Cross product - example
Recall that in 3-dimensional space the cross product of two non-collinear vectors [math]\displaystyle{ {\bf a} }[/math] and [math]\displaystyle{ b }[/math] is defined as a vector [math]\displaystyle{ {\bf c} = {\bf a} \times {\bf b} }[/math] such that [math]\displaystyle{ {\bf c} }[/math] is perpendicular to the plane spanned by [math]\displaystyle{ {\bf a} }[/math] and [math]\displaystyle{ {\bf b} }[/math], the length of [math]\displaystyle{ {\bf c} }[/math] is equal to the are of parallelogram with sides [math]\displaystyle{ {\bf a} }[/math] and [math]\displaystyle{ {\bf b} }[/math], and the triple [math]\displaystyle{ {\bf a}, {\bf b}, {\bf c} }[/math] is positively oriented.
Consider a tetrahedron in [math]\displaystyle{ \mathbb R^3 }[/math]. For each face of this tetrahedron, draw a vector beginning at some point of the face, pointing outwards, and with length equal to the area of the face. What is the sum of these four vectors?
Denote the vertices of the tetrahedron by [math]\displaystyle{ A, B, C, D }[/math]. Let also [math]\displaystyle{ {\bf a}, {\bf b}, {\bf c}, {\bf d} }[/math] be the vectors attached to the faces opposite to the vertices [math]\displaystyle{ A, B, C, D }[/math], respectively. Then it is obvious that [math]\displaystyle{ {\bf a} = \frac{1}{2} \overline{DC}\times \overline{DB} }[/math]
[math]\displaystyle{ {\bf a} = \frac{1}{2} \overline{DA}\times \overline{DC} }[/math]
[math]\displaystyle{ {\bf a} = \frac{1}{2} \overline{DB}\times \overline{DA} }[/math]
[math]\displaystyle{ {\bf a} = \frac{1}{2} \overline{AB}\times \overline{AC} }[/math]
Therefore
[math]\displaystyle{ {\bf a} + {\bf b} + {\bf c} + {\bf d} = \frac{1}{2} \left[ \overline{DC}\times \overline{DB} + \overline{DA}\times \overline{DC} + \overline{DB}\times \overline{DA} + \overline{AB}\times \overline{AC} \right] = \frac{1}{2} \left[ \overline{DC}\times \overline{DB} + \overline{DA}\times \overline{DC} + \overline{DB}\times \overline{DA} + \left(\overline{DB} -\overline{DA}\right)\times \left(\overline{DC} -\overline{DA}\right) \right] }[/math]
Denoting [math]\displaystyle{ {\bf u} =\overline{DA} }[/math], [math]\displaystyle{ {\bf v} =\overline{DB} }[/math], [math]\displaystyle{ {\bf w} =\overline{DC} }[/math], we further get
[math]\displaystyle{ {\bf a} + {\bf b} + {\bf c} + {\bf d} = \frac{1}{2} \left[ {\bf w}\times {\bf v} + {\bf u}\times {\bf w} + {\bf v}\times {\bf u} + \left({\bf v} -{\bf u} \right)\times \left({\bf w} -{\bf u} \right) \right] = \frac{1}{2} \left[ {\bf w}\times {\bf v} + {\bf u}\times {\bf w} + {\bf v}\times {\bf u} + {\bf v}\times {\bf w} - {\bf u}\times {\bf w} - {\bf v}\times {\bf u} + {\bf u}\times {\bf u} \right] = = \frac{1}{2} \left[ {\bf w}\times {\bf v} + {\bf v}\times {\bf w}\right] = \frac{1}{2} \left[ {\bf w}\times {\bf v} - {\bf w}\times {\bf v}\right] ={\bf 0}. }[/math]