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Cross product question

  1. Apr 20, 2010 #1
    1. The problem statement, all variables and given/known data

    Prove that If [tex]\vec{a}[/tex]x [tex]\vec{b}[/tex] = 0, then [tex]\vec{a}[/tex]is parallel to [tex]\vec{b}[/tex].

    2. Relevant equations

    3. The attempt at a solution
    I tried attempting the solution by using the following:

    [tex]\vec{a}[/tex] = [a1, a2, a3]
    [tex]\vec{b}[/tex] = [b1, b2, b3]

    When I took the cross product of a x b I got::

    [a2b3 - b2a3 a3b1 - b3a1, a1b2 - b1a2]

    and we can make this equal to 0, but I am confused here; I have no idea on how to prove that when a x b is 0, vectors a and b are parallel.

    any help is appreciated!
    Last edited: Apr 20, 2010
  2. jcsd
  3. Apr 20, 2010 #2
    Imagine the zero as the zero vector, defined in R3 as 0=[0,0,0]

    Edit: You could view the geometric interpretation. In which case you will have a parallellogram with area equal to zero.
    Last edited: Apr 20, 2010
  4. Apr 20, 2010 #3
    thanks, so this is what I end up with:

    a2b3 = b2a3.....(1)

    a3b1 = b3a1....(2)

    a1b2 = b1a2....(3)

    Does this suggest that the two vectors, a and b are parallel?
  5. Apr 21, 2010 #4
    Bump..any one?

  6. Apr 21, 2010 #5
    Its it simply

    Let [tex]\vec{a},\vec{b} \neq 0[/tex] and let

    [tex]\vec{a} \times \vec{b} = |\vec{a}| \cdot |\vec{b}| \cdot sin(\theta}) \cdot \mathrm{n}[/tex] be the definition of the cross product where [tex]0 \leq \theta \leq \pi [/tex]

    then for [tex]\theta = 0[/tex] [tex]\vec{a} \times \vec{b} = |\vec{a}| \cdot |\vec{b}| \cdot sin(0}) \cdot \mathrm{n} = 0[/tex]

    Last edited: Apr 21, 2010
  7. Apr 23, 2010 #6
    dot product? I don't understand it, could anyone please help..
  8. Apr 23, 2010 #7


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    Hi spoc21! :smile:
    Yes, because they become

    a2/a3 = b2/b3.....(1)

    a3/a1 = b3/b1....(2)

    a1/a2 = b1/b2....(3) :wink:
  9. Apr 25, 2010 #8
    Thanks tiny-tim So basically if I write the vectors in this form, I am showing that they are parallel, and this would be enough for the proof? Also, just for my knowledge could you please elaborate on this..
    Thank you very much :smile:
  10. Apr 25, 2010 #9


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    Hi spoc21! :smile:
    Yes, that's enough to do it. :wink:

    To elaborate …two vectors are parallel if one is a scalar times the other …

    and you can easily check that that means that the ratios of their coordinates must be the same. :smile:
  11. Apr 25, 2010 #10
    Thanks!! :smile:
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