# Frenet Serret: find equation for tangent, normal, and binormal at t=1

given the following:

position vector:

$$\vec{r}(t) = t \hat{\text{i}} + t^2 \hat{\text{i}} + \frac{2}{3} t^3 \hat{\text{k}}$$

unit tangent:

$$\vec{T}(t) = \bigg(\frac{1}{1+t^2}\bigg)\bigg(\hat{\text{i}} + 2 t \hat{\text{j}} + 2t^2\hat{\text{k}}\bigg)$$

and principle norm:

$$\vec{N}(t) = \bigg(\frac{1}{1+t^2}\bigg) \bigg(-2t\hat{\text{i}} + (1-2t^2)\hat{\text{j}} + 2t\hat{\text{k}}\bigg)$$

and binormal vector:

$$\vec{B}(t) = \bigg(\frac{1}{1+t^2}\bigg)\bigg(2t^2\hat{\text{i}} - 2t\hat{\text{j}} + \hat{\text{k}}\bigg)$$

FIND the equations at point "t=1" for the: (a) tangent (b) principle norm, and (c) binomial to the curve.

Let $$\vec{r}_0$$, $$\vec{T}_0$$, $$\vec{N}_0$$, and $$\vec{B}_0$$ denote the position, tangent, principal normal, and binomial vectors at the required point. Then:

$$\vec{r}_0 = \hat{\text{i}} + \hat{\text{j}} + \frac{2}{3}\hat{\text{k}}$$

$$\vec{T}_0 = \frac{1}{3}\bigg(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\bigg)$$

$$\vec{N}_0 = \frac{1}{3}\bigg(-2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\bigg)$$

$$\vec{B}_0 = \frac{1}{3}\bigg(2 \hat{\text{i}} -2\hat{\text{j}}+\hat{\text{k}} \bigg)$$

i'm ok with everything up to this point....here's where i start to get confused:

if $$\vec{A}$$ denotes a given vector while $$\vec{r}_0$$ and $$\vec{r}$$ denote, respectively, the position vectors of the initial point and an arbitary point of $$\vec{A}$$, then $$\vec{r} - \vec{r}_0$$ is parallel to $$\vec{A}$$ and so the equation of $$\vec{A}$$ is $$(\vec{r} - \vec{r}_0) \times \vec{A} = 0$$. (no problem with this part.) then:

$$\begin{matrix} \text{equation of tangent is} && (\vec{r} - \vec{r}_0) \times \vec{T}_0 = 0 \\ \text{equation of principle normal is} && (\vec{r} - \vec{r}_0) \times \vec{N}_0 = 0 \\ \text{equation of binomial is} && (\vec{r} - \vec{r}_0) \times \vec{B}_0 = 0 \\ \end{matrix}$$

How did they come up with these three equations?

I don't get it...

why would each of the 3 frenet-serret unit vectors be parallel to $$(\vec{r} - \vec{r}_0)$$?

I thought the binormal $$\vec{B}$$, tangent $$\vec{T}$$, and principle normal $$\vec{N}$$ form a right handed coordinate system at an arbitrary point according to: $$\vec{B} = \vec{T} \times \vec{N}$$

These equations are equations of 3 lines through $$\vec{r}_0$$ parallel to $$\vec{T}_0$$, to $$\vec{N}_0$$ and to $$\vec{B}_0$$; these lines are the "tangent line", the "principal normal line" and the "binormal line" to the curve at $$\vec{r}_0$$. No point except $$\vec{r}_0$$ lies on all three of these lines (they are indeed pairwise orthogonal, as you say); the $$\vec{r}$$ is the variable, aka $$\vec{r}=(x,y,z)$$ in all 3 equations, but you are not trying to solve all of these equations simultaneously.