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3D Geometric Algebra

  1. Dec 13, 2003 #1
    Consider 3D geometric algebra. Let all points on a line be given by the parametrization x=tu+y, in which the parameter runs from minus infinity to plus infinity.

    a. Show that for all points on the line we have
    x(wedge)u=y(wedge)u.


    b. Show that the vector d pointing from the origin onto a point on the line, such that d has the shortest length, satisfies
    d. u=0.


    c. Show no that this vector d is given by
    d=(y(wedge)u)(u)^(-1).


    d. Given two lines given by parametrizations
    s u1+y1 and
    r u2+y2,
    where the parameters s and r run from minus to plus infinity.
    Show that for the two lines to have an intersection we must have that
    (y1-y2)(wedge)(u1-u2)
    is proportional to
    u1)(wedge)(u2.



    I posted this because i wanted help first plotting the parametrization and i figured ill be asking questions about the rest of it later, thus i typed out the whole problem. I realize this plotting is hard to explain on an internet thread, but maybe some tips?
     
  2. jcsd
  3. Jan 16, 2004 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    let's just deal with a first.

    if x = u + ty

    then wedging with u on both sides gives you...?

    b is geometrically obvious by the triangle inequality amogst other things. Just think about descirbing the distance vectr from the origin to a point on the line in terms of travelling to the line first in the most direct manner then along the line

    c and d are left to you
     
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