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Equation of a plane parallel to a line and perpendicaular to another plane

  1. Sep 4, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the equation of the plane through the point (2, 5, 7) that is parallel
    to the line r = (3i + 2j - 2k) + t(i + 2j + 7k) and perpendicular to the plane 4x + 5y + 6z = 14
    It should be written in scalar equation form


    2. Relevant equations
    scalar equation of a plane through a point: a(x - x0) + b(y - y0) + c(z - Z0) = 0
    normal of a plane is (a, b, c)
    cross product of two parallel vectors = 0 (v x u = 0)



    3. The attempt at a solution

    what I have so far is __(x-2) + __(y-5) + __(z-7)
    but i can't figure out how I am supposed to find the normal of this plane...
     
  2. jcsd
  3. Sep 4, 2009 #2

    rock.freak667

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    The required normal passes through r = (3i + 2j - 2k) + t(i + 2j + 7k) and the normal of 4x + 5y + 6z = 14.

    So find the equation of the normal. You will now have a point on the normal and a point on the line r=(3i + 2j - 2k) + t(i + 2j + 7k).

    Just find the vector line joining these two points.
     
  4. Sep 4, 2009 #3
    k so the normal of the plane passes through line r and normal of plane "14"
    the plane im looking for is parallel to both of these and perpendicular to plane "14"
    so basically look for the normal of the line r?
    would the equation be (x-2) + 2(y-5) + 7(z-7) = 0?
     
  5. Sep 5, 2009 #4

    HallsofIvy

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    Incorrect, I misread the problem.

     
    Last edited: Sep 5, 2009
  6. Sep 5, 2009 #5

    VietDao29

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    No, it's wrong.. When working with parallel, and perpendicular property like this, one should think about using direction, lying (on the plane), and norm vector. It goes like this:

    Let [itex](\beta)[/itex] be our need-to-find plane, since its parallel to r, the vector (1, 2, 7) (the direction vector of r) must lie on it, right?

    And since its perpendicular to the plane "4x + 5y + 6z = 14", another lying-on-[itex](\beta)[/itex] vector must be the norm vector of the 4x + 5y + 6z = 14 plane. Right?

    Having 2 (non-parallel) vectors lying on [itex](\beta)[/itex], how can we find out the norm vector of it? Can you go from here? It's pretty simple. :)

    Well, no HoI, it's a plane parallel to a line, not a line parallel to a line. So this can be possible. :)
     
  7. Sep 5, 2009 #6

    HallsofIvy

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    Yes, thanks. I misread the problem.
     
  8. Sep 7, 2009 #7
    k sooo....
    i find the cross product to those two vectors to find the one thats perpendicular to those two which are on the plane making the orthogonal vector i find also perpedicular to the plane itself...right?
    so i got 23(x-2) - 22(y-5) - 3(z-7) = 0
    if it's right, dude thanks, i dunno why i dint see this before :0
    if it isn't, dude thanks, but there must be something wrong with me :eek:
     
  9. Sep 8, 2009 #8
    The equation of the plane is: Ax+By+Cz+D=0

    The normal vector of the plane is <A,B,C>.

    Since (2, 5, 7) lies on the plane substitute it in the equation so that:

    2A+5B+7C+D=0

    Because it is parallel of the line: r = (3i + 2j - 2k) + t(i + 2j + 7k)

    the dot product of the directional vector of the line <1,2,7> and the normal vector of the plane <A,B,C> is zero, because they are perpendicular to each other.

    <1,2,7>.<A,B,C>=0

    A+2B+7C=0

    Because it is perpendicular to the plane 4x + 5y + 6z = 14, the normal vectors of both planes are perpendicular, and their dot product is zero.

    <A,B,C>.<4,5,6>=0

    4A+5B+6C=0

    Now just solve the system of equations:

    2A+5B+7C+D=0

    A+2B+7C=0

    4A+5B+6C=0
     
  10. Sep 8, 2009 #9

    VietDao29

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    Well, you may have used the wrong formula for cross product, which should be:

    Say, a = (a1, a2, a3)
    b = (b1, b2, b3)

    [tex]\mathbf{c} = \mathbf{a} \times \mathbf{b} = \left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{array} \right|[/tex]

    (Be careful not to forget the (-1) factor, when finding the determinant of the above matrix)

    After finding the cross product of a, and b, you can always check your result by using Dot Product: (Since the Cross Product of a and b is perpendicular to both of them)

    Check to see if the 2 equations below hold:

    • ca = 0
    • cb = 0
     
    Last edited: Sep 8, 2009
  11. Sep 13, 2009 #10
    hihi...Vietdao..could u pls show me how to do this question?
    The line L has equation r=j+k+s(i-2j+k).The plane p has equation x+2y+3z+5.
    (a)Show that the line L lies in the plane p.
    (b)A second plane is perpendicular to the plane p,parallel to the line L and contains the point with position vector 2i+j+4k.Find the equation of this plane,giving your answer in the form ax+by+cz=d.

    Thank you very much..
     
  12. Sep 13, 2009 #11

    VietDao29

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    You should start a new thread instead of just posting in an old one like this.

    Well.. the forums' rules do not allow providing complete solutions though. So, may I ask what have you try, and how far have you got? So that I can know where you get stuck, and, hence, I can give you more accurate help. :)
     
  13. Sep 13, 2009 #12
    a)Line can be formed by two points. So show that two points of the line are also points on the plane.

    b)If the plane ax+by+cz=d is perpendicular to p then the dot product of the normal vector (a,b,c) and the normal vector to the plane p: (1,2,3) is zero:

    [tex](a,b,c) \cdot (1,2,3)=0[/tex]


    Try something and I will continue giving you hints.
     
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