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Conference rusure::math

Title:	Mathematics at DEC

Moderator:	RUSURE::EDP

Created:	Mon Feb 03 1986
Last Modified:	Fri Jun 06 1997
Last Successful Update:	Fri Jun 06 1997
Number of topics:	2083
Total number of notes:	14613

600.0. "Pythagorean Theorem in 3-Space" by CACHE::MARSHALL (beware the fractal dragon) Tue Oct 21 1986 01:32

    A "right tetrahedron" has a vertex with three right angles. If in
    a right tetrahedron A,B,C denote the areas of the three faces that
    share the "right vertex" and D denotes the area of the face opposite
    the right vertex, show that A^2 + B^2 + C^2 = D^2
                                                   
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600.1 dual space armwave... ENGINE::ROTH Tue Oct 21 1986 09:26 28

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600.1	dual space armwave...	ENGINE::ROTH		`Tue Oct 21 1986 09:26`	28
	The projection of the area D on the 3 coordinate planes will give the areas A, B, and C. That is A = D * cos(theta-a) = D * Ua B = D * cos(theta-b) = D * Ub C = D * cos(theta-c) = D * Uc where cos(theta-) are the dot products (components) of the unit vector normal to the surface of D, and the coordinate axes. So we have at once A^2 + D^2 + C^2 = D^2 (Ua^2 + Ub^2 + Uc^2) = D^2 This illustrates the idea of 'dual space' in linar algebra - the space of all vectors that are at right angles to all but one of each of the other vectors in the origional space. In 3 space this is the cross product, but it works in space of any dimension. It is also where 'covariant' and 'contravariant' transformations come from in tensor analysis. A practical use: Suppose you are writing shading and rendering software and want to know how to transform the normal vectors of a given surface when it is transformed by some transformation matrix. (consider what scaling one coordinate only does to the normal of a sphere for example). The notion of dual space gives the answer... - Jim

    The projection of the area D on the 3 coordinate planes will give
    the areas A, B, and C.  That is

	A = D * cos(theta-a) = D * Ua
	B = D * cos(theta-b) = D * Ub
	C = D * cos(theta-c) = D * Uc

    where cos(theta-*) are the dot products (components) of the unit vector
    normal to the surface of D, and the coordinate axes.

    So we have at once

	A^2 + D^2 + C^2 = D^2 * (Ua^2 + Ub^2 + Uc^2) = D^2

    This illustrates the idea of 'dual space' in linar algebra - the
    space of all vectors that are at right angles to all but one of
    each of the other vectors in the origional space.  In 3 space
    this is the cross product, but it works in space of any dimension.
    It is also where 'covariant' and 'contravariant' transformations
    come from in tensor analysis.

    A practical use:  Suppose you are writing shading and rendering software
    and want to know how to transform the normal vectors of a given surface
    when it is transformed by some transformation matrix. (consider what
    scaling one coordinate only does to the normal of a sphere for example).
    The notion of dual space gives the answer...

    - Jim