Calculating Angles Between Vectors: Formula and Help for Your Exam

[franky2727]

how is this done? say angles between 2i -3j -k and 4i+2k-3j got an exam coming up soon and can't find the formula for it anywhere help please?

 

[franky2727]

and also is the cross product just the determinant?

 

[Dick]

A.B=|A||B|cos(t) (dot product). And you can USE a determinant to find the cross product. I wouldn't think of it as being primarily a determinant, it's a vector.

 

[franky2727]

so angles between is just |A||B|cos(t) and cross product is just using the det so for A=2i-3j-k and B=4i+2k-3j i would have angle between of root 14 root 29 cos(T) would that just be the answer? still confused on that one and for the cross product would the answer be 12i-2j+16k?

 

[Dick]

Yes. A.B=sqrt(14)*sqrt(29)*cos(T). So now if you can figure out A.B then you can find cos(T) and then T. I don't get what you get for the cross product. For example for the i component I figure (-3)(2)-(-3)(-1)=(-9).

 

[franky2727]

not a clue where that determinants coming from, how do you do the determinant of a 3 by 2 matrix

 

[Dick]

not a clue where that determinants coming from, how do you do the determinant of a 3 by 2 matrix

There is no such thing as the determinant of a 3x2 matrix. Check out http://en.wikipedia.org/wiki/Cross_product

 

[franky2727]

ahah got ya! right now just give me 5 minutes to get stuck on finding A.B and i'l be right back!

 

[franky2727]

sorry i still have no clue where your plucking them numbers from for the determinant isn't the determinant AD-BC but on your answer you seen to have done AB-CD

 

[Dick]

What are the second and third rows of your determinant? I can guess what you are doing wrong. Notice the components in vector B are written in the order ikj. Not ijk. I was wondering if it's a typo, but you repeated it twice, so I'm guessing that's how it's written in the problem.

 

[franky2727]

ye just me being dum that's a typo ment to be ijk so it is AD-BC ye?

 

[Dick]

I get [2, - 3, - 1] x [4, 2, - 3]=[11, 2, 16], if that's the problem you are trying to solve...

 

[roam]

Well, if the two vectors u and v that are nonzero (in [tex]R^{2}[/tex] or [tex]R^{3}[/tex]) the angle between them is given by the formula;

[tex]\theta = cos^{-1}\left(\frac{u.v}{\left\|u\right\| \left\|v\right\|}\right)[/tex]

In your question you have u = [tex]\left(2, - 3, - 1\right)[/tex] & v = [tex]\left(4, 2, -3)[/tex]

But you have to put your equation system in the form of an argumented matrix!