Why would the norm of these vectors be 1?

Let a Cartesian coordinate system $uOx$ coincides with a vertical plane so that $Ou$ is the horizontal axis and $Ox$ is the axis oriented vertically upwards (see Fig. 1). We are looking for the smooth curves connecting two fixed points $A = (u_0, x_0)$ and $B = (u_1, x_1)$ with $u_0, u_1 \geq 0$, $u_0 \neq u_1$, and $0 \leq x_1 < x_0$ so that a bead $M$ sliding with initial speed $v_0 \geq 0$ downward from $A$ and accelerated by gravity will slip with a nonlinear kinetic friction to $B$ in the least time $T$. We suppose for the beginning that there exists a solution represented by a sufficiently smooth curve $\gamma$ and an arbitrary point $M$ lying on $\gamma$. Let $\mathbf{\tau}$ be the unit tangent vector to $\gamma$ and $M$, $\mathbf{v}$ be the velocity vector of the bead $M$, $\mathbf{\nu}$ be the unit normal vector to $\gamma$ at $M$, $\mathbf{g}$ be the acceleration gravity vector, $\mathbf{f}_\mu$ be the friction force, $\mathbf{f}_\nu$ be the normal component of the constraint reaction force, $\theta$ be the slope angle of the tangent, and $\mathbf{i}$ and $\mathbf{j}$ be the unit vectors of the Cartesian coordinate system $uOx$.

The position of a particle $M$ relative to the coordinate system $uOx$ is determined by the position vector $\mathbf{r}$ (see Fig. 1). The particle $M$ is moving from $A$ to $B$, so its position vector $\mathbf{r}$ is a function of time $t$, i.e.,....

Hi, I am having a lot of problems understanding this excerpt. The quote is an introduction, and later on they said that $\lVert\mathbf{\tau}\rVert=\lVert\mathbf{\nu}\rVert=\lVert\mathbf i\rVert=\lVert\mathbf j\rVert=1$. Why? Also, why is the velocity vector $\mathbf v=-v \cos\theta\ \mathbf i -v\sin\theta\ \mathbf j$?

• Do you know what a unit vector is? – Kyle Kanos Jul 8 '15 at 13:52
• Yes I do @KyleKanos I know very basic vectors, but not sure of the concepts of norm and inner products – WilliamKin Jul 8 '15 at 13:56
• What math courses have you had? Calculus? Linear algebra? – Kyle Kanos Jul 8 '15 at 14:12
• @KyleKanos I study calculus, until the calculus of variations. However, I am fairly new to linear algebra. Do you mind explaining the above equations? I will try to understand thank you! – WilliamKin Jul 8 '15 at 14:14
• @KyleKanos in my experience, calculus is typically a high school subject whereas linear algebra isn't covered until college. – David Z Jul 8 '15 at 14:27

The introductory quote defines $\tau$ and $\nu$ to be unit vectors. That means that their magnitude is 1. $\mathbb{i}$ and $\mathbb{j}$ are unit vectors along $x$ and $y$ axes. That's a standard notation. Some of the comments are stressing the fact that the basic understanding of vectors is missing. I have a feeling that you've jumped into a 'complicated' vector resolution problem directly without doing some basic exercises.
That is the reason for your difficulty in understanding $\mathbf{v}$. First, read this page on how to resolve vectors. It's simple, but you've to be sure of what you're doing. Then, read this pdf on circular motion. Notice there how $\hat{r}$ is written. I believe, you will get your answer.
A note: if you extend the dotted line, towards left of $M$, notice that the vector $\mathbf{v}$ makes an angle $\theta$ with it on the left side. Use this along with resolution of vectors, and you'll have your answer.