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Equation of line parallel to plane and intersaction with other line

  1. May 18, 2008 #1
    1. The problem statement, all variables and given/known data


    I have one problem which seems not so difficult:

    -Find the equation of line which passes throught the point M(1,0,7), parallel of the plane 3x-y+2z-15=0 and it intersects the line [tex]\frac{x-1}{4}=\frac{y-3}{2}=\frac{z}{1}[/tex]

    2. Relevant equations

    3. The attempt at a solution

    The equation of the line will be: [tex]\frac{x-1}{a_1}=\frac{y}{a_2}=\frac{z-7}{a_3}[/tex]

    So we need to find [tex]\vec{a}(a_1,a_2,a_3)[/tex] and we need three conditions in the system.

    The first condition is [tex]\vec{a} \circ \vec{n}=0[/tex] or [tex](a_1,a_2,a_3)(3,-1,2)=0[/tex] or [tex]3a_1-a_2+2a_3=0[/tex].

    The second condition is the intersection of two lines, and it is:


    What about the third condition?
  2. jcsd
  3. May 18, 2008 #2


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    Staff Emeritus
    Science Advisor

    There is no third condition. You cannot determine a1, a2, and a3 uniquely. Any multiple of a given a1, a2, and a3 will also determine the same line. From the two equations you have you can solve for two of the numbers as functions of the third. Choose that third as you please, say equal to 1, and solve for the other 2.
  4. May 18, 2008 #3
    Other way (I think it is very similar to what you did):
    r = (1,3,0) + t (4,2,1)
    P = (1,0,7)
    find direction vector: r - P
    and you know n.(r-P) = 0

    So, only one condition .. (and only one unknown)
  5. May 18, 2008 #4
    rootX what is r, and what is P?
  6. May 18, 2008 #5
    r is (1,3,0) + t (4,2,1) ... a line
    P is (1,0,7) .. a point


    Should work, shouldn't it?
  7. May 19, 2008 #6
    It wouldn't work, since by your opinion you will find point, product of the intersection of line and plane... And we'll need to find parallel vector to the line which satisfies the above conditions...
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