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Really easy question about perpendicular lines in 3D?

  1. Sep 29, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the line through (3, 1, -2) that intersects and is perpendicular to the line x = -1 + t, y = -2 + t, z = -1 + t.


    2. Relevant equations
    line l through P (x1, y1, z1) and Q (x2, y2, z2) has the following form

    x = x1 + (x2 - x1)t
    y = y1 + (y2 - y1)t
    z = z1 + (z2 - z1)t

    3. The attempt at a solution

    Alright so I'm pretty sure that this is a REALLY easy problem, but I just started vector calculus and I keep mixing up my steps and my variables. first of all, I'm trying to solve what the point of intersection would be, and I was trying to use the equations above, but then I wasn't really sure what I was doing...can anyone help to point me in the right direction? thank you.
     
  2. jcsd
  3. Sep 29, 2007 #2
    you would use those equations but you need the point Q assuming (3,1,-2) is point P.

    Have you learned the dot product? and that if the dot product of 2 vectors =0 then they are perpendicular.
     
  4. Sep 29, 2007 #3

    D H

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    First thing, if you are going to describe your line parametrically use a parameter other than 't'. How are you going to distinguish it from the 't' used in the parametric description of the target line?

    Can you find a plane on which the desired line lies? If you find the right plane (hint) you will have the problem nearly licked.
     
  5. Sep 30, 2007 #4
    OK so I read over the advice you guys gave me and I'm still a little confused, so what I went ahead and figured out was that

    the equation of a line l(t) = a + tv, where
    x = x1 + at
    y = y1 + bt
    z = z1 + ct
    and a = (x1, y1, z1) and v = (a, b, c)
    where the line points in the direction of v and passes through the tip of a

    so, according to the equations for the original line, x = -1 + t, y = -2 + t, z = -1 + t, the line points in the direction of vector v which = (1, 1, 1)

    then, we need to find the equation of the line which is perpendicular to that vector, so that

    x = x2 + (x3 - x2)p
    y = y2 + (y3 - y2)p
    z = z2 + (z3 - z2)p
    where (3, 1, -2) = (x2, y2, z2)

    according to the logic i used to find the direction of vector v, shouldn't the direction of the perpendicular vector c then be (x3 - x2, y3 - y2, z3 - z2) or (x3 - 3, y3 -1, z3 +2)?

    which means that, when the dot product is used to show that v dot c = 0, we get

    (1)(x3 - 3) + (1)(y3 - 3) + (1)(z3 + 2) = 0

    but once I got here, I realized I didn't know how to solve for x3, y3 or z3, so I wasn't sure if my steps were even right to begin with, or where to go from here

    anyways, sorry for my incompetence, but if someone could give me another little nudge in the right direction! thanks!
     
  6. Sep 30, 2007 #5
    oh boy oh boy, i still need help! please!
     
  7. Sep 30, 2007 #6

    D H

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    I'll expand on my hint: Find the plane normal to the line x = -1 + t, y = -2 + t, z = -1 + t that contains the point (3, 1, -2).
     
  8. Sep 30, 2007 #7
    does that mean finding an equation for the plane? what you're saying makes sense, but I don't really know anything about finding the equations of planes...
     
  9. Sep 30, 2007 #8
    OK, so I looked ahead in the book and figured out how to solve for the equation of a plane, so using the fact that the line points in the direction of the vector (1, 1, 1) = (A, B, C), the equation of the plane to which that vector is perpendicular would be

    A(x - x0) + B(y - y0) + C(z - z0) = 0, where (x0, y0, z0) = (3, 1, -2), the point through which our new line must go, thus

    1(x - 3) + 1(y - 1) + 1(z + 2) = x + y + z - 2 = 0

    then, I plugged in the equations for the original line

    x = -1 + t, y = -2 + t, z = -1 + t

    under the assumption that in order for the lines to intersect, x, y and z must be equal, therefore

    (-1 + t) + (-2 + t) + (- 1 + t) -2 = 3t - 6 = 0
    then t = 2, and I plugged that back into the original equations to find that x1 = 1, y1 = 0, and z1 = 1

    Then, I used the equations

    x = x0 + (x1 - x0)p
    y = y0 + (y1 - y0)p
    z = z0 + (z1 - z0)p

    and found that
    x = 3 - 2p, y = 1 - p, and z = -2 + 3p

    (I used p rather than t to show that I'm using a different parameter than in the line perpendicular to this one). Is my reasoning logical???

    P.S. Thanks for the help!
     
  10. Sep 30, 2007 #9
    ^^ subquestion

    assuming I did everything right, do I use a different parameter for my new equation, or do I have to use the same one in order for the two lines to intersect?
     
  11. Oct 1, 2007 #10
    ^^ anyone that can help confirm my work???
     
  12. Oct 1, 2007 #11

    HallsofIvy

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    You can use whatever parameter you want! In fact, if you are trying to determine WHERE two lines intersect, you had better use different parameters. If you think of the parameter as "time" you can think of the equations as describing something moving along that line. Two lines can can intersect without the "objects" arriving at the intersection at the SAME time!
     
  13. Oct 1, 2007 #12
    great! that helps me understand the issue of parameters
    does my logic look okay too?
     
  14. Oct 1, 2007 #13

    HallsofIvy

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    Yes. Now set the x equations equal to one another, the y equations, the z equations. Notice that this will give you three equations in the two unknown quantities p and t. You can solve two of the equations for p and t and then check to see if they satisfy the third- and there is no reason to think they will! In 3 dimensions, in general, two linear do NOT intersect but are "skew".
     
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