Linear Functions View as Surfaces or Vectors I am familiar(on a basic level) with concepts of inner product, dual space and linear functionals. Reviewing my understanding of simple linear functions of the type $Y=ax+by+cz$ I see that this can be represented as an inner product between $Y=<(x,y,z),(a,b,c)>$ over the reals. 
While I understand $Y=ax+by+cz$ to be a 4D surface where we go to points $(x,y,z)$ and check $Y$'s value, I want to better understand how it can be seen in terms of single vectors and the inner product space. 
The inner product can be written as $<a,b>=||a||||b||cos \theta$, so now I am taking my vector $(x,y,z)$ and projecting it on the vector $(a,b,c)$ and scaling it times the magnitude of $(a,b,c)$. This vector projection picture seems quite different to me that the picture of the surface earlier. I'm having difficulty seeing them both as the same thing. I would like some explanation as to how they can be seen as the same or how to better understand this view of things. Is there something deep here?
 A: [I’m sure that this has been covered many times before here, but I can’t find a suitable duplicate at the moment.]
Working in $\mathbb R^3$, fix the vector $\mathbf n\ne0$ and scalar $d$. The equation $\mathbf n\cdot\mathbf x=d$ describes some surface in $\mathbb R^3$. Multiplying both sides of this equation by $\mathbf n/\lVert\mathbf n\rVert^2$ gives $$\left({\mathbf n\over\lVert\mathbf n\rVert}\cdot\mathbf x\right){\mathbf n\over\lVert\mathbf n\rVert} = \left({d\over\lVert\mathbf n\rVert}\right){\mathbf n\over\lVert\mathbf n\rVert}.$$ The left-hand side of this equation is just the formula for the orthogonal projection of $\mathbf x$ onto $\mathbf n$, that is, the component of $\mathbf x$ that’s parallel to $\mathbf n$. On the other hand, the right-hand side is a constant multiple of $\mathbf n$, so the vectors that satisfy this equation all have the same projection onto $\mathbf n$; the quantity $\left\lvert d/\lVert\mathbf n\rVert\right\rvert$ is the length of this projection.  
Now let $\mathbf x_0$ be some point on this surface. For any other point $\mathbf x$ on it, we have $\mathbf n\cdot\mathbf x=\mathbf n\cdot\mathbf x_0$, or $\mathbf n\cdot(\mathbf x-\mathbf x_0)=0$. That is to say, all of the displacement vectors to other points on this surface are orthogonal to $\mathbf n$, so the surface described by this equation is a plane in $\mathbb R^3$. This is easiest to see when $d=0$, in which case the origin lies on the surface. The vector $\mathbf n$ is usually called a normal to/of this plane. What’s more, the quantity $\left\lvert d/\lVert\mathbf n\rVert\right\rvert$ we examined in the previous paragraph is the perpendicular distance of the plane from the origin.   
Note that none of the above really depended on the dimension of the ambient space, so all of it holds for $\mathbb R^n$: in general, the equation $\mathbf n\cdot\mathbf x=d$ describes an $(n-1)$-dimensional hyperplane in $\mathbb R^n$. In particular, it works in $\mathbb R^2$, where it’s easy to draw some diagrams to help you understand the geometrical relationships described above.
