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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: $\vec{B} = \vec{T} \times \vec{N}$

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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.

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