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.