the old basis.
Once you have the transformation matrix T (or rather, it's inverse) you are set to go with any vector transformation.
See
here for a good video on this.
-Dan
Note: The link gets you to the video but it's starting halfway through. I don't know why.