Inuition behind orthogonality of Euclidean vectors Let  $a=(x_1,x_2,x_3)$ and $b=(y_1,y_2,y_3)$ be Euclidean vectors.
These vectors are said to be orthogonal iff $\displaystyle \sum_{i=1}^n x_iy_i=0$.
I have a hard time understanding how this formalisation captures the intuitive notion of orthogonality. How does it?
 A: 
We proceed to develop an intuitive understanding of the inner product by appealing to a geometric picture.  We then proceed to apply straightforward vector algebra.


First, we draw two vectors $\vec A$ and $\vec B$ that originate at the origin.  The vector $\vec A-\vec B$ is the vector from $\vec B$ to $\vec A$.  Therefore, the three vectors $\vec A$, $\vec B$, and $\vec A-\vec B$ form a triangle.
From the Law of Cosines, we have
$$|\vec A|^2+|\vec B|^2-2|\vec A|\,|\vec B|\cos(\theta)=|\vec A-\vec B|^2 \tag 1$$
where $\theta\in [0,\pi]$ is the angle between $\vec A$ and $\vec B$.  
But note that
$$\begin{align}
|\vec A-\vec B|^2=&(A_1-B_1)^2+(A_2-B_2)^2+(A_3-B_3)^2\\\\
&=\color{red}{(A_1^2+A_2^2+A_3^2)}+\color{green}{(B_1^2+B_2^2+B_3^2)}-2\color{blue}{(A_1B_1+A_2B_2+A_3B_3)}\\\\
&=\color{red}{|\vec A|^2}+\color{green}{|\vec B|^2}-2\color{blue}{(\vec A\cdot \vec B)}\tag2
\end{align}$$
Using $(1)$ in $(2)$ yields
$$\vec A\cdot \vec B=|\vec A|\,|\vec B|\cos(\theta)\tag 3$$

Finally, note from $(3)$ that if $\vec A\cdot \vec B=0$, then $\cos(\theta)=0$, so that $\theta =\pi/2$,  and thus $\vec A$ and $\vec B$ are orthogonal.

A: The crucial insight comes (without any trigonometry) from the plane. Recall that two lines are orthogonal if and only if their slopes are negative reciprocals. (You can see this, if you want, by considering the right triangles formed by the vectors $\mathbf x=(a,b)$ and $\mathbf y=(-b,a)$. As you can see, the right angle makes the triangles congruent. 

But then we observe that $\mathbf x\cdot\mathbf y = a(-b)+b(a)=0.$ 
Algebraically, we can deduce that dot product $0$ implies perpendicularity. Algebraic properties of the dot product tell us that $\|\mathbf x+\mathbf y\|^2 = \|\mathbf x\|^2 + \mathbf y\|^2 \iff \mathbf x\cdot\mathbf y=0$. The converse of the Pythagorean Theorem will then tell us that this means $\mathbf x$ and $\mathbf y$ are perpendicular (orthogonal).
A: The dot product
$$
\displaystyle \sum_{i=1}^n x_iy_i
$$
can be shown to be equal to
$$
\| x\| \| y\| \cos\theta
$$
where $\theta$ is the angle between $x$ and $y$. Then it becomes clear that if the dot product is $0$ then $\theta = \pi/2$, which means, geometrically, that $x$ and $y$ are orthogonal (perpendicular).
A: You have used $$\vec{a} \cdot \vec{b} = \sum_{i=1}^n x_i y_i$$
but $$\vec{a} \cdot \vec{b} = \left\| \vec{a} \right\| \left\|\vec{b} \right\| \cos \theta$$
where $\theta$ is the angles between $\vec{a}$ and $\vec{b}$, so $\vec{a} \cdot \vec{b} = 0$ either when $\vec{a} = \vec{0}$, or $\vec{b} = \vec{0}$ or $\theta = \pi/2 \pm n\pi$...
