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Is vector division possible?

  1. Jun 27, 2013 #1
    hi guys,
    i have a basic question on vector division..can i divide a vector by another vector to get the third vector.. i know, if it is scalar product then i can not do the reverse engineering but what happens with the vector products?
    like v=wxr
    can i get r(both direction and quantity) by dividing v by w?

    i got p=F/A, where 3 of them are vectors...
    so how is it possible ?
  2. jcsd
  3. Jun 27, 2013 #2

    D H

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  4. Jun 27, 2013 #3
    Hmmm... let's see what we can do.

    Suppose you have a dot product a.x = A where a and x are vectors and A is some number.

    Now, given a and A, what can we say about x?

    Well a.x = |a||x|cos(theta), where theta is the angle between a and x.

    So |a||x|cos(theta) = A

    For a non-zero, we can say |x|cos(theta) = A/|a|

    Frankly, that's as much as we can say given a dot product. You'll need extra information about x in order to do the "division".

    Now, given the direction of x, you can find theta and hence the length there by finding x.
    In 2D, if you're given the length then you can find theta using the above and you're done.
    For higher dimensions however, knowing just the length won't be enough but if you know in addition which plane it lies in then you're good.

    What about the cross product? Well, suppose a X y = b, where a, y and b are vectors.

    Given a and b, what can we say about y?

    Well, for b non-zero we know that y is orthogonal to b, so we at the very least we have a plane in which to find y.

    Also |a||y|sin(theta) = |b| where theta is the angle between a and y.

    Hence using this, if you know either the length of y or the angle it makes with a, then you can effectively do the division.

    Lastly if b is zero the you have one of the following cases:
    1) a is zero, so y could be any vector
    2) a is non-zero and hence y points in the same direction as a (so knowing the length of y will determine your solution)

    So yeah, I wouldn't really say you can do "division" in the classical sense, but you can at least make some progress. Hope that helps :)
    Last edited: Jun 27, 2013
  5. Jun 27, 2013 #4


    Staff: Mentor

    Not necessarily. y is also orthogonal to the zero vector, so if a and y happen to have the same direction, then b will be 0. In this case you don't have a plane to work with.
  6. Jun 27, 2013 #5
    Ah yes, sorry I missed the special case where b is zero.
  7. Jun 27, 2013 #6


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    No, because there are many different vectors, r, such that wxr= v for given v and w?

    I have no idea what your "p", "F", and "A" are.
  8. Jun 27, 2013 #7

    D H

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    It's not just b being zero. Given your [itex]\vec a \times \vec y = \vec b[/itex] where [itex]\vec a[/itex] and [itex]\vec b[/itex] are orthogonal to one another, suppose you find some [itex]\vec y[/itex] that satisfies this expression. Then any other vector [itex]\vec y' = \vec y + \alpha \vec a[/itex] where [itex]\alpha[/itex] is an arbitrary scalar will satisfy [itex]\vec a \times \vec y' = \vec b[/itex].

    One solution is [itex]\vec y = \frac{\vec b \times \vec a}{||\vec a||^2}[/itex]. As noted above, this isn't unique.

    Another problem is that this only works with vectors in ℝ3 because the cross product is special to ℝ3. It doesn't generalize. It's really hard to call this "division".

    Yet another problem: What if [itex]\vec a[/itex] and [itex]\vec b[/itex] are not orthogonal to one another? What is [itex]\vec b / \vec a[/itex] in this case?

    Yet another problem: What's the multiplicative identity? How can you call any operation division if there is no multiplicative identity?
    Last edited: Jun 27, 2013
  9. Jun 27, 2013 #8
    Oh, I'm not doing this from anywhere near a mathematically rigorous point of view (far to many subtleties as you've mentioned), wouldn't know where to begin with constructing anything sensible. I'm simply stating the best you can do with the information you've been given (and even then there isn't much).

    It may not seem particularly useful, but when dealing with angular momentum or Electromagnetism it can be a good starting point for some problems.
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