1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple Cross Product Equation Question

  1. Jan 27, 2015 #1

    RJLiberator

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    This is a general question about the equation.
    So, I know that the cross product requires a vector in at least 3 dimensions crossed with another.

    Here is the formula that I am using:
    uxv = Inline6.gif

    My problem is the negative/positive sign orientation in front of the y element and z element.
    I've seen equations for this where there exists only addition between the x,y, and z elements in a cross product. And my teacher tried to explain it to me, and I softly understood, but I want to make sure I know what's going on.

    Why does the cross product equation sometimes have all positive signs between elements, but the one that I use (and works) has a negative and then a positive sign?


    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. Jan 27, 2015 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    It's not clear from what source you are deriving these different formulas for the cross product.

    Here is an article with various formulas for computing it:

    http://en.wikipedia.org/wiki/Cross_product

    I tend to favor the formula which uses the expansion of the 3 x 3 formal determinant. It is easy to remember and expand, and the negative signs take care of themselves.
     
  4. Jan 27, 2015 #3

    RJLiberator

    User Avatar
    Gold Member

    Ah, wiki. Why did I not think of that? :p.

    This appears to have my answer. I will post back if I get confused further.

    Thank you.
     
  5. Jan 27, 2015 #4

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Wiki has some surprisingly good math articles, particularly on the practical aspects of calculation. You can always check what's there with Wolfram's MathWorld site:

    http://mathworld.wolfram.com/
     
  6. Jan 27, 2015 #5


    there is a better video that uses matrix(kin of like craters rule) somewhere. I'll find it later
     
  7. Jan 27, 2015 #6

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    It's Cramer's Rule, not craters rule. Cramer's Rule is used to solve a system of linear equations using determinants.

    The method used to expand the determinant of a 3 x 3 matrix is actually called the Rule of Sarrus. See

    http://en.wikipedia.org/wiki/Rule_of_Sarrus
     
  8. Jan 27, 2015 #7
    I am typing on a tablet.
     
  9. Jan 27, 2015 #8

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Think of the three cyclic permuations XYZ, YZX, ZXY, obtained by taking the letter in the first position and looping it around to be last. So, starting from XYZ we take the X and move it around to last place, shoving the other two to the left---giving YZX. Similarly, we then take the Y and loop it around to be last, to get ZXY. Those orders give you the first parts of each of the three terms:
    [tex] \vec{u} \times \vec{v} = \underbrace{\vec{e}_x ( u_y v_z \cdots)}_{XYZ} +
    \underbrace{\vec{e}_y(u_z v_x \cdots)}_{YZX} + \underbrace{\vec{e}_z (u_x v_y \cdots)}_{ZXY} [/tex]
    Now just complete each bracket ##( )## by putting in the interchange of the first term, with a - sign:
    [tex] \vec{u} \times \vec{v} = \vec{e}_x (u_y v_z - u_z v_y)
    + \vec{e}_y ( u_z v_x - u_x v_z) + \vec{e}_z (u_x v_y - u_y v_x) [/tex]
    Now there are ##+## signs in front of each of the terms.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Simple Cross Product Equation Question
  1. Cross product question (Replies: 9)

Loading...