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Proof for cross product (please)

  1. Mar 25, 2009 #1
    TO be very honest i only know the definition of cross product of two vectors but i would like to prove.........................in fact all text books just define the cross product
    how can i derive it starting from the first principles i,e, i must start with something and finally arrive at the expression given below .......... anybody who can help me out with this ......
    If a and b are two vectors and alpha is the angle between them
    then i would like to "mathematically" "show that" "a cross b"=(mod)a* (mod)*b*sin alpha *n(cap) where n is the unit vector perpendicular to the plane containing both a and b

    this is killing me every day because i have no idea about how to prove it .........
    how to start proving and from where ........ this is an other big question ....
    any website that gives the detailed proof or anyone here who can do it
  2. jcsd
  3. Mar 25, 2009 #2


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    You might start with the triple product:
    a.) it is the volume of the parallelepiped formed with the three vectors, and
    b.) it is the determinant of the matrix with the three vectors as row vectors (or column vectors).
    You can then use the volume = base times height to get the cross product in terms of the area of the base.

    This may not be any easier however. I would start with showing that rotations respect the cross product:
    [tex] \mathbf{R}(\vec{u}\times\vec{v}) = (\mathbf{R}\vec{u})\times (\mathbf{R}\vec{v})[/tex]
    where [itex]\mathbf{R}[/itex] is an arbitrary rotation matrix.

    You can then easily show that the cross product:
    [tex] a\hat{\imath} \times (b\cos(\theta)\hat{\imath} + b\sin(\theta)\hat{\jmath})=ab\sin(\theta)\hat{k}[/tex]
    and any other cross product of two vectors is but a rotated version of this one.

    The rotational covariance of course is the tricky part. Use the fact that rotations are linear transformations and that the cross product is bilinear and then show this rotational covariance holds for the nine possible cross products of the standard basis vectors.

    There's probably a more elegant shorter demonstration but this is the first thing I could think of.
  4. Mar 25, 2009 #3


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    Now that I think about it, the standard proof is probably a matter of resolving one of the two vectors in a cross product into orthogonal and parallel components relative to the other.

    e.g. given vectors u and v:
    [tex] \vec{v}=\vec{v}_\perp + \vec{v}_{||}[/tex]
    [tex] \vec{v}_{||} = \vec{u}\frac{\vec{v}\bullet\vec{u}}{\vec{u}\bullet\vec{u}}[/tex]
    solve for the perpendicular component and then work out the magnitude of the cross product.

    To my thinking the cross product is first defined by the magnitudes times the sine of the angle form. You then would show that it is bilinear and anti-symmetric and then are able to work out the cross products of the standard unit basis vectors. Then via bilinearity one may expand the vectors into components and work out the component cross products.

    The hardest part is I think the demonstration that [itex] \vec{u}\times (\vec{v}+\vec{w}) = \vec{u}\times \vec{v} + \vec{u}\times \vect{w}[/itex],
    using just the "magnitudes times sine of the angle" definition. Hmmm....
  5. Mar 25, 2009 #4
    We want i x i = 0 and j x j = 0 and i x j = k and j x i = -k. Now just use basic algebra:

    (a i + b j) x (c i + d j) = a c (i x i) + a d (i x j) + bc (j x i) + b d (j x j) = a c (0) + a d (k) + bc (-k) + b d (0)
    =(a d - b c) k
  6. Mar 25, 2009 #5
    no no the proof is even simpler .......... i was shown once by my lecturer ...... he now wants me to crack my head and recaptulate it.........
    he started like this
    (vector A cross VectorB)=sqrt(something) ............. i just want to know that something............. anyone who can remember this and tell me what that "something" is?........and just proceed on the same lines and get the proof
    and yes My lecturer proceeded on the same lines and arrived at the last step that was(i don remember)probably this
    (vector A cross VectorB)=sqrt( (a11+a22+a33)^2 * (b11+b22+b33)^2 *(1-cos^2@) k(cap) )
    hence it was from there one could define the cross product
    (a11+a22+a33)^2 * (b11+b22+b33)^2 *(1-cos^2@) becomes mod A*mod B*sin@
    please think about this and get the proof
  7. Mar 26, 2009 #6


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    It's not clear to me what it is you want to prove. The "proof" of cross product? You prove a statement, not an operation. What statement do you want to prove?
  8. Mar 26, 2009 #7
    No sir , i just showed u the attempts i made in proving, failing although
    Yes,i want to know "The 'proof' of cross product"
  9. Mar 26, 2009 #8


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    As Halls points out, you are asking us to derive something from first principles without saying what your first principles are.
    The definition as most textbooks give it looks like a perfectly good first principle to me. On the other hand, you can take the requirement that a x b be a vector perpendicular to both a and b (in an orientation preserving sense) as a first principle.
    Or you can take the requirement that you can define an angle [itex]\theta[/itex] between two vectors a and b through
    [tex]\sin\theta = (\mathbf a \times \mathbf b) / (|a| \cdot |b|),
    \cos\theta = (\mathbf a \cdot \mathbf b) / (|a| \cdot |b|)
    as a first principle.

    Also, you cannot write
    [tex]\mathbf a \times \mathbf b = \sqrt{\cdot}[/tex] (*)
    because the left hand side is a vector, so the right hand side should be vector. If (*) being true is your "first principle" then you need to first define what should be in the place of the dots and how taking a "square root" of it produces a vector.

    Finally, you still haven't made clear what you want to prove.
    Can you prove multiplication of numbers to me?
    What is the proof that blue?
  10. Mar 26, 2009 #9
    oh i am sorry if you have encountered some frustration.............
    Fine my question plain..........
    product of two vectors should give u a vector ...............
    now my question is as to how one could arrive at the expression for the croos product of two vectors which is (mod)a* (mod)*b*sin alpha *n(cap) where n is the unit vector perpendicular to the plane containing both a and b ,@ is the angle between those two vectors a &b.........
  11. Mar 26, 2009 #10


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    So you mean derive the "coordinate" formula from the definition in terms of length and angle between the vectors?

    Then do what confinement suggested: look at the basis vectors.

    If [itex]\vec{u}\times\vec{v}[/itex] has length [itex]|\vec{u}||\vec{v}|sin(\theta)[/itex] where [itex]\theta[/itex] is the angle between the vectors, and the direction is perpedicular to both, according to the "right hand rule", since [itex]\vec{i}[/itex], [itex]\vec{j}[/itex], and [itex]\vec{k}[/itex] all have length 1 and are at right angles to each other (so that [itex]sin(\theta)= 1[/itex]) it is easy to see that [itex]\vec{i}\times\vec{j}= \vec{k}[/itex], [itex]\vec{j}\times\vec{k}= \vec{i}[/itex], and [itex]\vec{k}\times\vec{i}= \vec{j}[/itex]. Of course it follows also from that definition that [itex]\vec{i}\times\vec{i}= \vec{j}\times\vec{j}= \vec{k}\times\vec{k}= \vec{0}[/itex] and that the cross product is associative, distributive, and anti-commutative.

    Given all that, [itex](a\vec{i}+ b\vec{j}+ c\vec{k})\times(u\vec{i}+ v\vec{j}+ w\vec{k})[/itex][itex]= a\vec{i}\times(u\vec{i}+ v\vec{j}+ w\vec{k})+ b\vec{j}(u\vec{i}[/itex]
    [itex]+ v\vec{j}+ w\vec{k})[/itex][itex]+ c\vec{k}(u\vec{i}+ v\vec{j}+ w\vec{k})[/itex]
    [itex]= au\vec{i}\times\vec{i}+ av\vec{i}\times\vec{j}+ aw\vec{i}\vec{k}[/itex]
    [itex]+ bu\vec{j}\times\vec{i}+ bv\vec{j}\times\vec{j}+ bw\vec{j}\times\vec{k}[/itex][itex]+cu\vec{k}\times\vec{i}+ cv\vec{k}\times\vec{j}+ cw\vec{k}\times\vec{k}[/itex]
    [itex]= au\vec{0}+av\vec{k}-aw\vec{j}- bu\vec{k}+ bv\vec{0}+bw\vec{i}[/itex]
    [itex]+ cu\vec{j}- cv\vec{i}+ cw\vec{0}[/itex][itex]= (bw- cv)\vec{i}+ (cu-aw)\vec{j}+(av-bu)\vec{k}[/itex]
    Last edited by a moderator: Mar 26, 2009
  12. Mar 27, 2009 #11
    I'm guessing that what yagami light is trying to prove is that

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


    [tex]v \times u = Det\left[ \begin{array}{ccc} i & j & k \\ v_x & v_y & v_z \\ u_x & u_y & u_z \end{array} \right][/tex]

    produces the same answer.
  13. Mar 28, 2009 #12
    yes i am trying to prove this u x v = |v| |u| sin (theta)
    the determent form is obtained only after i know what the cross product is
    i,e first prove the cross product which is this |v| |u| sin (theta) and then u can apply the same in the form of a determinant
  14. Mar 28, 2009 #13


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    OK, so you want to prove that u x v = |u| |v| sin(theta).

    What is your definition of cross product?
  15. Mar 31, 2009 #14
    i have defined the cross product clearly ........ not my cross product
    i just want to know if this can be proved algebraically ......
    are definitions not proved ............
    are they defined simply ...........
    you first prove and then form it as a definition or a theorem
    i finally want to know if someone can help me .......by proving the cross product
  16. Mar 31, 2009 #15
    Hi, yagami light. Forgive us, but I think we might be struggling a bit with your English. You are using the phrase "proving the cross product" but that is a meaningless sentence fragment.

    Referring to my previous post, I don't think you're trying to "prove the cross product," you are just trying to show that the two ways of calculating it give the same numerical answer. (One uses the determinant mnemonic, and the other uses a sine.)

    Why not start by limiting yourself to pairs of vectors in the xy plane? Try to make some progress there first, and see what you get.
  17. Apr 1, 2009 #16


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    That is the point of a definition.
    To talk about the cross product of two vectors, you first have to define in some way what it means. Then you can prove things about it, like your definition is equivalent to some other way of calculating it.
    That is what we are trying to get across to you the whole time.

    You can define
    u x v = |u| |v| sin(theta) n
    where theta is the angle between u and v and n is a normal vector perpendicular to u and v, and then prove that this is equivalent to the determinant calculation.

    Or you can define u x v by some set of properties, for example
    "the vector which is perpendicular to u and v in a right-hand sense, whose magnitude is proportional to the magnitudes of u and v and equal to |u| |v| if u and v are perpendicular, while zero when u and v are parallel <and possibly some other properties>"
    and then prove that these properties uniquely determine the cross product and the magnitude is given by |u| |v| sin(theta).
  18. Apr 1, 2009 #17


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    I believe that is what I showed in the post just above yours.
  19. Apr 14, 2011 #18
    Just for the record, the cross product is NOT associative.




    Regardless, associativity is not needed to prove the equivalence of:

    [tex]\mathbf{a}\mathbf{b}sin(\theta)\vec{n}=\mathbf{a}\times\mathbf{b}= det\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \\
    a_1 & a_2 & a_3 \\
    b_1 & b_2 & b_3 \end{array}\right|[/tex]

    For a more rigorous proof see: http://en.wikiversity.org/wiki/Cross_product
    Last edited: Apr 14, 2011
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