Why is it called "Orthogonal Projection"? Why not just "Projection"? Right now, we are learning decomposing vectors, but something I don't understand is the names given to this stuff

For instance, in the text, the parallel component of y is said to be the orthogonal projection of y onto u. This makes no sense to me. Why is the word "orthogonal" even in there in the first place? I think I understand why they use "orthogonal" for z, but it makes no sense to me when they could just call it "orthogonal"
 A: There are many ways to "project" onto a subspace. This particular projection yields an orthogonal decomposition.
Given vectors $v$ and $w$ ($w \not=0$), $a = \mathrm{proj}_w(v) = \frac{v \;\cdot\; w}{|w|^2}w$ and $b = v - a = v - \mathrm{proj}_w(v)$. We have that 


*

*$v=a+b$

*$a$ is parallel with $w$

*$a$ and $b$ are orthogonal (just check that $a \;\cdot\; b=0$).


There are other "projections" for which $v=a+b$ and $a$ is parallel with $w$. For example: Let $a=w$ and $b=v-w$. Then $v=w+(v-w)=a+b$ and $a=w$ is parallel with itself (i.e $w$). But in general $v-w$ and $w$ are not orthogonal.
A: As Bill's answer explains, we can decompose every vector in the original space by using the projection map. This lends to an intuitive geometric interpretation of orthogonal projections.
If $p:V\to W$ is an $\color{Green}{orthogonal}$ projection down to a subspace, the fibers (pre-images) of every point $w\in W$ is perpendicular to the base (the subspace $W$). With $\dim V=2$ and $\dim W=1$:
$\hskip 0.4in$ 
The base (black line) and fibers (gray lines) can in general be viewed as higher-dimensional planes.
A: Consider $$\begin{pmatrix} 1&1\\ 0&0\end{pmatrix}$$ it projects all vectors to the $x$-axis. But it doesn't project vectors onto the $x$-axis orthogonally. The vector $(0,1)^T$ is not sent to its orthogonal projection $(0,0)^T$ but is sent to $(1,0)^T$. Vectors are "diagonally" projected onto the $x$-axis.
A: We can actually project with respect to any vector in a basis, as follows:
If $v_1,\ldots, v_n$ is a basis for your vector space, then the projection of vector $a_1 v_1 + \ldots + a_n v_n$ onto $a_n$ with respect to this basis is $a_n v_n$. Note that the projection depends on the choice of basis.
If you were to take an orthogonal basis containing $u$, then the definition of orthogonal projection would coincide with the above, more general, definition of projection.
A: Because line along which your original point travel to projective plane is orthogonal to projective plane. In this context synonym for orthogonal is perpendicular, which I think is more common and better undertstood word.
A: I took linear algebra with this same textbook and went through the same dilemma. I decided that the word "orthogonal" in orthogonal projection is referring to the way some vector v is being projected onto a subspace W. If you imagine the projection occurring as a drawn out process, with the point represented by v moving directly toward W on the shortest path, then the motion of the point is orthogonal to W. This would be in contrast with a "non-orthogonal," or "diagonal" projection, in which the projection of the point is not orthogonal to W. Hope this helps—it worked for me!
A: Consider the example of a vector in 3-D projected onto a 2-D plane (let's call it base).
For the projection to be orthogonal, the vector and its projection onto the base must lie in a plane perpendicular to the base i.e if you imagine the vector to be a series of points, each of these should fall perpendicularly onto the base as shown in the pic below (sorry for the bad drawing).
Orthogonal Projection
Now imagine that the plane(and the vector) are rotated slightly to the left, both the vector and its projection still lie on the plane, but the plane is no longer perpendicular to the base. The projection is not orthogonal in this case.
Hope this helps ...
