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Help needed finding vector, parametric, symmetric equation

  1. Mar 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the vector, parametric and symmetric equations of a line that intersect both line 1 and line 2 at 90°.

    line 1:
    x = 4 + 2t
    y = 8 + 3t
    z = -1 - 4t

    line 2:
    x = 7 - 6t
    y = 2+ t
    z = -1 + 2t

    2. Relevant equations

    not sure. I am not asking for the answer this question, but rather a hint as to how to go about solving this. However, this doesn't mean that I would hate seeing the answer, because it will help me see if I am taking the hint the right way.

    3. The attempt at a solution
     
  2. jcsd
  3. Mar 30, 2012 #2

    HallsofIvy

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    I would start by finding the formula for all planes perpendicular to the first line in the form 2x+ 3y- 4z= C. Then find C such that that plane is perpendicular to the second line.
     
  4. Mar 30, 2012 #3
    now this is something that none of my teachers have explained to me: what does C do exactly? I have been told that it affects how the graph looks, but I don't understand how. and are you saying that C should be the same in both equations? or negative reciprocals?
     
  5. Mar 30, 2012 #4
    and another question: for the equation you gave above (2x+ 3y- 4z= C), shouldn't the normal vector go where you put 2, 3, and 4? or is this sort of equation different from scalar equations?
     
  6. Mar 30, 2012 #5

    LCKurtz

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    Here's a couple more suggestions how you might work that problem. You can find the normal direction by crossing the two direction vectors. Call that direction ##\vec n##. Then take a general point on the first line, with parameter ##t##, add to it a constant times your normal vector ##c\vec n## and set that equal to a general point on your second line with parameter ##s##. Looking at the components of that equation will give you three equations in the unknowns ##t,\ s,\ c##. Once you know ##t## and ##s## you will know the two points your line must go through.

    Alternatively you can treat it as a calculus problem. Minimize the square of the distance between general points on the first and second lines. Get ##s## and ##t## that way.
     
  7. Mar 31, 2012 #6
    so your saying this:

    (t+w) x cn = (q+s)

    where w is a point on line 1
    where q is a point on line 2

    where t, c, and s are unknowns, or the symmetric equations for x, y, or z?

    And I am guessing that the t is the unknown from the parametric equations I gave for line 1,
    s is the unknown from the parametric equation for line 2 if they are unknowns.
     
    Last edited: Mar 31, 2012
  8. Apr 1, 2012 #7

    LCKurtz

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    That is very confused. Your two lines are$$
    \vec R_1 = \langle 4,8,-1\rangle + t \langle 2,3,-1\rangle$$ $$
    \vec R_2 = \langle 7,2,-1\rangle + s\langle -6,1,2 \rangle$$Now calculate the normal vector and try what I suggested.
     
  9. Apr 2, 2012 #8
    I understood that part, but what I don't get is what your doing with the normal vector. from your earlier explanation, it seems as though [4,8,-1] is being multiplied by the normal.
     
  10. Apr 2, 2012 #9

    LCKurtz

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    "Your" is a possessive pronoun. If you mean "you are" you use the contraction "you're".

    Have you calculated the normal yet? I'm waiting for you to actually try my suggestion.
     
  11. Apr 2, 2012 #10
    dude, I have spent the last three weeks doing quite a bit of finding normals:
    [2,3,-4] x [-6,1,2] = [(3)(2)-(1)(-4),(-4)(-6)-(2)(2),(2)(1)-(-6)(3)]
    = [6+3,20+4,2+18]
    = [9,24,20]

    And there is the normal.

    but as I said, the one thing I don't understand from the explanation you gave is how you use the normal vector (I am very weak in word problems, and I have tried to change that with little avail)
     
  12. Apr 2, 2012 #11

    LCKurtz

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    Apparently you are a little weak on your arithmetic too, since you have 2 of the 3 components wrong. You should get [10,20,20]. And to keep things simple you can use ##\vec n =\langle 1,2,2\rangle##. Now you want ##\vec R_1 +c\vec n = \vec R_2##. This gives$$
    \langle 4,8,-1\rangle + t \langle 2,3,-4\rangle + c\langle 1,2,2\rangle = \langle 7,2,-1\rangle + s\langle -6,1,2 \rangle$$This says some point on the first line plus some multiple of the normal has to get you some point on the second line. You have three unknowns ##t, c, s## and three equations by equating components.
     
  13. Apr 2, 2012 #12
    okay, ill try that out and get back to you.
     
    Last edited: Apr 3, 2012
  14. Feb 28, 2013 #13
    Does it matter whether you use (1) R1-R2=cn or (2) R1+cn=R2 (the form which you've outlined).
    I ask this because I've seen other individuals use (1), and final answers vary. As (1) provides you with s=1, t=-1 and n=1, and I used (2) and which provides me with s=-1...so which method is right? or are they both right?
    (Please don't mind my notation).
     
  15. Feb 28, 2013 #14

    LCKurtz

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    This thread is almost a year old and the OP disappeared long ago. Also it is not clear to whom you are replying or to what you are following up. If you have a topic you wish to discuss please start a new thread.
     
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