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Finding point-normal equations

  1. Oct 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Say that I have a problem along the lines of: Find the point-normal equation for the line L.

    L=u1+k*u2, where k is ℝ.

    And then I'm given u1 and u2, which both are 2×1 vectors.

    Then L will also be a 2×1 vector (which consists of a11 and a21), which I guess is ok (since it gives the two coordinats in R2).

    However the point-normal equation is in the form of (for some constants a and b and some points x0 and y0 on the line L):

    a(x-x0)+b(y-0)=0.

    So here comes my question: In the vector L: Are a11 or a21 x?


    2. Relevant equations
    -


    3. The attempt at a solution
    See above.
     
  2. jcsd
  3. Oct 9, 2013 #2

    LCKurtz

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    Let ##u_1=\langle a, b\rangle## and ##u_2=\langle d_1,d_2\rangle##. Then your line is ##\langle x,y\rangle = \langle a, b\rangle + k\langle d_1,d_2\rangle##. Eliminate the ##k## between the ##x## and ##y## equations and put it in your required form. Note that ##(a,b)## is a point on the line.
     
  4. Oct 9, 2013 #3
    Two questions:

    1. (Perhaps a dumb question) Is ##\langle x,y\rangle## another way of writing a 2×1 matrix where x=a11 and y=a21? (I'm really new to linear algebra...)

    2. If I'm correct about question 1, then I still don't see how I can "translate" ##\langle x,y\rangle = \langle a, b\rangle + k\langle d_1,d_2\rangle## to the form of

    a(x-x0)+b(y-y0)=0

    (That was what my original question was all about.)

    Thanks for the help LCKurtz!
     
  5. Oct 9, 2013 #4

    LCKurtz

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    It does't matter whether you use a 2x1 or 1x2 matrix in this problem. You have two entries either way.

    Do what I suggested. Write the equations for ##x## and ##y## from the given matrix equation and eliminate the ##k## parameter. Try it.
     
  6. Oct 9, 2013 #5
    I'm not sure what you mean with "eliminate the ##k## parameter".

    But I guess this is a start?

    ##x=a+kd_1##

    and

    ##y=b+kd_2##

    now let me put ##x## and ##y## into the "form" of the point-normal equation. Which I will call:

    ##p(x-x_0)+q(y-y_0)=0##, where ##p## and ##q## are constants.

    We know that ##x_0=a## and ##y_0=b## are solutions because ##k## could be ##0##.

    so

    ##p((a+kd_1)-a)+q((b+kd_2)-b)=0##

    ##p(kd_1)+q(kd_2)=0##

    ##k(pd_1+qd_2)=0##

    ##pd_1+qd_2=0##

    where ##(p,d)## is the normal to ##(x,y)##. But how do you get ##(p,d)##?

    Does the calculations look ok LCKurtz?
     
    Last edited: Oct 9, 2013
  7. Oct 9, 2013 #6

    LCKurtz

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    Not quite. From ##x=a+kd_1,~y=b+kd_2## you eliminate ##k##. Solve each equation for ##k## and set them equal:$$
    k = \frac {x-a}{d_1} = \frac {y-b}{d_2}$$Put that last equation in the form$$
    (?)(x-a) + (??)(y-b) = 0$$and you will have your form.
     
  8. Oct 9, 2013 #7
    I'm a bit lost. :/ Any more tips?
     
  9. Oct 10, 2013 #8
    I know that the ##(?)## ##(??)## are the normal to the line ##\langle x,y\rangle = \langle a, b\rangle + k\langle d_1,d_2\rangle##. However what I don't know is how you should obtain the normal for that line...
     
  10. Oct 10, 2013 #9

    HallsofIvy

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    You say you are taking linear algebra but the problems you are having are all much more basic algebra matters.

    [tex]\frac{x-a}{d_1}= \frac{y- b}{d_2}[/tex]

    An obvious thing to do is to multiply both sides by [itex]d_1d_2[/itex] to get rid of the fractions:
    [tex]d_2(x- a)= d_1(y- b)[/tex]
    [tex]d_2(x- a)- d_1(y- b)= 0[/tex]
     
  11. Oct 10, 2013 #10
    I feel like such an idiot now, that was my initial thought but I dismissed it because it felt like it was too easy to be the right answer. :redface:
     
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