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Vector Triple product dilemma

  1. Nov 5, 2013 #1
    I am currently going through the book Introduction Of Electrodynamics by Griffiths. I have come across vector triple product which is stated as follows in the book:

    $$\textbf{A} \times (\textbf{B} \times \textbf{C})=\textbf{B}(\textbf{A}\cdot \textbf{C})-\textbf{C}(\textbf{A}\cdot \textbf{B})$$
    The author then states a few more vector products and says that they can be derived using the vector triple product. One of them I am unable to derive is :
    $$(A\times B)\cdot (C\times D)=(A\cdot C)(B\cdot D)-(A\cdot D)(B\cdot C)$$

    I don't see how I can derive the above using the vector triple product. :confused:

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. Nov 5, 2013 #2

    Ray Vickson

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    First, prove that
    [tex]\mathbf{U} \times \mathbf{V} \cdot \mathbf{W}
    = \mathbf{U} \cdot \mathbf{V} \times \mathbf{W} [/tex]
    for any three vectors U, V, W. (Note: UxV.W means (UxV).W, etc). In other words, in a mixed vector-scalar product you can interchange the 'x' and the '.'. It is pretty easy to prove, just by expanding out both things and comparing the results.

    Next, prove that
    [tex] \mathbf{U} \times ( \mathbf{V} \times \mathbf{W})
    = (\mathbf{U} \times \mathbf{V}) \times \mathbf{W},[/tex] just by using ##\mathbf{R} \times \mathbf{S} = -\mathbf{S} \times \mathbf{R}## and applying the vector triple product you are already given.

    Now apply these two facts to your problem.
     
  4. Nov 5, 2013 #3
    First interchange the signs of '×' and '.'

    $$(A\times B)\cdot (C\times D) = A\times B \times C \cdot D$$

    Which gives

    $$[(A \cdot C)B - (B \cdot C)A] \cdot D $$

    And finally,

    $$[(A \cdot C)(B \cdot D) - (B \cdot C)(A \cdot D)] $$
     
  5. Nov 5, 2013 #4
    Thanks, didn't notice that it was a scalar triple product. :)

    I did not require this in proving what I wanted to but still I would like to know how would you prove this. Here's how I tried:

    Applying the vector triple product on LHS:
    $$V(U\cdot W)-W(U\cdot V)$$

    The RHS is:
    $$-W \times (U \times V)$$
    $$=-(U(W\cdot V)-V(W \cdot U))$$
    $$=V(W \cdot U)-U(W\cdot V)$$

    They don't turn out to be the same. :confused:
     
  6. Nov 5, 2013 #5

    D H

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    That's just wrong, Ray. The cross product is not associative.
     
  7. Nov 5, 2013 #6

    D H

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    You can use both the scalar triple product and vector triple product to prove this identity. Denote ##U=C \times D##. With this substitution, ##(A \times B) \cdot (C \times D) = (A \times B) \cdot U##. See if you can take it from here, using the scalar triple product and then the vector triple product.
     
  8. Nov 5, 2013 #7
    I did exactly the same to prove it. I stated it in my previous reply.

    $$(A \times B)\cdot U=A\cdot (B\times (C\times D))$$
    $$=A\cdot(C(B\cdot D)-D(B\cdot C))=(A\cdot C)(B\cdot D)-(A\cdot D)(B\cdot C)$$

    I have a few more doubts regarding vector proofs but they involve the del operator, should I post them here even though it is a precalculus section?
     
  9. Nov 5, 2013 #8

    D H

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    Sorry, I didn't see that post. I saw your post where you trued (but failed) to prove the associativity of the cross product.

    [EDIT]: That should be tried, not trued.

    The cross product is not associative in general. Since the cross product is anti-commutative, ##(A \times B) \times C = - C \times (A \times B) = C \times (B \times A)##. Now this is in a form where you can use the vector triple product rule: ##(A \times B) \times C = C \times (B \times A) = B (A \cdot C) - A (B \cdot C)##. The difference between ##A \times (B \times C)## and ##(A \times B) \times C## is thus ##A \times (B \times C) - (A \times B) \times C = C (A \cdot B) - A (B \cdot C)##. This is zero if both ##A \cdot B## and ##B \cdot C## are zero or if C is parallel to A (i.e., ##C=\alpha A## where alpha is a scalar). The difference is non-zero otherwise, and hence the two forms are not equal to one another in general.

    Note that the above means that ##U \times (V \times U)## and ##(U \times V) \times U## are equal to one another. This comes up quite often.


    It's best to post those questions in the calculus section.
     
    Last edited: Nov 5, 2013
  10. Nov 5, 2013 #9
    Thanks a lot DH for such wonderful tips. I have made a note of these. :smile:
     
  11. Nov 5, 2013 #10

    D H

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    A couple more points on this. One is that since ##U \times (V \times U) = (U \times V) \times U## the parentheses can be omitted in this case: It's okay to write ##U \times V \times U##. Note that writing ##U \times V \times W## is *not* okay. Parentheses are needed in the general case.

    The other is that one reason this comes up quite often is that it is closely related to the component of V that is perpendicular to U, which I'll designate as ##V_{\perp_U}##. One way to calculate the component of V that is normal to U is to subtract the projection of V onto U from V. The projection is given by ##\frac {U (V \cdot U)} {U^2}##, and thus the normal component is ##V_{\perp_U} = V - \frac {U (V \cdot U)} {U^2} = \frac{V (U \cdot U) - U (V \cdot U)} {U^2}##. The numerator is ##U\times (V \times U) = U \times V \times U##. Thus ##V_{\perp_U} = \frac {U \times V \times U}{U^2}##. If U is a unit vector, this reduces to ##U \times V \times U##.
     
  12. Nov 5, 2013 #11
    Thank you once again D H! :)
     
  13. Nov 5, 2013 #12

    Ray Vickson

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    You are right, sorry. Anyway, you don't need anything like that.
     
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