When we find the norm of a vector, why don't we square $\vec{i}$ and $\vec{j}$? And what would it even mean to square them? It's been too long since I learned linear algebra, so apologies for the basic questions. Here's a snippet from my textbook: 

Why is it that in the calculation of $||\vec{v}||$ we don't square the $\vec{i}$ and $\vec{j}$ terms even though they're in the definition of $\vec{v}$ right above? 
Also, what does it mean to "square" a unit vector (or a vector in general)? I'm reading conflicting things online. Does $(\vec{i})^2$ denote calculating the dot product $<\vec{i},\vec{i}>$ and getting $1$ as the answer? Or would $(\vec{i})^2$ mean that we take the projection of $\vec{i}$ onto itself?
 A: In general we have that
$$|\vec v|^2=\vec v\cdot \vec v=(a\vec i+b\vec j)\cdot (a\vec i+b\vec j)=\\=a^2(\vec i\cdot \vec i) +2ab(\vec i\cdot \vec j) +b^2(\vec j\cdot \vec j)=a^2+ b^2$$
therefore we can use 
$$|\vec v|^2=a^2+ b^2$$
A: There is no such thing as squaring a vector. You can take dot product or cross product, but still that has little meaning for magnitude (which is a scalar representing the length of a vector).
So turns out the only inputs we need are the scalars representing the distance in each coordinate. Which will give us the magnitude,a scalar representing the "joint" distance of all coordinates
A: A vector squared is its magnitude squared (or its dot product with itself), by definition. Together with associativity of multiplication (and other "obvious" properties), this produces a very interesting and useful system called geometric algebra.
In this system, indeed $\vec i\,^2=\vec j\,^2=1$, so they can be ignored. But there are other terms in $\vec v\,^2$ :
$$\vec v=\frac{dx}{dt}\vec i+\frac{dy}{dt}\vec j$$
$$\vec v\,^2=\left(\frac{dx}{dt}\vec i+\frac{dy}{dt}\vec j\right)\left(\frac{dx}{dt}\vec i+\frac{dy}{dt}\vec j\right)$$
$$=\frac{dx}{dt}\;\vec i\left(\frac{dx}{dt}\vec i+\frac{dy}{dt}\vec j\right)+\frac{dy}{dt}\;\vec j\left(\frac{dx}{dt}\vec i+\frac{dy}{dt}\vec j\right)$$
$$=\left(\frac{dx}{dt}\right)^2\;\vec i\,^2+\left(\frac{dx}{dt}\frac{dy}{dt}\right)(\vec i\,\vec j+\vec j\,\vec i)+\left(\frac{dy}{dt}\right)^2\;\vec j\,^2$$
$$=\lVert\vec v\rVert^2=\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2$$
so we must have $\vec i\,\vec j=-\vec j\,\vec i$. This product of perpendicular vectors is called a bivector. But you can read about that elsewhere; I don't want to go off-topic.
