Grassmann Algebra

GeometricInterpretations.nb 25
L � �� � x1�e1 � x2�e2���� � y2�e2 � y3�e3�
L � �� � x1�e1 � x2�e2���� � z1�e1 � z3�e3�
L � �� � y2�e2 � y3�e3���� � z1�e1 � z3�e3�
Graphic of a line through 3 points, one in each coordinate plane.
Information required to express a line in a 3-plane
As with a line in a 2-plane, we find that a line in a 3-plane is expressed with the minimum
number of parameters by expressing it as the product of two points, each in one of the
coordinate planes. In this form, there are just 4 independent scalar parameters (coordinates)
required to express the line.
Checking the invariance of the description of the line
We can use GrassmannAlgebra to explore the invariance of how a line is expressed. Again,
suppose we are in a 3-plane.
�3
��, e1, e2, e3�
Define a line L as the exterior product of two points, one in the � � e1 � e2 coordinate plane
and one in the � � e2 � e3 coordinate plane.
L � �� � x1�e1 � x2�e2���� � y2�e2 � y3�e3�;
Declare the coordinates as scalars.
DeclareExtraScalars��x1, x2, y2, y3��;
Verify that the intersection of the line with the first coordinate plane does indeed give a point
congruent to the first point.
P1 � ��L ��� � e1 � e2�� �� ToCongruenceForm
�� � e1 x1 � e2 x2� y3
�������������������������������� ������������������
�
Next determine the (weighted) point in the line in the third coordinate plane (the coordinate
plane which did not figure in the original definition of the line).
P2 � ��L ��� � e1 � e3�� �� ToCongruenceForm
� �x2 � y2� � e1 x1 y2 � e3 x2 y3
�������������������������������� �������������������������������� ���������
�
We can now confirm that the exterior product of these two points is still congruent to the
original specification of the line by expanding the quotient of the two expressions and showing
that it reduces to a scalar.
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