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Determinant, dot product, and cross product

  1. Feb 20, 2010 #1
    I am reading through David Widder's Advanced Calculus and he abbreviates a determinant


    \left( \begin{array}{cccc}

    r_{1} \ s_{1} \ t_{1}\\

    r_{2} \ s_{2} \ t_{2}\\

    r_{3} \ s_{3} \ t_{3}\\

    \end{array} \right)

    And refers to it by (rst). He then states that expanding by the minors of a given
    column, we have:


    (rst) = r \cdot (s \times t) = s \cdot (t \times r) = t \cdot (r \times s)


    Now, I worked it out by looking at the cofactors of r_1, r_2, and r_3 which are the components of the vector (s x t) and confirmed it holds. But how can I see
    this property without having to do that? Is there another way to see this say algebraically or geometrically? Some more intuitive way, perhaps?
  2. jcsd
  3. Feb 20, 2010 #2
    this is just the coordinate formula for the volume of the parallelepiped spanned by the row vectors - up to a sign.
  4. Feb 21, 2010 #3
    Indeed. The triple scalar product can be shown geometrically to be the signed volume of the parallelepiped spanned by the three vectors, while the determinant can be shown to equal the same signed volume independently, and thus give you a component based formula for the triple scalar product. In some texts, this is how the determinant is defined, and the component based formula is derived from that definition.
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