1. ## vectors

$\displaystyle \bold{r}(t) = t^{2} \bold{i} - 4t \bold{j} -t^{2} \bold{k}$

$\displaystyle \bold{v}(t) = 2t \bold{i} -4 \bold{j} -2t \bold{k}$

$\displaystyle r(t) = \sqrt{2t^{4} + 16t^{2}}$

$\displaystyle v(t) = \sqrt{8t^{2} + 16}$.

Find the cosine of the angle between $\displaystyle \bold{r}$ and $\displaystyle \bold{v}$. For what values of $\displaystyle t$ is $\displaystyle \bold{r}$ perpendicular to $\displaystyle \bold{v}$? Parallel?

So $\displaystyle \cos \theta = \frac{ \bold{r}(t) \cdot \bold{v}(t)}{r(t) \cdot v(t)}$

$\displaystyle \cos \theta = \frac{4t^{3} + 16t}{2t \sqrt{4t^{4} + 40t^{2} + 64}}$

So $\displaystyle t = 0$ or $\displaystyle t = \pm 2i$ for $\displaystyle \bold{r} \perp \bold{v}$.

But for $\displaystyle \bold{r} \parallel \bold{v}$ I get $\displaystyle t = 0$. But this is undefined. So there are no values of $\displaystyle t$ for which $\displaystyle \bold{r}$ and $\displaystyle \bold{v}$ are parallel?

2. Originally Posted by heathrowjohnny
$\displaystyle \bold{r}(t) = t^{2} \bold{i} - 4t \bold{j} -t^{2} \bold{k}$

$\displaystyle \bold{v}(t) = 2t \bold{i} -4 \bold{j} -2t \bold{k}$

$\displaystyle r(t) = \sqrt{2t^{4} + 16t^{2}}$

$\displaystyle v(t) = \sqrt{8t^{2} + 16}$.

Find the cosine of the angle between $\displaystyle \bold{r}$ and $\displaystyle \bold{v}$. For what values of $\displaystyle t$ is $\displaystyle \bold{r}$ perpendicular to $\displaystyle \bold{v}$? Parallel?

So $\displaystyle \cos \theta = \frac{ \bold{r}(t) \cdot \bold{v}(t)}{r(t) \cdot v(t)}$

$\displaystyle \cos \theta = \frac{4t^{3} + 16t}{2t \sqrt{4t^{4} + 40t^{2} + 64}}$

So $\displaystyle t = 0$ or $\displaystyle t = \pm 2i$ for $\displaystyle \bold{r} \perp \bold{v}$.

But for $\displaystyle \bold{r} \parallel \bold{v}$ I get $\displaystyle t = 0$. But this is undefined. So there are no values of $\displaystyle t$ for which $\displaystyle \bold{r}$ and $\displaystyle \bold{v}$ are parallel?
To me all your calculations seem to be OK. You only forgot to cancel the t when calculating the $\displaystyle \cos(\theta)$.

If t = 0 then $\displaystyle \vec r = 0$ and $\displaystyle \vec v = (0, -4, 0)$. So $\displaystyle \vec r$ is the nullvector which is parallel and perpendicular to any other vector.

3. Originally Posted by heathrowjohnny
$\displaystyle \bold{r}(t) = t^{2} \bold{i} - 4t \bold{j} -t^{2} \bold{k}$

$\displaystyle \bold{v}(t) = 2t \bold{i} -4 \bold{j} -2t \bold{k}$

$\displaystyle r(t) = \sqrt{2t^{4} + 16t^{2}}$

$\displaystyle v(t) = \sqrt{8t^{2} + 16}$.

Find the cosine of the angle between $\displaystyle \bold{r}$ and $\displaystyle \bold{v}$. For what values of $\displaystyle t$ is $\displaystyle \bold{r}$ perpendicular to $\displaystyle \bold{v}$? Parallel?

So $\displaystyle \cos \theta = \frac{ \bold{r}(t) \cdot \bold{v}(t)}{r(t) \cdot v(t)}$

$\displaystyle \cos \theta = \frac{4t^{3} + 16t}{2t \sqrt{4t^{4} + 40t^{2} + 64}}$

So $\displaystyle t = 0$ or $\displaystyle t = \pm 2i$ for $\displaystyle \bold{r} \perp \bold{v}$.

But for $\displaystyle \bold{r} \parallel \bold{v}$ I get $\displaystyle t = 0$. But this is undefined. So there are no values of $\displaystyle t$ for which $\displaystyle \bold{r}$ and $\displaystyle \bold{v}$ are parallel?
The vectors are perpendicular when t = 0.

They will be parallel when $\displaystyle \bold{r}(t) = \lambda \bold{v}(t)$, where $\displaystyle \lambda$ is a non-zero scalar.

So you need the following three equations to be satisfied:

$\displaystyle t^2 = \lambda 2t$ ... (1).

$\displaystyle -4t = -4 \lambda$ ... (2).

$\displaystyle -t^2 = -\lambda 2t$ ... (3) (and this is equivalent to (1)).

From (2): $\displaystyle t = \lambda$. Sub this into (1):

$\displaystyle \lambda^2 = 2\lambda^2 \Rightarrow \lambda = 0$ which is no good since $\displaystyle \lambda$ is a non-zero scalar.

So there is NO value of t such that the vectors are parallel,

Edit: unless the trivial solution of both vectors being the zero vector is granted.

4. But doesn't this imply that $\displaystyle \bold{r}$ is the null vector as earboth said?

5. Originally Posted by heathrowjohnny
But doesn't this imply that $\displaystyle \bold{r}$ is the null vector as earboth said?
Yes. Unfortunately.

I don't like it. But that's the case.