guys I gotta be honest, I've taken notes on everything in the last two sections for this but I'm not sure how to find a basis for a subspace that is a lone plane/line etc.. a full explanation would really help..
all I can think of is adding two random planes to (a), taking the determinent and if its not 0 find out its a basis for R3? but I have a feeling thats not right, also is there another way?
Best Answer
I'll demonstrate with the first one. I like to think of basis vectors via parameters/constraints/equations. In your first problem, we have
$$3x-2y+5z = 0.$$
Or equivalently,
$$5z = -3x+2y \quad\Longrightarrow \quad z = -\frac{3}{5}x+\frac{2}{5}y.$$
So what does this tell us? We have two independent values, $x$ and $y$, which uniquely determine $z$. So any vector on the plane is of the form $(x,y,-\frac{3}{5}x+\frac{2}{5}y)$.
To extract basis vectors, what you want to do is to collect your parameters. So in the above, we have two parameters: $x$ and $y$. We can collect them in two different vectors:
$$(x,y,-\frac{3}{5}x+\frac{2}{5}y) = \left(1,0,-\frac{3}{5}\right)x+\left(0,1,\frac{2}{5}\right)y.$$
Your basis vectors are then the separated vectors $\left(1,0,-\frac{3}{5}\right)$ and $\left(0,1,\frac{2}{5}\right)y.$ Can you see how to do the others by following this train of logic?