Trouble Understanding Geometric Algebra: Seeking Guidance

In summary: That clears it up for me.In summary, the line segment from y to tx is given by y- tx and that must be perpendicular to x (I would have used "tx" but the vector from 0 to any point on that line, in particular x where t= 1 is parallel to tx and so perpendicular to any line perpendicular to tx)- their dot product must be 0: (y-tx). x= 0. Multiplying that out, y.x- t x.x= y.x- t|x||2= 0 (If we had used tx instead of x, we would now have ty.x- t2|x||2= t(y.x- t|x||2)=
  • #1
matheinste
1,068
0
This example appears in a set of notes entitled Geometric Algebra.

I cannot follow the first half of the example. Is the reasoning incorrect.

View attachment Doc1.doc

Thanks. Matheinste.
 
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  • #2
A lot of people won't open your .doc file because it could potentially contain a virus. Try turning it into a pdf (there are loads of free sites online which will do this for you), you'll probably get a better response.
 
  • #3
Thanks for the advice Flux = Rad.

View attachment Doc1.pdf

This appears in a set of files on the web making up a book called Geometric Algebra. Is there something wrong with the question or answer to the first problem or am I being stupid.

Matheinste.
 
  • #4
Where, specifically, are you lost?
 
  • #5
matheinste said:
This appears in a set of files on the web making up a book called Geometric Algebra. Is there something wrong with the question or answer to the first problem or am I being stupid.

I don't see anything wrong with the question or answer to the first problem.

I'm niot sure what you mean.
 
  • #6
I have to agree with the others. It seems clear to me. Exactly where are you. You are given a line through O and x so that any point can be written as the vector xt for some number t. You are given a point y not on the line and asked to find t so that the line through y and tx is perpendicular to the original line.

Okay, the line segment from y to tx is given by y- tx and that must be perpendicular to x (I would have used "tx" but the vector from 0 to any point on that line, in particular x where t= 1 is parallel to tx and so perpendicular to any line perpendicular to tx)- their dot product must be 0: (y-tx). x= 0. Multiplying that out, y.x- t x.x= y.x- t|x||2= 0 (If we had used tx instead of x, we would now have ty.x- t2|x||2= t(y.x- t|x||2)= 0 and cancel that t leading to the same equation). Solving for t, t= y.x/||x||2. In particular the point is tx= (y.x/||x||2)x. To find the length, put that into ||y- tx||.

All I've really done is repeat what was said in your pdf file. If you need more, tell us what is confusing you.
 
  • #7
Hello all. I see my mistake now, Just a basic slip in algebraic manipulationn and misinterpreting y as a point rather than a vector. So the answer to my question is yes--- I was being stupid.

Thanks Matheinste.
 

1. What is geometric algebra?

Geometric algebra is a mathematical framework that extends traditional vector algebra to include both scalars and higher-dimensional geometric entities, such as points, lines, planes, and volumes. It allows for the representation and manipulation of geometric objects in a more intuitive and concise way compared to traditional methods.

2. Why is geometric algebra useful?

Geometric algebra is useful because it provides a unified and powerful language for solving problems in geometry, physics, and engineering. It allows for the representation of complex geometric concepts in a simpler and more elegant way, making it easier to understand and apply in various fields.

3. How is geometric algebra different from traditional vector algebra?

Geometric algebra differs from traditional vector algebra in that it allows for the representation of both scalars and higher-dimensional geometric objects using a single mathematical framework. Traditional vector algebra only deals with vectors and matrices, while geometric algebra can represent a wider range of mathematical entities.

4. Is it necessary to have a strong mathematical background to understand geometric algebra?

While a strong mathematical background can certainly be helpful in understanding geometric algebra, it is not necessary. The concept of geometric algebra can be understood by anyone with a basic understanding of vectors and matrices. With practice and patience, anyone can develop a solid understanding of geometric algebra.

5. Are there any practical applications of geometric algebra?

Yes, there are many practical applications of geometric algebra in various fields such as computer graphics, robotics, physics, and engineering. Geometric algebra has been used to solve problems in computer vision, control systems, and even quantum mechanics. Its versatility and ability to represent complex geometric concepts make it a valuable tool in many areas of science and technology.

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