Vector spaces: how to prove the linear combination of $V_1$, $V_2$, and $V_3$ solve $z = ax+by$ + c

Question

I am stuck while making an assignment and need your help. Could someone help me what I am doing wrong? I am trying to comprehend it.

Problem 3.8.5: Let a, b, c be real numbers. Consider the equation z = ax+by+c. Prove that there are three 3-vectors v0, v1, v2 such that the set of points [x, y, z] satisfying the equation is exactly {v0 + α1 v1 + α2 v2 : α1 ∈ R, α2 ∈ R} (Hint: Specify the vectors using formulas involving a, b, c.)

We have z = ax + by + c, but want [x,y,z], so that's why we get ax + by + c - z = 0

Correct?

So, what I also know is, is that the standard generators are like:

v1 = [1,0,0]
v2 = [0,1,0]
v3 = [0,0,1]

Right?

So I assumed:

a * [v1]
b * [v2]
c * [v3]

Which gives me:

a1+b0+c0 = x
a0+b1+c0 = y
a0+b0+c1 = z

Correct?

This would make:

ax + by + z

But, I have no c now... There is clearly something I am doing wrong, but I have no clue what exactly. Could someone help me, please? I want to understand it.

Question 2:

That's because I didn't understand this part:

{v0 + α1 v1 + α2 v2 : α1 ∈ R, α2 ∈ R}

Could someone explain this as well?

Answer 1

$z = ax+by+c$ defines a plane

In standard form $ax+by - z= -c$

Vector form of a plane is $p + \mathbf u s + \mathbf v t$

Where $p$ is a point in the plane, and $\mathbf u, \mathbf v$ are independent vectors in the plane.

$N = (a, b, -1)$ is the normal vector perpendicular to the plane. (This is just the coefficients of x,y,z when the equation is in standard form.)

$v_1 = (1,0,a)$ would be a vector in the plane.

$v_2 = (0,1,b)$ would be an independent vector in the plane.

There are infinitely many combinations.

$v_0$ needs to be a point in the plane. $(0, 0,c)$ will do.