Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dot Product

  1. Apr 20, 2007 #1
    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.
  2. jcsd
  3. Apr 20, 2007 #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.
  4. Apr 20, 2007 #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.

  5. Apr 20, 2007 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Where, specifically, are you lost?
  6. Apr 20, 2007 #5

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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

    I'm niot sure what you mean.
  7. Apr 21, 2007 #6


    User Avatar
    Science Advisor

    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.
  8. Apr 21, 2007 #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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook