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Simple Vector Problem

  1. Jan 17, 2008 #1
    1. The problem statement, all variables and given/known data
    Find a plane that is perpendicular to the line [tex] \vec{L}(t) = (5,0,2)t + (3,-1,1)[/tex] and passes through the point [tex] (5,-5,0) [/tex]

    2. Relevant equations
    The equation of the plane that P through [tex] (x_{0},y_{0},z_{0}) [/tex] that has a normal vector [tex] \vec{n} = A \vec{i} + B \vec{j} + C\vec{k} [/tex] is:
    [tex] A(x - x_{0}) + B(y - y_{0}) + C(z - z_{0}) = 0 [/tex]
    that is, [tex] (x,y,z) \in P [/tex]

    3. The attempt at a solution
    [tex] \vec{L}(t) = (5t + 3, -1, 2t + 1) [/tex]
    let [tex] t = 1 \Rightarrow \vec{L}(1) = (8,-1,3) = (A,B,C)[/tex]
    [tex](x_{0},y_{0},z_{0}) = (5,-1,0) [/tex]

    [tex] A(x - x_{0}) + B(y - y_{0}) + C(z - z_{0}) = 0 [/tex]
    [tex](8)(x - 5) + (-1)(y +1) + (3)(z - 0) = 0 [/tex]
    [tex]8x - 40 - y -1 + 3z = 0; [/tex]
    [tex]8x - y + 3z = 41 [/tex]
    Last edited: Jan 17, 2008
  2. jcsd
  3. Jan 17, 2008 #2
    The correct answer is [tex] 5x +2z = 25[/tex]

    What am I doing wrong?
  4. Jan 17, 2008 #3
    I have to assume that you haven't yet finished your corrections, because the L(t) functions seem different. Let us know when you've updated it correctly in your OP please.

    OK, I'm assuming that your second post indicates that you are now thinking that you have the original problem correctly set up.

    My first question is - Where are you getting the L(t) in your attempt at a solution? It looks to me as if it's different from the one in the problem.
    Last edited: Jan 17, 2008
  5. Jan 17, 2008 #4
    Alright, the I am finished with the corrections.
  6. Jan 17, 2008 #5
    From the problem, L(1) = (8,-1,3). Where are you getting the second L(t) that appears in your solution?
    Last edited: Jan 17, 2008
  7. Jan 17, 2008 #6
    I don't see what I have written that is wrong.
  8. Jan 17, 2008 #7
    Then there's something weird in the software or your formatting. I see in your OP a value of L(1) = (-1,5,4).

    When I look at it in quote mode it is different than what I see when looking at the post itself.

    It looks like you are assuming that the value L(1) =(8,-1,3) is the normal vector. It's not. You need to find a vector that is parallel to the line. That's not the same as a point on the line.

    Think about what it means if (8,-1,3) and (3,-1,1) both lie on the line.
    Last edited: Jan 17, 2008
  9. Jan 17, 2008 #8
    So do I take the cross product of (8,-1,3) and (3,-1,1) to find the normal vector?
    Last edited: Jan 17, 2008
  10. Jan 17, 2008 #9


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    You don't need to do that. In fact, you don't need to do any calculation at all. If you write a plane as A(x-x0)+ B(y- y0)+ C(z- z0)= 0, then [itex]A\vec{i}+ B\vec{j}+ C\vec{k}[/itex] is perpendicular to the plane. You know that (5, -5, 0) is a point in the plane so you immediately know x0, y0, and z0.

    You know that the line [tex] \vec{L}(t) = (5,0,2)t + (3,-1,1)[/tex], which has "direction vector" [itex]5\vec{i}+ 2\vec{k}[/itex], is to be perpendicular to the plane so you immediately know A, B, and C.
    Last edited: Jan 17, 2008
  11. Jan 17, 2008 #10
    You need a vector that's parallel to the line. The direction vector from (3,-1,1) to (8,-1,3) fits. Notice what you get, and compare it to what HallsofIvy says.
    Last edited: Jan 17, 2008
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