# 3D Geometric Algebra

1. Dec 13, 2003

### scariari

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. Jan 16, 2004

### matt grime

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