Is this a Norm? How is the following formula calculated, assuming $a$ and $b$ are $n$-dimensional vectors?
$\parallel \overrightarrow{a} - \overrightarrow{b}\parallel^2$
 A: as $(\vec{a}-\vec{b}).(\vec{a}-\vec{b})$
A: The standard (pythagorian) norm for vectors in $\mathbb{R}^n$ is defined as
$$||\vec{x}|| \equiv \sqrt{\vec{x}\cdot\vec{x}}$$
where the dot-product of two vectors $\vec{x} = (x_1,x_2,\ldots,x_n)$ and $\vec{y} = (y_1,y_2,\ldots,y_n)$ is defined as 
$$\vec{x}\cdot\vec{y} \equiv x_1y_1 + x_2y_2 + \ldots + x_n y_n$$

Applying the definitions above with $\vec{x} = \vec{a}-\vec{b}$ gives us
$$||\vec{a}-\vec{b}||^2 = (a_1-b_1)^2 + (a_2-b_2)^2 + \ldots + (a_n-b_n)^2$$
where $a_i,b_i$ are the $i$'th component of the vector $\vec{a}$ and $\vec{b}$ respectively.
A: Define a vector to be the difference:
$$ \mathbf{c} = \mathbf{a} - \mathbf{b}.$$
And the take the dot product of the difference, $\mathbf{c} = (c_1, c_2, c_3, \dots, c_n)$, with itself:
$$
\| \mathbf{a} - \mathbf{b} \|^{2} = \mathbf{c}\cdot\mathbf{c}, \\
\qquad = c_{1}^{2} + c_{2}^{2} + c_{3}^{2} + \cdots + c_n^{2}.
$$
A: If your question is
$$\text{ Is } d(a,b) = \| a - b\|^2 \text{ a norm on } \mathbb R^n?$$
then the answer is no as the triangle inequality can be violated. For example, let $\vec e$ be any unit vector. Then
$$d(0\vec e,2\vec e) = 2^2 \color{red}{>} 1^2 + 1^2 = d(0\vec e,1\vec e) + d(1\vec e,2\vec e)$$
