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The Proof Of Cross Product?

  1. Oct 6, 2012 #1
    Hi, I was studying Maths the subject of Vector and found two important rule which are "dot" and "cross" product. I proved the first by myself with the ([itex]c^{2}[/itex]= [itex]a^{2}[/itex]+[itex]b^{2}[/itex]-2.[itex]\|a\|[/itex].[itex]\|b\|[/itex].cosθ) However, the I couldn't make out how to prove the other one. I've been so thoughtful about it and need help. Thanks..
  2. jcsd
  3. Oct 6, 2012 #2
    Just what exactly are you trying to prove? The formula for the cross product? Something else?
  4. Oct 6, 2012 #3
    If it is the algebraic proof of the cross product you are looking for, then I'm sure you could google it and find it quite simply (maybe even on wikipeida), but know it's not nearly as a nice as the dot product's derivation.
  5. Oct 6, 2012 #4
    Take two vectors a=(a1,a2,a3) and b=(b1,b2,b3).
    So now a×b should be perpendicular to both b and a.
    So then (a×b).b=0 and (a×b).a=0...
  6. Oct 22, 2012 #5
    The last reply helped me little bit to visualize the proof but I could still not reach the exact proof of Cross Product. I just got [itex]0=a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3}[/itex] and [itex]0=b_{1}x_{1}+b_{2}x_{2}+b_{3}x_{3}[/itex] However, I don't know how to find [itex]x_{1},x_{2},x_{3}[/itex]. [itex]\stackrel{\rightarrow}{A}\times[/itex][itex]\stackrel{\rightarrow}{B}[/itex][itex]=\stackrel{\rightarrow}{C}[/itex] or [itex]\stackrel{\rightarrow}{B}\times[/itex][itex]\stackrel{\rightarrow}{A}[/itex][itex]=\stackrel{\rightarrow}{C}[/itex] ??
  7. Oct 24, 2012 #6
    2 equations, 3 unknowns, so there are infinitely many solutions. Make your solution obey the right-hand rule ;O
  8. Oct 24, 2012 #7
    I just need to find the mathematical proof, I mean the origin of this rule. I don't want the reproduction of this rule. I just want the exact proof.
  9. Oct 24, 2012 #8
    I reiterate: what do you want to prove? The cross product is a definition. You can prove some property of the cross product, but you've yet to state what property you want to prove.
  10. Oct 24, 2012 #9
    I meant if there is such a proof that we do for finding dot product with ([itex]c^{2}=a^{2}+b^{2}-2abcos\theta[/itex]) ??
    I'm looking for that kind of stuff. I've been striving for so long time to find out how it might be proved in that way.
  11. Oct 24, 2012 #10
    That's not a proof. A proof does not do what you seem to think it does. That is a formula that can be used to motivate a definition.

    Regardless, I think you should be looking at some form of the law of sines for a triangle. Consider a triangle with sides ##a, b, c## and angles opposite those sides ##A, B, C##. The law of sines says that the area of this triangle is

    $$A = \frac{1}{2} ab \sin C = \frac{1}{2} b c \sin A = \frac{1}{2} ca \sin B$$

    Put this into vector form, and you should see the cross product staring you in the face.
  12. Oct 26, 2012 #11
    Okay, I got your point. But what you said just gives us the magnitude of the result of a vector calculated by cross product. It's not what I've wanted, thanks a lot though. I've been still working on it to find the proof I've wanted.
  13. Oct 26, 2012 #12
    What do you understand the cross product to mean and why do you think that any such proof as you seek actually exists?

    Of course you can prove the sine formula if you start with a different definition of the cross product.

    However as with any proof it depends where you start from (ie what are your original definitions)
    Last edited: Oct 26, 2012
  14. Oct 26, 2012 #13
    For proof of why you have to combine the vector components in such and such ways to get a vector perpendicular to both and of the prescribed magnitude, one method is showing that the cross product is distributive, see figure 8 on page 9 and the text below it:


    Then the cross products of the unit basis vectors are calculated using the definition, after which you can just expand the cross product of two vectors and calculate the result, see for example:

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