Calculating Angle Between 3D Vectors: A & B

[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?

Thanks for any advice,
Karla

 

[Hubert]

Draw a right triangle.

Edit-Nevermind, I misread the question.

 

[JasonRox]

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

 

[robphy]

Dot's the answer.

 

[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 can't find any info for 3d? And what is R^2 exactly?

 

[Hootenanny]

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 can't find any info for 3d? And what is R^2 exactly?

\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?

 

[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.

 

[D H]

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

 

[Hootenanny]

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}}}

 

[Karla]

Thanks very much guys :)

 

[Office_Shredder]

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

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

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

 

[Hootenanny]

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

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

 

[robphy]

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}}}

This should read

\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}}}

 

[Office_Shredder]

hootenanny can't catch a break :D

 

[robphy]

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

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}}

That's why: Dot[/color]'s the answer.

 

[Hootenanny]

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}}

That's why: Dot[/color]'s the answer.

Damn it! 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