# Projections and planes

I'm taking a Signals class as an electrical engineering student and am need of help understanding some concepts related to linear algebra (I think that's what we're doing, but correct me if I'm wrong).

Given 3 column vectors (using Matlab notation):

v1 = [3 1 1 -1]'

v2 = [1 -2 0 1]'

v3 = [-1 1 5 3]'

and a vector s = [1 6 2 7]'

I need to find the angle between s and the plane defined by v2 and v3. The solution says to take the projections of s onto v2 and v3 (let's call them f2 and f3, respectively), add them, and then project s onto the combination h = f2 + f3.

I have little understanding of the concepts underlying this procedure and the class notes don't mention anything about planes or projecting onto them.

Furthermore, I don't understand how adding the two projection vectors f2 and f3 constitutes a plane. Based on my knowledge of Calc 3, it doesn't. It should just form a vector parallel to the plane of interest I believe. This leads me to wonder why I can't just add vectors v2 + v3 and then project s onto the resultant vector to get the same result as the solution. I actually tried, and didn't get the correct answer, so I'm puzzled.

I'll present the formula I use for determining the coefficient of the projection vector. If I want to determine the coefficient of the projection f2 of s onto v2 then I compute

dot(v2, s)/(norm(v2)^2).

• f2 $+$ f3 doesn't "constitute a plane", but span(f2,f3) is a plane. – user137731 Oct 27 '16 at 2:08

## 1 Answer

It looks like you might be confusing the span of a set of vectors, which is the set of all linear combinations of them, with their sum, which is a specific linear combination with all coefficients equal to $1$, which is, of course, another vector (that spans a line). Given two vectors $\mathbf v_1$ and $\mathbf v_2$, their span consists of all linear combinations $a_1\mathbf v_1+a_2\mathbf v_2$, which forms a plane. Their sum $\mathbf v_1+\mathbf v_2$ is in their span, of course, but there are many other vectors in it as well.

It might be instructive to rewrite the formula that you have for the (orthogonal) projection of a vector $\mathbf w$ onto the vector $\mathbf v$: $${\mathbf w\cdot\mathbf v\over\|\mathbf v\|^2}\mathbf v=\left(\mathbf w\cdot{\mathbf v\over\|\mathbf v\|}\right){\mathbf v\over\|\mathbf v\|}=\|\mathbf w\|\cos\theta\,\mathbf u,$$ where $\theta$ is the angle between the two vectors and $\mathbf u$ is a unit vector in the direction of $\mathbf v$ as in the following diagram.

The angle between a vector $\mathbf w$ and a plane is the minimum angle between $\mathbf w$ and vectors in the plane, which is achieved when those planar vectors are parallel to the projection $\pi\mathbf w$ of the vector onto the plane. (This holds for subspaces of any dimension, by the way, not just for planes.) This angle $\theta$ therefore satisfies $$\cos\theta={\|\pi\mathbf w\|\over\|\mathbf w\|}={\mathbf w\cdot\pi\mathbf w\over\|\mathbf w\|\,\|\pi\mathbf w\|}.$$

$\pi\mathbf w$ will be some linear combination of $\mathbf v_1$ and $\mathbf v_2$, of course, but the procedure given for finding this projection only works if the two spanning vectors $\mathbf v_1$ and $\mathbf v_2$ are orthogonal. If they aren’t, then you can’t find $\pi\mathbf w$ by adding up the individual projections onto the two vectors: their contributions overlap, so some parts of the net projection might get overcounted.

You can see that the sum of projections is not in general equal to the projection onto a sum by considering the projection of the vector $(1,2,3)$ onto the $x$-$y$ plane in $\mathbb R^3$. This is, of course, $(1,2,0)=(1,0,0)+(0,2,0)$, the sum of the projections onto the $x$- and $y$-axes, which are just the respective coordinates times a unit vector. If we take these unit vectors as our $\mathbf v_1$ and $\mathbf v_2$, respectively, it’s clear that the projection onto $\mathbf v_1+\mathbf v_2=(1,1,0)$ is going to be some multiple of this vector, which isn’t in the same direction as the correct answer. The resulting angle will therefore be too big. The situation is exactly the same for an arbitrary plane in $\mathbb R^3$ or a higher-dimensional space.

Algebraically, we have $$\pi\mathbf w={\mathbf w\cdot\mathbf v_1\over\|\mathbf v_1\|^2}\mathbf v_1+{\mathbf w\cdot\mathbf v_2\over\|\mathbf v_2\|^2}\mathbf v_2.\tag{1}$$ For this to be a scalar multiple of $\mathbf v_1+\mathbf v_2$, which is what you get by projecting onto this sum, the coefficients on the right-hand side must be equal, which is not going to be true in general.