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The logic behind the cross product with units

  1. Sep 16, 2008 #1
    The cross product equals to the area of the parallelogram defined by the two vectors (at least in R^3). So if working on vectors which units

    v_1 = (1,2,3)m
    v_2 = (3,4,5)m

    it correctly returns the according area. However, if used to get a vector perpendicular to each of the vectors the resulting vectors has the unit m^2.

    How does this fit together? Whats the term for the "dimension" of this vector anyway, since it's actual dimension is 3 (am I mistaken here?) as in the field of R^3.

    Thanks for your help, PF
  2. jcsd
  3. Sep 17, 2008 #2


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    I'm not clear what your question is. Yes, v_1, v_2, and their cross product all have dimension 3. But I don't know what you mean by "the term for the dimension".

    If the "m" on v_1 and v_2 means "meters" and denotes that the components are in meters, yes, their cross product will have units of m^2, an area. General vectors, of course, either don't have units, or if in a specific application, units appropriate to that application. In that case the cross product of two vectors will have "square units" since it is a multiplication- just as the dot product of two such vectors will have "square units".

    Yes, you can think of the cross product of two vectors as representing an area: here the area "between" the two vectors. In fact, if you are given a surface represented by a vector equation, [itex]\vec{r}= f(u,v)\vec{i}+ g(u,v)\vec{j}+ h(u,v)\vec{k}[/itex] with the two parameters u and v, then the derivatives with respect to u and v, [itex]\vec{r}_u= f_u(u,v)\vec{i}+ g_u(u,v)\vec{j}+ h_u(u,v)\vec{k}[/itex] and [itex]\vec{r}_v= f_v(u,v)\vec{i}+ g_v(u,v)\vec{j}+ h_v(u,v)\vec{k}[/itex] are vectors tangent to the surface and their cross product is a vector normal to the surface and its length is the "differential of surface area".
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