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?

Question

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?

Answer 1

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

$$|\vec v|^2=a^2+ b^2$$

Answer 2

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

Answer 3

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.