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Cross product!

  1. Oct 18, 2009 #1
    Why is the cross product defined this way

    (blah)i + (blah)j + (blah)k = u

    and used this way

    (blah)i - (blah)j + (blah)k = u
     
  2. jcsd
  3. Oct 18, 2009 #2
    Well; simply make the second blah in the first line a -blah,- it remains a " blah" doesn't it? :)

    The " trick" you mention makes use of a determinant to calculate aXb .
     
  4. Oct 18, 2009 #3

    HallsofIvy

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    Making the second coeffient "-" makes it fit a nice mnemonic:
    expanding a determinant along the top row
    [tex]\left|\begin{array}{ccc}a & b & c \\ d & e & f\\ g & h & i\end{array}\right|= a\left|\begin{array}{cc}e & f \\ h & i\end{array}\right|- b\left|\begin{array}{cc}d & f \\ g & i\end{array}\right|+ c\left|\begin{array}{cc}d& e \\ g & h\end{array}\right|[/tex]

    If [itex]\vec{u}= a\vec{i}+ b\vec{j}+ c\vec{k}[/itex] and [itex]\vec{v}= x\vec{i}+ y\vec{j}+ z\vec{k}[/itex], then writing [itex]\ve{u}\times \vec{v}= (bz-cy)\vec{i}+ (az- cx)\vec{j}+ (ay- bx)\vec{k}[/itex] as [itex](bz-cy)\vec{i}- (cx- az)\vec{j}+ (au= bx-)\vec{k}[/itex] makes it clearer that we can think of it as
    [tex]\vec{u}\times\vec{v}= \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ a & b & c \\ x & y & z\end{array}\right|[/tex]
     
  5. Oct 18, 2009 #4
    Thanks ivy!

    We basically got the formula thrown at us because it is useful when finding torques. I see that cross products are also going to be covered in our linear algebra class, but we have not yet covered determinants in the class so cross products are a little out of my reach in terms of gaining a full understanding. I think your explanation nudged me in the right direction.
     
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