Cross product - example - Revision history

Alexandk at 21:31, 8 March 2022

2022-03-08T21:31:33Z

Alexandk at 03:51, 8 March 2022

2022-03-08T03:51:06Z

Alexandk at 03:50, 8 March 2022

2022-03-08T03:50:05Z

Alexandk at 03:49, 8 March 2022

2022-03-08T03:49:16Z

Alexandk at 03:48, 8 March 2022

2022-03-08T03:48:51Z

Alexandk at 03:43, 8 March 2022

2022-03-08T03:43:47Z

Alexandk at 03:43, 8 March 2022

2022-03-08T03:43:30Z

Alexandk at 03:42, 8 March 2022

2022-03-08T03:42:51Z

Alexandk at 03:42, 8 March 2022

2022-03-08T03:42:18Z

Alexandk at 03:41, 8 March 2022

2022-03-08T03:41:38Z

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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

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 * 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>
-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?
+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?'''
 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

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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
-* 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>
 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?

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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
-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 {b} = - {\bf b} \times {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>
 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?

← Older revision		Revision as of 03:48, 8 March 2022
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−	Recall that in 3-dimensional space the cross product of two non-collinear vectors <math> {\bf a} </math> and <math> b </math> is defined as a vector <math> {\bf c} = {\bf a} \times {\bf b} </math> such that <math> {\bf c} </math> is perpendicular to the plane spanned by <math> {\bf a} </math> and <math> {\bf b} </math>, the length of <math> {\bf c} </math> is equal to the are of parallelogram with sides <math> {\bf a} </math> and <math> {\bf b} </math>, and the triple <math> {\bf a}, {\bf b}, {\bf c} </math> is positively oriented.	+	Recall that in 3-dimensional space the cross product of two non-collinear vectors <math> {\bf a} </math> and <math> b </math> is defined as a vector <math> {\bf c} = {\bf a} \times {\bf b} </math> such that <math> {\bf c} </math> is perpendicular to the plane spanned by <math> {\bf a} </math> and <math> {\bf b} </math>, the length of <math> {\bf c} </math> is equal to the are of parallelogram with sides <math> {\bf a} </math> and <math> {\bf b} </math>, and the triple <math> {\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> \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>
		+
		+	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>

	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?		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?

← Older revision		Revision as of 03:43, 8 March 2022
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	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
		+
	<math> {\bf a} = \frac{1}{2} \overline{DC}\times \overline{DB} </math>		<math> {\bf a} = \frac{1}{2} \overline{DC}\times \overline{DB} </math>

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 <math> {\bf a} = \frac{1}{2} \overline{DC}\times \overline{DB} </math>
-<math> {\bf a} = \frac{1}{2} \overline{DA}\times \overline{DC} </math>
+<math> {\bf b} = \frac{1}{2} \overline{DA}\times \overline{DC} </math>
-<math> {\bf a} = \frac{1}{2} \overline{DB}\times \overline{DA} </math>
+<math> {\bf c} = \frac{1}{2} \overline{DB}\times \overline{DA} </math>
-<math> {\bf a} = \frac{1}{2} \overline{AB}\times \overline{AC} </math>
+<math> {\bf d} = \frac{1}{2} \overline{AB}\times \overline{AC} </math>
 Therefore

← Older revision		Revision as of 03:42, 8 March 2022
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	Denoting <math> {\bf u} =\overline{DA} </math>, <math> {\bf v} =\overline{DB} </math>, <math> {\bf w} =\overline{DC} </math>, we further get		Denoting <math> {\bf u} =\overline{DA} </math>, <math> {\bf v} =\overline{DB} </math>, <math> {\bf w} =\overline{DC} </math>, we further get

−	<math> {\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>	+	<math> {\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> = \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>

← Older revision		Revision as of 03:41, 8 March 2022
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	<math> {\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> {\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> = \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>		<math> = \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>