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Vector geometry help

  1. Jan 28, 2005 #1
    I have a question about solving the following problem
    "Find the line through (3,1,-2) that intersects and is perpendicular to the line x=-1+t, y=-2+t, z=-1+t."

    My question is regarding the intersection part, I know how to find the equation of the line that is perpendicular to another line but not both perpendicular and intersecting.

  2. jcsd
  3. Jan 28, 2005 #2
    I'd think your forgeting that the dot product of two perpendicular vectors is 0 think about it...
  4. Jan 28, 2005 #3
    dot product....
    the answer is as easy as you can think


    why the hell you need dot product?
  5. Jan 28, 2005 #4
    If two lines are perpendicular to each other, don't you think they're already intersecting?
  6. Jan 28, 2005 #5
    And that line is perpendicular to x=-1+t, y=-2+t, z=-1+t?

    No, two perpendicular lines in R^3 don't necessarily intersect.
    Last edited: Jan 28, 2005
  7. Jan 28, 2005 #6
    Wouldn't (3,1,-2)+t(1,1,1) be correct for the line perpendicular to x=-1+t, y=-2+t, z=-1+t since the vector <1,1,1> is coming from that?
  8. Jan 28, 2005 #7
    A direction vector of (3,1,-2)+t(1,1,1) is (1,1,1), and a direction vector of x=-1+t, y=-2+t, z=-1+t is (1,1,1). Two lines are perpendicular if and only if their direction vectors are perpendicular. Surely (1,1,1) is not perpendicular to (1,1,1)?
  9. Jan 28, 2005 #8
    So, how would you go about finding a line that is perpendicular?
  10. Jan 28, 2005 #9
    Ah.. I over-looked that.
  11. Jan 28, 2005 #10
    The dot product between their direction vectors should be 0.
  12. Jan 28, 2005 #11


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    Note: there exist an infinite number of lines (in 3D) through a given point and perpendicular to a given line. Perhaps part of the problem is that there are too many answer. You should know that, in 2D, a perpendicular to <a, b> is <b, -a> since that way <a, b>.<b,-a>= ab-ab= 0. In 3d, the simplest thing to do is just take one of the components 0 so that the problem goes back to 2D.

    To find a perpendicular to <1, 1, 1> try <0, a, b> What must a, b be?
    Of course, <a, 0, b> and <a, b, 0> would work as well.
  13. Jan 28, 2005 #12
    the line has the vector <1,1,1> as its direction
    dot with the vector that lies on the perp line c...
    \vect{c} = <c_x, c_y, c_z>[/tex]
    [tex]<1,1,1> dot <c_x, c_y,c_z> = 0 [/tex]
    gives you
    c_x = -c_y - c_z [/tex]
    [tex]c_y free [/tex]
    [tex]c_z free

    subracting the vector that points to the point on the line of perpendicularity (r), and the vector the point we are trying to make a line through gives us the vector in the direction of the line which is c...
    <r_x, r_y, r_z> - <3,1,-2> = <-c_y-c_z, c_y, c_z>
    [tex]r_x - 3 = -c_y - c_z [/tex]
    [tex]r_y - 1 = c_y [/tex]
    [tex]r_z + 2 = c_z [/tex]

    however the terminal point of vector r must lie on the origal line..there fore..

    r_x = t -1
    r_y = t - 2
    r_z = t - 1 ...substute...

    (t-1) - 3 = -c_y - c_z [/tex]
    [tex] (t-2)-1 = c_y [/tex]
    [tex](t - 1) + 2 = c_z [/tex]

    which we can easily solve for t...t = 2 put find the vector c by substuting in the t value...you'll get...
    [tex] \vect{c} = <-2, -1, 3> [/tex]
    now we can form the line equation
    x = -2t +3
    y = -t + 1
    z = 3t -2

    done...check the dot product its perp
    Last edited: Jan 28, 2005
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