1. ## messy vectors

$\displaystyle \bold{r}(t) = (\cosh t) \bold{i} + (\sinh t) \bold{j} + t \bold{k}$

$\displaystyle \bold{v}(t) = (\sinh t) \bold{i} + (\cosh t) \bold{j} + \bold{k}$

So we want to do the same thing: find the cosine of the angle between $\displaystyle \bold{r}$ and $\displaystyle \bold{v}$.

So: $\displaystyle \cos \theta = \frac{2e^{2t} + 2e^{-2t} + t}{\sqrt{4e^{4t} + 8 + 2e^{2t} + 4e^{-4t} + 2e^{-2t} + 2t^{2}e^{2t} + 2t^{2}e^{-2t} + t^{2}}}$

But the numerator is always positive, so $\displaystyle \bold{r}$ is never perpendicular to $\displaystyle v$.

For $\displaystyle \bold{r}$ to be parallel to $\displaystyle \bold{v}$, both the numerator and denominator have to equal each other. This is messy to solve. But I don't think there are values of $\displaystyle t$ for which $\displaystyle \bold{r}$ is parallel to $\displaystyle \bold{v}$. Is this correct?

2. Originally Posted by heathrowjohnny
$\displaystyle \bold{r}(t) = (\cosh t) \bold{i} + (\sinh t) \bold{j} + t \bold{k}$

$\displaystyle \bold{v}(t) = (\sinh t) \bold{i} + (\cosh t) \bold{j} + \bold{k}$

So we want to do the same thing: find the cosine of the angle between $\displaystyle \bold{r}$ and $\displaystyle \bold{v}$.

So: $\displaystyle \cos \theta = \frac{2e^{2t} + 2e^{-2t} + t}{\sqrt{4e^{4t} + 8 + 2e^{2t} + 4e^{-4t} + 2e^{-2t} + 2t^{2}e^{2t} + 2t^{2}e^{-2t} + t^{2}}}$

But the numerator is always positive, so $\displaystyle \bold{r}$ is never perpendicular to $\displaystyle v$.

For $\displaystyle \bold{r}$ to be parallel to $\displaystyle \bold{v}$, both the numerator and denominator have to equal each other. This is messy to solve. But I don't think there are values of $\displaystyle t$ for which $\displaystyle \bold{r}$ is parallel to $\displaystyle \bold{v}$. Is this correct?
Use the idea from my other post - the scalar business:

$\displaystyle \cosh t = \lambda \sinh t$ .... (1)

$\displaystyle \sinh t = \lambda \cosh t$ .... (2)

$\displaystyle t = \lambda$ ... (3)

Sub (3) into (1) and (2):

$\displaystyle \cosh t = t \sinh t$ .... (1')

$\displaystyle \sinh t = t \cosh t$ .... (2')

Sub (2') into (1'):

$\displaystyle \cosh t = t^2 \cosh t$ which has $\displaystyle t = \pm 1$ as its only solutions.