# 3D Vector Help

1. Nov 18, 2006

### Karla

Hi,

If you have 2 3d vectors a and b, i.e (1,2,3) and (3,4,5) how would you calculate the angle between them?

Karla

2. Nov 18, 2006

### Hubert

Draw a right triangle.

3. Nov 18, 2006

### JasonRox

As far as I'm concerned, it is the same thing as doing it in R^2.

4. Nov 18, 2006

### robphy

5. Nov 19, 2006

### Karla

Any none cryptic ways for working this out? How can the dot product be used? I know a formula for 2d angle calculating but i cant find any info for 3d? And what is R^2 exactly?

6. Nov 19, 2006

### Hootenanny

Staff Emeritus
$\Re^2$ is (real) 2-space, or 2D. $\Re^3$ would be 3-space (3D) etc. None of the above post's were cryptic in the slightest (except the first one in which the author acknowledged was erroneous). What is the formula for calculating the cosine of the angle between any two vetors?

Last edited: Nov 19, 2006
7. Nov 19, 2006

### Karla

Sorry, cryptic probably means i don't understand them, For the angle between 2 vectors in the past i have used,

Cos Theta = (AxBx + AyBy) / |a|*|b| then I simply did Cos-1(Theta)

I attempted to alter this formula to work with 3d vectors, but it went very wrong.

Thanks for any help, i do appreciate it.

8. Nov 19, 2006

### D H

Staff Emeritus
Use the inner product:
$$\cos(\theta) = \frac{\vec a \cdot \vec b}{|\vec a|\,|\vec b|}$$

9. Nov 19, 2006

### Hootenanny

Staff Emeritus
And if we rewrite this in co-ordinate form (as yours is above) we have;

$$\cos\theta = \frac{x_{a}x_{b} + y_{a}y_{b} + z_{a}z_{b}}{\sqrt{x_{a}^{2}+y_{a}^{2}}\cdot\sqrt{x_{b}^{2}+y_{b}^{2}}}$$

Last edited: Nov 19, 2006
10. Nov 19, 2006

### Karla

Thanks very much guys :)

11. Nov 19, 2006

### Office_Shredder

Staff Emeritus
I think you have one too many vectors in there...

12. Nov 19, 2006

### Hootenanny

Staff Emeritus
Good catch, duly corrected. I think my fingers got carried away there

13. Nov 19, 2006

### robphy

$$\cos\theta = \frac{x_{a}x_{b} + y_{a}y_{b} + z_{a}z_{b}}{\sqrt{x_{a}^{2}+y_{a}^{2}+z_{a}^{2}}\cdot\sqrt{x_{b}^{2}+y_{b}^{2}+z_{b}^{2}}}$$

14. Nov 19, 2006

### Office_Shredder

Staff Emeritus
hootenanny can't catch a break :D

15. Nov 19, 2006

### robphy

While this is the best, most succinct answer, I just want to make it clear that it really is the dot-product that answers the question:
$$\cos(\theta) = \frac{\vec a \cdot \vec b}{\sqrt{\vec a \cdot \vec a} \sqrt{\vec b \cdot \vec b}}$$

16. Nov 21, 2006

### Hootenanny

Staff Emeritus
Damn it! :grumpy: Sorry guys; I'm gona have to stop working at the same time as posting! It lucky that not too many of these errors crop up in my work