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A little concept problem with those vectors

  1. Jun 13, 2006 #1
    A little concept problem with those vectors :)

    Ok, so I got two objects which somehow interacts to each other, or I have two electomagnetic waves which interact in that same way.
    I know the angle between those two vectors, and I know lets say the result vector of that impact, the module of this isn't important for me(!), and I also know that if f,g are two vectors
    So if I know, or I need to know whereto the result is pointing, how do I proceed.

    If The Above Stantment ^^^ Are a missunderstuding of a primary concept, I wouldn't be surprise :), also excuse my kinda bad english, I might have done some mistaqes up there :).

  2. jcsd
  3. Jun 13, 2006 #2
    the magnitude of the vector product of 2 vectors is given by

    [tex]\left|u\times v\right|=\left|u\right|\left|v\right|\sin\theta[/tex]

    where theta is the direction.

    also the vector product in 2 dimensions can be given as


    hope this helps

    Edit: Actually theta is the angle between u and v.
    Last edited: Jun 13, 2006
  4. Jun 13, 2006 #3
    ooops messed up on the latex command

    it should say

    [tex]\left|u\times v\right|=\left|u\right|\left|v\right|\sin\theta[/tex]:surprised
  5. Jun 13, 2006 #4


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    Staff: Mentor

    I don't think I understand the OP's question. Are you asking about adding vectors, taking their dot product, or taking their cross product as NEWO shows?

    Also, EM waves do not generally interact. They certainly don't bounce off of each other or something.

    When you add or dot or cross vectors, you do that based on the coordinate system that they are represented in. If you are using rectangular coordinates, then you add the (x,y,z) components of each vector to get the resultant vector's (x,y,z) values. If you want the dot product in a rectangular coordinate system, you use the forumula that you posted, and the result is a scalar, not a vector. It has only magnitude and no direction. [Hey, maybe that's what the OP was asking about....] If you want the cross product, the resultant vector will be at right angles to the two other vectors, as determined by the right-hand rule.
  6. Jun 14, 2006 #5
    @NEWO: I know that formula , but I was not seeking for a product there, to be more specific I would give you a problem in this category as an example:

    Two billiard balls come with velocity1 and velocity2 to each other, the angle that the velocitys vectors form is a right angle(90o), after the "bounce" the first ball only changes direction by 30o , but the secound one remains at rest.now the question is v1/v2=? vx-velocity of ball x.

    So in this kind of problem like @berkman sayd very well , I want to know, what kind of formula or mathematical procedure can give me the RELATIVE angle , when I know the MODULE of two vectors, and the angle between them.

    @berkman, I don't know what right-hand rule is :D. Hope is not too lame :).
  7. Jun 14, 2006 #6


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    Staff Emeritus
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    Gold Member

    The consine of the acute angle between two vectors is given by;

    [tex]\cos\theta = \frac{|\vec{a}\cdot\vec{b}|}{|\vec{a}|\cdot |\vec{b}|}[/tex]

    Where [itex]|\vec{a}\cdot\vec{b}|[/itex] is the scalar poduct.
  8. Jun 14, 2006 #7


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    Science Advisor

    Your English is excellent- with one correction. The word for the length of a vector is "modulus", not "module".

    I personally prefer to use components but: The two vectors being added are the sides of a parallelogram. The sum is a diagonal of that parallelogram and divides it into two congruent triangles. If the angle between the two vectors is [itex]\theta[/itex] then other angle in the parallelogram, opposite the sum vector, is [itex]\pi- \theta[/itex]. Taking f and g as the lengths of the two vectors, by cosine law, using |v| to represent the length of a vector, v, the length of the sum vector, squared, is [itex]|f|^2+ |g|2- 2|f||g| cos(\pi-\theta)[/itex].
    Since [itex]cos(\pi- \theta)= cos(\theta)[/itex], the length of the sum is given by [itex]\sqrt{|f|^2+ |g|^2- 2|f||g|cos(\theta)}[/itex], just as you have.

    To find the angle made by the sum vector, use the sine law:
    In any triangle, with sides a, b, c, and corresponding angles A, B, C, we have [itex]\frac{sin A}{a}= \frac{sin B}{b}= \frac{sin C}{c}[/itex].
    Here, we know that one angle is [itex]\pi-\theta[/itex] and, of course, [itex]sin(\pi- \theta)= sin(\theta)[/itex]. If [itex]\alpha[/itex] is the angle the sum vector makes with vector f, then [itex]\frac{sin(\theta)}{|f+g|}= \frac{sin(\alpha)}{|g|}[/itex] so [itex]sin(\alpha)= sin(\theta)\frac{|g|}{|f+g|}[/itex] and, thus, [itex]\alpha= cos^{-1}(sin(\theta)\frac{|g|}{|f+g|})[/itex].
  9. Jun 14, 2006 #8
    @HallsOfIvy : Thank You Very Much :), I finally understood this thing, the main problem is that what you did up there is (and it has to be) very rare, I haven't seen that kind of explanation before. Thank you All again !

    PS: One more question , why is cos^(-1) in your finale phase? Type error, or I don't got it already? :D
    Last edited: Jun 14, 2006
  10. Jun 15, 2006 #9
    Yes that was a typo.It should be[itex]\alpha= cos^{-1}(sin(\theta)\frac{|g|}{|f+g|})[/itex]

    A simpler formula would be,

    where the symbols have their meanings as Hallsofivy's post.
    Note that in this case you don't have to actually find the resultant to calculate the angle.
    Another little typo there. You can figure out what :)
    Last edited: Jun 15, 2006
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