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Vector equations of lines

  1. Sep 21, 2004 #1
    hi, im having a bit of trouble with this problem

    Find the parametric equations for the line through the point(0,1,2) that is perpen dicular to the line x=1+t, y=1-t, z=2t and intersects this line.

    first i seen that i have a direction vector from the given parametric eqations which is
    vector_v= <1,-1,2> and i can get the parallel line to that by taking <1,1,0>,which i will call vector_r, also from the parametric equations.

    then to find a line perpendicular to this i take (vector_v) x (vector_r)

    (vector_v) x (vector_r) = <-2,2,2>=vector_n

    taking point(0,1,2) and vector_n i get

    <0,1,2> + t<-2,2,2> = (0-2t)i+(1+2t)j+(2+2t)k =

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

    i was wondering if this is how you would go about solving this problem, if not a push in the right direction would be great.

    thanks
     
  2. jcsd
  3. Sep 21, 2004 #2

    mathwonk

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    well you are looking for a point on the given line that, when joined to the given point, makes a vector perpendicular to the given line.

    you have a general point on the given line as (1+t, 1-t,2t) so you know how to subtract it from the given point and then how to dot the result with the direction vector of the given line i guess. that should do it.
    sorry i ignored everything after your first three lines.
     
  4. Sep 22, 2004 #3
    ok, i just want to make sure i got this, when i subtracted the point (0,1,2) from (1,1,0) which came from the parametric equation. and got <-1,0,2>

    then i doted that with the direction vector from the parametric equation which is <1,-1,2> and got <-1,0,4>.

    is that right so far? and how would i get a parametric equation from just one vector?

    thanks
     
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