Grassmann Algebra

GeometricInterpretations.nb 32
L � � ��ae�c f� bg� e1 � bf�a g� ce� e2 � cg��b e� af� e3�
�a bef�g � c� e1 � e2
�a ceg��b � f� e1 � e3
�b cfg��e � a� e2 � e3;
To verify that this line lies in both planes, we can take its exterior product with each of the
planes and show the result to be zero.
����1 � L, �2 � L��
�0, 0�
In the special case in which the planes are parallel, their intersection is no longer a line, but a
bivector defining their common 2-direction.
� Example: The osculating plane to a curve
� The problem
Show that the osculating planes at any three points to the curve defined by:
P � � � ue1 � u 2 �e2 � u 3 �e3
intersect at a point coplanar with these three points.
� The solution
The osculating plane � to the curve at the point P is given by ��P � P �
� P �
, where u is a scalar
parametrizing the curve, and P �
and P �
are the first and second derivatives of P with respect to u.
��P � P �
� P �
;
P � � � ue1 � u 2 �e2 � u 3 �e3;
P �
P �
� e1 � 2�u e2 � 3�u 2 �e3;
� 2�e2 � 6�u e3;
We can declare the space to be a 3-plane and the parameter u (and subscripts of it) to be a
scalar, and then use GrassmannSimplify to derive the expression for the osculating plane
as a function of u.
�3; DeclareExtraScalars��u, u_ ��; ����
2 � � e1 � e2 � 6u� � e1 � e3 � 6u 2 � � e2 � e3 � 2u 3 e1 � e2 � e3
Now select any three points on the curve P1 , P2 , and P3 .
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