# 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.