# Dot product with 3 dimensions, confused on concept, easy question i think

1. Oct 5, 2005

### mr_coffee

Hello everyone! I'm confused on what i'm suppose to do here, I think i might got it though but i need to make sure...
Here is the problem and my work:
http://show.imagehosting.us/show/764032/0/nouser_764/T0_-1_764032.jpg
he let r(t) = f(t) i + g(t) j + h(t) k. So if i multiply this by itself, won't that make the vectors go away? because isn't (i)(i) = (j)(j) = (k)(k) = 1?
Thanks!

Last edited by a moderator: Apr 21, 2017
2. Oct 5, 2005

### whozum

In the very first line, g(t) needs a j unit vector.

For part (b), remember to apply the chain rule.

3. Oct 5, 2005

### big man

Well if it's the same vector, then essentially it's the dot product of two vectors that are parellel, that is, the angle between them is 0. So it will be the magnitude of the vector squared. That's how I see it anyway...

4. Oct 5, 2005

### mr_coffee

Thanks for the replies, I see big man, that makes sense but he isn't asking what |r(t) dot r(t)| he is asking r(t) dot r(t). Also Whozum, Thanks for picking that up in a, but is that right, the dot product of the same vector, will the unit vectors just be all one and ur just left with the functions of f(t) + g(t) + h(t)? Somthing seems odd there...also why would the chain rule apply to part b? Don't you apply the chain rule when you have somthing like, f(g(x))? or (x-3x^3)^(2)?

5. Oct 5, 2005

### Pyrrhus

For the first part,

$$\vec{r}(t) \cdot \vec{r}(t) = |\vec{r}(t)|^{2} = c^2$$

for the 2nd part

$$\frac{d}{dt} \vec{r}(t) \cdot \vec{r}(t) = \vec{r}'(t) \cdot \vec{r}(t) + \vec{r}(t) \cdot \vec{r}'(t) = 2 \vec{r}'(t) \cdot \vec{r}(t) = 0$$

$$\vec{r}'(t) \cdot \vec{r}(t) = 0$$

which basicly means both vectors are orthogonal.

6. Oct 5, 2005

### mr_coffee

Thanks cyclovenom!! right when u posted that I Just found the exact same solution in the book :)