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Cross Product Proof #1

  1. Mar 6, 2012 #1
    The cross product between two vectors [itex]\vec{A}[/itex]x[itex]\vec{B}[/itex] and [itex]\vec{C}[/itex] is given by the following equation:
    ([itex]\vec{A}[/itex]x[itex]\vec{B}[/itex])x[itex]\vec{C}[/itex]=([itex]\vec{A}[/itex].[itex]\vec{C}[/itex])[itex]\vec{B}[/itex]-([itex]\vec{B}[/itex].[itex]\vec{C}[/itex])[itex]\vec{A}[/itex]

    Well, as I'm sure you know, proving something is true is different than proving how something is true. In this proof, I will not only prove it holds up but I also will demonstrate how it came to be.

    What is given is [itex]\vec{A}[/itex],[itex]\vec{B}[/itex] and [itex]\vec{C}[/itex], which are vectors belonging to ℝ3, or mathematically:
    [itex]\vec{A}[/itex],[itex]\vec{B}[/itex],[itex]\vec{C}[/itex][itex]\in[/itex]ℝ3
    Let's define the coordinates of [itex]\vec{A}[/itex],[itex]\vec{B}[/itex] and [itex]\vec{C}[/itex] as the following:
    [itex]\vec{A}[/itex]=(a1,a2,a3)
    [itex]\vec{B}[/itex]=(b1,b2,b3)
    [itex]\vec{C}[/itex]=(c1,c2,c3)
    Note that by the definition of cross product, [itex]\vec{A}[/itex]x[itex]\vec{B}[/itex] is simultaneously perpendicular to [itex]\vec{A}[/itex] and [itex]\vec{B}[/itex], or:
    ([itex]\vec{A}[/itex]x[itex]\vec{B}[/itex]).[itex]\vec{A}[/itex]=0
    ([itex]\vec{A}[/itex]x[itex]\vec{B}[/itex]).[itex]\vec{B}[/itex]=0
    Let [itex]\vec{A}[/itex] and [itex]\vec{B}[/itex] be linearly independent, and so they both define a certain plane [itex]\Omega[/itex]. Using the same chain of thought, ([itex]\vec{A}[/itex]x[itex]\vec{B}[/itex])x[itex]\vec{C}[/itex] is simultaneously perpendicular to [itex]\vec{A}[/itex]x[itex]\vec{B}[/itex] and [itex]\vec{C}[/itex], or:
    (([itex]\vec{A}[/itex]x[itex]\vec{B}[/itex])x[itex]\vec{C}[/itex]).([itex]\vec{A}[/itex]x[itex]\vec{B}[/itex])=0
    (([itex]\vec{A}[/itex]x[itex]\vec{B}[/itex])x[itex]\vec{C}[/itex]).[itex]\vec{C}[/itex]=0
    Hence, we can conclude that ([itex]\vec{A}[/itex]x[itex]\vec{B}[/itex])x[itex]\vec{C}[/itex] is a linear combination of [itex]\vec{A}[/itex] and [itex]\vec{B}[/itex], and so:
    ([itex]\vec{A}[/itex]x[itex]\vec{B}[/itex])x[itex]\vec{C}[/itex]=λ[itex]\vec{A}[/itex]+μ[itex]\vec{B}[/itex] [itex]\in[/itex][itex]\Omega[/itex]
    We can start by noting the resemblance of this equation with the one given, in which λ is a scalar given by -[itex]\vec{B}[/itex].[itex]\vec{C}[/itex] and μ by [itex]\vec{A}[/itex].[itex]\vec{C}[/itex], as we'll soon show. For the next step, let's write the coordinates of ([itex]\vec{A}[/itex]x[itex]\vec{B}[/itex])x[itex]\vec{C}[/itex] in respect to λ and μ on the LHS/RHS and in respect to the coordinates of [itex]\vec{A}[/itex],[itex]\vec{B}[/itex] and [itex]\vec{C}[/itex] on the RHS/LHS, like so (skipping intermediary steps):
    (a1λ+b1μ,a2λ+b2μ,a3λ+b3μ)=(c3(a3b1-a1b3)-c2(a1b2-a2b1),c1(a1b2-a2b1)-c3(a2b3-a3b2),c2(a2b3-a3b2)-c1(a3b1-a1b3))
    Now we have three equations with only 2 unknowns (λ,μ), which algebraically means that there's no degree of freedom in the system. These equations may not be easy on the eyes but with strong motivation, they're feesable. And they go as follows (from x and y coordinates):
    λ=c3(a3b1/a1-b3)-c2(b2-a2b1/a1)-μ(b1/a1)
    [itex]\Rightarrow[/itex] c1(a1b2-a2b1)-c3(a2b3-a3b2)=a2c3(a3b1/a1-b3)-a2c2(b2-a2b1/a1)-μ(a2b1/a1)+μb2[itex]\Leftrightarrow[/itex]μ(b2-a2b1/a1)=-c3(a2a3b1/a1-b3a2+a2b3-a3b2)+a2c2(b2-a2b1/a1)+c1(a1b2-a2b1)[itex]\Rightarrow[/itex]μ=a1c1+a2c2+a3c3=[itex]\vec{A}[/itex].[itex]\vec{C}[/itex]
    Replacing μ in the first line, we get:
    λ=c3(a3b1/a1-b3)-c2(b2-a2b1/a1)-(a1c1+a2c2+a3c3)(b1/a1)
    Which computed gives:
    λ=-(b1c1+b2c2+b3c3)=-[itex]\vec{B}[/itex].[itex]\vec{C}[/itex]
    Note that initially we thought of [itex]\vec{A}[/itex] and [itex]\vec{B}[/itex] as linearly independent but if we hadn't, that would be saying that [itex]\vec{A}[/itex]=κ[itex]\vec{B}[/itex] and such would mean that:
    κ([itex]\vec{B}[/itex]x[itex]\vec{B}[/itex])x[itex]\vec{C}[/itex]=(0,0,0)
    And so:
    [itex]\vec{A}[/itex]=(μ/λ)[itex]\vec{B}[/itex]
    With κ=μ/λ.

    If you find any incongruence in this resolution (or any doubt), let me know.
     
    Last edited: Mar 7, 2012
  2. jcsd
  3. Mar 7, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi Mathoholic!! Welcome to PF! :smile:

    Yes, that's fine …

    you're saying that it has to be in the plane of A and B,

    and (if they're not parallel) that means it has to be a linear combination of A and B.

    But it also has to be perpendicular to C, so (aA + bB).C = 0, so a/b = -(B.C)/(A.C).
     
  4. Mar 7, 2012 #3
    Re: Welcome to PF!

    Yes, I understood your simplification. But there's a reason I chose the first equation and not the second one, mainly because the second equation doesn't quite tell you what's λ and μ, it only shows that:
    λ/μ=-([itex]\vec{B}[/itex].[itex]\vec{C}[/itex])/([itex]\vec{A}[/itex].[itex]\vec{C}[/itex])

    The factor (-1) could either be from [itex]\vec{B}[/itex].[itex]\vec{C}[/itex] or [itex]\vec{A}[/itex].[itex]\vec{C}[/itex] and that's significant to the result. I only assumed that it belong to the plane [itex]\Omega[/itex] because that vector is equipolent. Finally, when you have:

    α/β=4

    You cannot assume that α is 4 and that β is 1. There's ∞1 possibilities of that ratio happening (degree of freedom: 1). So, in the first case there's a possibility that:

    λ=±[itex]\vec{A}[/itex].[itex]\vec{C}[/itex]±1 [itex]\wedge[/itex] μ=±[itex]\vec{B}[/itex].[itex]\vec{C}[/itex]±1 (alternately)
     
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