Can You Plot the Parametrization for 3D Geometric Algebra?

[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?

 

[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 traveling to the line first in the most direct manner then along the line

c and d are left to you

 

[M98Ranger]

Yes, it is possible to plot the parametrization for 3D geometric algebra. However, it may be difficult to explain the process in a written response as it involves visualizing and manipulating geometric objects in three dimensions.

To plot the parametrization x=tu+y, you can start by choosing values for the parameter t and then calculating the corresponding values for the vector x. For example, if t=0, then x=y. This means that for every value of t, the vector x will lie on the same line as the vector y, but at a different position along the line.

To visualize this, you can plot the vector y as a fixed point on the line and then plot x for different values of t to see how it moves along the line. You can also plot multiple points on the line by varying the values of t and connecting them to create a visual representation of the entire line.

Now, let's move on to the given statements and their proofs:

a. To show that for all points on the line we have x(wedge)u=y(wedge)u, we can use the properties of the wedge product, which is a multiplication operation in geometric algebra. Since the vectors x and y lie on the same line, they are parallel and can be expressed as x=ay and y=bx for some scalars a and b. Therefore, we have x(wedge)u=(ay)(wedge)u=a(y(wedge)u) and y(wedge)u=(bx)(wedge)u=b(x(wedge)u). Since a and b are scalars, they can be pulled out of the wedge product, giving us a(y(wedge)u) and b(x(wedge)u). Since x=y, we have a=b, and therefore, a(y(wedge)u)=b(x(wedge)u). This proves that x(wedge)u=y(wedge)u for all points on the line.

b. To 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, we can use the properties of the dot product, which is another multiplication operation in geometric algebra. The dot product between two parallel vectors is equal to the product of their lengths. Since d is the vector pointing from the origin onto a point on the line, its length is the shortest possible distance from the origin to the line.