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Question regarding vector (cross) product

  1. Feb 15, 2010 #1
    Yo guys, I was wondering:

    Why does the cross product of two vectors produce a third vector which is perpendicular to both the original vectors?

    I've come across it in math and it's one of those things I've accepted... but I'm not quite sure why it happens.
  2. jcsd
  3. Feb 15, 2010 #2
    As you may know, the definition of u x v, is the determinant of the following matrix:

    [tex]\hat{x}, \hat{y}, \hat{z}[/tex]
    [tex]u_{1}, u_{2}, u_{3}[/tex]
    [tex]v_{1}, v_{2}, v_{3}[/tex]

    Now before going ahead, let me show you another thing which is not really related but will help me answer you easily.

    Suppose a=[tex]a_{1}\hat{x}+a_{2}\hat{y}+a_{3}\hat{z}[/tex] and b=(b1,b2,b3)

    Then a.b=a1*b1+a2*b2+a3*b3. So you can say that dot product replaces the unit vectors x,y,z with their corresponding components from b (or from a).

    Using this you easily get that (u x v).u is the determinant of

    [tex]u_{1}, u_{2}, u_{3}[/tex]
    [tex]u_{1}, u_{2}, u_{3}[/tex]
    [tex]v_{1}, v_{2}, v_{3}[/tex]

    But the determinant of a matrix with two equal rows is zero. Therefore (u x v).u=0, which means u x v is perpendicular to u. The same goes for v.
  4. Feb 15, 2010 #3
    Thanks for the response elibj123, that's made things a lot clearer. However, I don't quite understand why the dot product replaces the unit vectors.

    Is this the reason? :

    a = a1[tex]\hat{i}[/tex] + a2[tex]\hat{j}[/tex] + a3[tex]\hat{k}[/tex]
    b = b1[tex]\hat{i}[/tex] + b2[tex]\hat{j}[/tex] + b3[tex]\hat{k}[/tex]

    When the dot product of these are taken, the unit vectors become:


    and these are all equal to 1, and therefore the unit vectors disappear and appear to be replaced by b.
  5. Feb 15, 2010 #4

    D H

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    That is *not* the definition of the cross product. That is just an incredibly bad (but very widely used) abuse of notation. Defining something in terms of abuse of notation is just a bad idea.

    The cross product was originally defined to reflect the quaternion product, in which i2=j2=k2=ijk=-1. This leads to the anti-commutative expressions ij=k, ji=-k, jk=i, kj=-i, ki=j, ik=-j. Gibbs and Heaviside thought quaternions were overly cumbersome and a bit artificial (we live in a three dimensional world). The two independently developed vector analysis, stealing useful chunks of Hamilton's quaternions along the way.

    Another way to look at the cross product is that the cross product of two vectors u and v is defined as having magnitude equal to uvsinθ, where u and v are the magnitudes of the two vectors and θ is the angle between the two vectors. The cross product vector points in a direction normal to both u and v by definition. Which way it points is a bit ambiguous; there are two choices. The right-hand rule removes this ambiguity.
  6. Feb 15, 2010 #5


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    I like the definition in terms of the Levi-Civita symbol. In every expression below, there's a sum over all indices that appear exactly twice, but I will not write any summation sigmas for those. (Einstein's summation convention).

    [tex](x\times y)_i=\varepsilon_{ijk}x_jy_k[/tex]

    This definition makes it extremely easy to prove the orthogonality:

    [tex]x_i(x\times y)_i=\varepsilon_{ijk}x_ix_jy_k=0[/itex]

    The last step is based on this extremely useful result (excercise): If [itex]S_{ij}=S_{ji}[/itex] and [itex]A_{ij}=-A_{ji}[/itex], then [itex]S_{ij}A_{ij}=0[/itex].
  7. Feb 15, 2010 #6


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    It is not a question of why. It is just a matter of definition.
  8. Feb 16, 2010 #7
    I think the notation will forgive me for being so abusive, I certainly didn't want to harm its feelings.
  9. Feb 16, 2010 #8


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    I think you should be worrying about a law suit!:wink:

    I once had a teacher, in a graduate course, who, when a student presenting a proof said 'by abuse of notation', told him "Let's not be that abusive"!
    Last edited by a moderator: Feb 16, 2010
  10. Feb 16, 2010 #9


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    That is one definition. It is never safe to assume that everyone learns the same definition when there are several equivalent definitions of the same thing. Another equivalent definition is <a, b, c> X <x, y, z>= <bz- cy, cx- az, ay- bx>, which is the same as elibj123's without the "abusive" notation.

    From that, you can calculate that
    <a, b, c>.<bz- cy, cx- az, ay- bx>= abz- acy+ bcx- abz+ acy- bcx= (abz-abz)+ (acy-acy)+ (bcx- bcx)= 0

    <x, y, z>.<bz- cy, cx-az, ay- bx>= bxz- cxy+ cxy- ayz+ ayz- bxz= (bxz- bxz)+ (cxy- cxy)+ (ayz- ayz)= 0.

    Since both dot products are 0, <bz- cy, cx- az, ay- bx> is perpendicular to both <a, b, c> and <x, y, z>.
    Last edited by a moderator: Feb 22, 2010
  11. Feb 20, 2010 #10


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    I don't think the determinant notation is abusive. It's just necessary to clearly state that we're extending the definition of "determinant" to apply to matrices that have vectors on one of the rows (or one of the columns). We don't even have to change the formula that defines the determinant

    [tex]\det A=\sum_P s(P) A_{1,P1}\cdots A_{n,Pn}[/tex]

    The sum is over all permutations P, and s(P) is the sign of the permutation P (i.e. s(P)=1 if P is even and s(P)=-1 if P is odd). The reason why things work out so nicely is that the definition never multiplies two matrix elements from the same row or two from the same column.

    I highly recommend learning to use the Levi-Civita symbol though. It makes a lot of proofs involving cross products extremely easy.
  12. Feb 22, 2010 #11
    Code (Text):
    I think you should be worrying about a law suit!
    Wow, I'm new here and already I've discovered a maths professor with a sense of humour.

    I might even get to like this place.

    Mess1n, this is a good question you have posted in the mathematics section of a physics forum.

    The answers given have been mathematical, rather than physical so I will try to provide a physical explanation or motivation.

    In mathematics there are many vector algebras, not all of which possess a cross product.

    The vector algebra used by physicists and engineers use a particular vector algebra, because it is useful to have such an operation.

    It is useful to have one because it describes effects, observable in the physical world, which are in some way additional to the original physical effect or quantity under consideration.

    For example

    Consider a physical vector quantity, force, and the statement

    "A force can have no effect at right angles to its line of action"

    This is true insofar as the horizontal component can exert no effect (force) in the vertical direction.

    But what about moment?

    This is another effect that a force can exert, but which is certainly not a force.
    By saying that moment is not a force we are saying that it cannot be found in the list of all possible force vectors that can be generated by linear combination of the horizontal and vertical components. Mathematically, moment is not in the vector space of forces.

    However the moment effect does exist and we can represent it by a 'vector' generated by a
    'cross product' of distance and the generating force.

    It should be noted that this 'vector' is fundamentally different from the generating force with its own properties and effects.

    So the motivation is to chose a mathematics (vector algebra) which correctly represents the full picture.
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