Difference between revisions of "Cross product - example"
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Recall also that the definition of the cross product can be extended to arbitrary vectors in <math> \mathbb R^3 </math> so that this operation is | Recall also that the definition of the cross product can be extended to arbitrary vectors in <math> \mathbb R^3 </math> so that this operation is | ||
| − | * anti-symmetric, i.e. <math> {\bf a} \times {b} = - {\bf b} \times {a} </math> (and consequently <math> {\bf a}\times {\bf a} = {\bf 0} </math> | + | * anti-symmetric, i.e. <math> {\bf a} \times {\bf b} = - {\bf b} \times {\bf a} </math> (and consequently <math> {\bf a}\times {\bf a} = {\bf 0} </math>; |
| − | * linear, i.e. <math> {\bf a}\times (\lambda {\bf b} + \mu {\bf c}) = \lambda ({\bf a}\times {\bf b}) + \mu ({\bf a} \times {\bf c}) </math> | + | * linear, i.e. <math> {\bf a}\times (\lambda {\bf b} + \mu {\bf c}) = \lambda ({\bf a}\times {\bf b}) + \mu ({\bf a} \times {\bf c}) .</math> |
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| + | Consider a tetrahedron in <math> \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?''' | ||
| + | [[File:Tetrahedron.png|alt=Tetrahedron and vectos|thumb|Tetrahedron]] | ||
Denote the vertices of the tetrahedron by <math> A, B, C, D </math>. Let also <math> {\bf a}, {\bf b}, {\bf c}, {\bf d} </math> be the vectors attached to the faces opposite to the vertices <math> A, B, C, D </math>, respectively. Then it is obvious that | Denote the vertices of the tetrahedron by <math> A, B, C, D </math>. Let also <math> {\bf a}, {\bf b}, {\bf c}, {\bf d} </math> be the vectors attached to the faces opposite to the vertices <math> A, B, C, D </math>, respectively. Then it is obvious that | ||
Latest revision as of 17:31, 8 March 2022
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.
Recall also that the definition of the cross product can be extended to arbitrary vectors in [math]\displaystyle{ \mathbb R^3 }[/math] so that this operation is
- anti-symmetric, i.e. [math]\displaystyle{ {\bf a} \times {\bf b} = - {\bf b} \times {\bf a} }[/math] (and consequently [math]\displaystyle{ {\bf a}\times {\bf a} = {\bf 0} }[/math];
- linear, i.e. [math]\displaystyle{ {\bf a}\times (\lambda {\bf b} + \mu {\bf c}) = \lambda ({\bf a}\times {\bf b}) + \mu ({\bf a} \times {\bf c}) . }[/math]
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 b} = \frac{1}{2} \overline{DA}\times \overline{DC} }[/math]
[math]\displaystyle{ {\bf c} = \frac{1}{2} \overline{DB}\times \overline{DA} }[/math]
[math]\displaystyle{ {\bf d} = \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] = }[/math]
[math]\displaystyle{ = \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]