What is the Euclidean inner product intuition? The Euclidean inner product $<x,y>$ of the vectors $x , y \in \mathbb{R}^n$ is defined by:
$\langle x,y\rangle = x_1y_1 + x_2y_2 + x_3y_3 + ... + x_ny_n$
I am unable to find the intuition behind this? Why do we need this inner product? What happens to the vectors under equation and what does the scalar answer tell us?
P.S. I have already seareched it, most the articles take it into geometric terms and introduce $\cos\theta$. Can anybody help me visualize thing without angular stuff?
Edit: Question addressed here uses the angle to explain geometric dot product. I am wondering if there exist some explanation for dot product in Vector Space.
 A: You are probably familiar with the following formula involving
a vector $x = (x_1, \ldots, x_n) \in \mathbb R^n$
and $n$ constants $a_1, \ldots, a_n$ that are not all zero: 
$$
a_1 x_1 + a_2 x_2 + \cdots + a_n x_n = 0.
$$
This formula describes an $(n-1)$-dimensional hyperplane in $\mathbb R^n.$
Let's call that hyperplane $A.$ That is, any vector $x$ that satisfies the formula lies in the hyperplane $A.$
If we define a vector $a = (a_1, \ldots, a_n),$
then another way to write the formula is
$$  \langle x, a\rangle = 0. $$
Now if we choose an arbitrary vector $x \in \mathbb R^n,$
it may happen that $\langle x, a\rangle \neq 0.$
If we have two such vectors, let's say $x'$ and $x'',$
such that $\langle x', a\rangle > 0$ and $\langle x'', a\rangle > 0,$
then $x'$ and $x''$ will be on the same side of the hyperplane $A.$
But if $\langle x', a\rangle < 0 < \langle x'', a\rangle$ then the vectors are on opposite sides.
For a given vector $x,$ suppose you find a vector $y$ such that
$x - \langle x, a\rangle y$ is in the hyperplane $A.$
Then $x' - \langle x', a\rangle y$ also is in the hyperplane $A$
for any other vector $x'.$
That is, $\langle x, a\rangle$ tells you how many times the length of $y$
you have to travel in the direction of $y$ to get from $x$ to the hyperplane,
or in other words, $\langle x', a\rangle$ is a kind of measurement of the distance of $x$ from the hyperplane (measured in some particular units in some particular direction).
All of this works without any "angles" (unless you consider "parallel" to be a "zero angle").
You don't even need 
$(x_1, \ldots, x_n)$ to be coordinates over an orthonormal basis,
although if the basis is orthonormal then other nice results follow.
For example, 
$$\sqrt{\langle x, x\rangle} = \sqrt{x_1^2 + x_2^2 + \cdots + x_n^2},$$
which is the length of $x$ (according to the Pythagorean Theorem)
if the basis of $(x_1, \ldots, x_n)$ is orthonormal.
Admittedly, to speak of "normals" one must have a concept of things being perpendicular, which starts to sound like we're dealing with angles again.
But they're right angles, which are especially simple to work with.
A: $<x,y>$ is positive if and only if the angle between $x$ and $y$ is less than 90 degrees. $<x,y>$ is sort of like a measure of how pointed in the same direction $x$ and $y$ are. 
The geometry of least squares problems and separating hyperplanes is useful for gaining intuition about inner products.
A: The intuition is gathered from the following, given by the Law of Cosines
$$\| a -b\|^{2} = \| a\|^{2} + \| b\|^{2} - 2\| a\|^{2} \| b\|^{2}\cos(\theta) $$
$$ \| a-b\|^{2}  = (a-b)^{2}\cdot(a-b)^{2} $$
$$ \| a-b\|^{2}  =  a\cdot a - a\cdot b - b\cdot a + b \cdot b $$
useful to note that $a \cdot a = \| a\| $
$$ \| a-b\|^{2}  =  \|a \| - a\cdot b - b\cdot a + \| b\|$$
further more we have $ a \cdot b = b\cdot a $
$$ \| a-b\|^{2}  =  \|a \| - 2 a\cdot b + \| b\|$$
so we have
$$\| a -b\|^{2} = \| a\|^{2} + \| b\|^{2} - 2\| a\|^{2} \| b\|^{2}\cos(\theta) $$
$$ \| a\|^{2} - 2 a \cdot b + \| b\|^{2} = \|a\|^{2} + \|b\|^{2} - 2\|a\| \|b\| \cos(\theta) - 2 a \cdot b = -2 \|a \| \|b\| \cos(\theta) $$
yielding finally
$$a \cdot b  = \| a \| \| b\| \cos(\theta) $$
Now suppose that 
$$ \|a\| = \|b\| =1 $$
$$ a \cdot b = 1 \cdot 1 \cos(\theta)  = \cos(\theta)$$
ok when is this negative. We have a unit circle here for vectors
$$ a \cdot b = \cos( \theta) < 0 \implies \frac{n\pi}{2} < \theta < \frac{3n\pi}{2} $$
that is we are in one part of the circle. However, it goes on forever.
Note from lamar
