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Cross product

  1. Aug 2, 2008 #1
    I have not yet taken Linear Algebra, so perhaps that is why I don't have an answer to this question. But:

    Given a vector a <1,-1,0> and a vector b <-1,1,0>, I would suspect (by the right hand rule) that taking a X b, I would get a vector pointing out of the computer screen (assuming that the computer screen is the xy plane). However, when I take the actual cross product, I get:

    i j k
    1 -1 0
    -1 1 0

    (0-0)i-(0-0)j+(1-1)k, giving me the vector 0i+0j+0k. Why is this?
  2. jcsd
  3. Aug 2, 2008 #2

    D H

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    Because b=-a = (-1)*a. In other words, a and b are co-linear and hence their cross product is zero.
  4. Aug 2, 2008 #3
    Just to add onto what the previous poster said, the cross product between two vectors a and b is defined as a*b =|a||b|sinx(n^), where x is the angle between the vectors and n^ is a unit vector, n. Since they are collinear, the angle between them is 180 degrees. So, sin180 = 0, making the whole product 0.
  5. Aug 2, 2008 #4
    So the right hand rule does not apply to vectors that are co-linear? I suppose that that was my main question.
  6. Aug 2, 2008 #5


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    You know the following rule for cross product: -aXb = -(aXb) = aX(-b).

    Now this reduces it to -(aXa). What is aXa?
  7. Aug 2, 2008 #6

    D H

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    The right-hand rule doesn't even make sense for vectors that are co-linear. Take your example, [itex][1,-1,0]\times[-1,1,0][/itex] What makes you think the result would be coming out of the screen? Why not along [1,1,0], or for that matter, in any direction of the form [a,a,b]? The problem is that two co-linear vectors do not define a plane. Fortunately, the cross product of a pair of co-linear vectors is identically zero, so the fact that the direction is not well-defined is not a problem.
  8. Aug 3, 2008 #7

    Well, I guess I'm not seeing this. I can't see how the right hand rule doesn't work. I put my hand at 315 degrees for vector a, and then I curl it until it points in the direction of vector b, at 135 degrees. And my thumb is pointing out of the computer screen.
  9. Aug 3, 2008 #8
    aXa = 0i+0j+0k. I don't know those rules though, could you explain why they make sense?
  10. Aug 3, 2008 #9
    If a and b are not colinear, you can only rotate a through one plane to put it on top of b.

    If a and b *are* colinear, there is an infinite number of planes that you can rotate a through to put it onto b: any plane containing both vectors.

    So how do you know which plane to put your "hand" in when applying the rule? You seem to think it should be the xy plane, but there is no good reason for this. It could just as well be any other plane containing the two vectors.
  11. Aug 3, 2008 #10
    One way of thinking about the cross product is that |a x b| is the area of the parallelogram whose sides are a and b. If a and b are colinear, we no longer have a parallelogram but a degenerate case of a parallelogram, i.e. a line segment. Since the area of a line segment is zero, |a x b| = 0 if a and b are colinear. Thus a x b = 0-> (where 0-> denotes the zero vector)
  12. Aug 3, 2008 #11
    BoundbyAxioms, the zero vector has an arbitrary direction. It is undefined (the direction).
  13. Aug 4, 2008 #12


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    No, not necessarily. Since those vectors are diametrically opposite, you can turn the fingers of your right hand from one to the other either way, with the thumb pointing in either direction. Try it!
  14. Aug 4, 2008 #13
    Actually, there are infinite possible directions for the thumb to be pointing!
  15. Aug 5, 2008 #14
    Ok now I get it, that's what was confusing me. Thank you for your help, everyone.
  16. Aug 6, 2008 #15


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    By the way, the cross product of vectors is a matter of vector calculus and is not normally dealt with in Linear Algebra.
  17. Aug 6, 2008 #16
    Ah, ok. Good to know. I know cross products from my vector calculus class, not quite sure why I thought I should ask it in a linear algebra forum.
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