Results 1 to 4 of 4

Math Help - Intersection of orthogonal planes

  1. #1
    Member
    Joined
    Jan 2011
    Posts
    87

    Intersection of orthogonal planes

    Hello, I have been trying to tackle vectors over the weekend, but cant seem to get this one right. I have two lines, s(1,2,1) and (9,6,0) + t(0,1,-1). The orthogonal vector is v(-3,1,1) found using cross product. How can i find the equation of a line say P parallel to v(-3,1,1) which intersects both lines? Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by Oiler View Post
    Hello, I have been trying to tackle vectors over the weekend, but cant seem to get this one right. I have two lines, s(1,2,1) and (9,6,0) + t(0,1,-1). The orthogonal vector is v(-3,1,1) found using cross product. How can i find the equation of a line say P parallel to v(-3,1,1) which intersects both lines? Thanks.
    1. You've 3 lines:

    x,y,z)=s \cdot (1,2,1)" alt="l_1x,y,z)=s \cdot (1,2,1)" />

    x,y,z)=(9,6,0) + t \cdot (0,1,-1)" alt="l_2x,y,z)=(9,6,0) + t \cdot (0,1,-1)" />

    and a line l_3 which passes through a point of l_1 in the direction of (-3, 1, 1).

    x,y,z)= s \cdot (1,2,1) + r \cdot (-3,1,1)" alt="l_3x,y,z)= s \cdot (1,2,1) + r \cdot (-3,1,1)" />

    2. Line l_3 passes through a point of l_2 :

    s \cdot (1,2,1) + r \cdot (-3,1,1) = (9,6,0) + t \cdot (0,1,-1)

    Solve for (r, s, t).

    3. For confirmation only: I've got (r, s, t) = \left( -\frac{21}{11}, \frac{36}{11}, -\frac{15}{11} \right)
    Last edited by earboth; January 31st 2011 at 03:12 AM. Reason: corrected!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    An alternative:

    Generic point of the first line l_1:

    P=(s,2s,s)\;(s\in\mathbb{R})

    Generic point of the second line l_2:

    Q=(9,6+t,-t)\;(t\in\mathbb{R})

    Solving the system:

    \begin{Bmatrix} \vec{PQ}\cdot (1,2,1)=0\\\vec{PQ}\cdot (0,1,-1)=0\end{matrix}

    we obtain s_0,t_0 , that provide two points P_0,Q_0 of the line l which intersects and it is orthogonal to l_1 and l_2 .


    Fernando Revilla
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jan 2011
    Posts
    87
    Hey earboth, shouldn't line 3, just be (1,2,1) + r(-3,1,1) since adding a s in front of (1,2,1) would make it a plane ? which would then be parallel to r(-3,1,1) instead of being orthogonal ? Thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Angle between planes and line of intersection of planes.
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 6th 2011, 01:08 PM
  2. Intersection of 3 Planes
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 27th 2009, 07:00 AM
  3. Cross products and orthogonal planes/vectors
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: May 18th 2009, 12:02 PM
  4. Orthogonal Projections On Planes?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 27th 2008, 05:23 AM
  5. Intersection of three planes
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: March 24th 2008, 02:04 PM

Search Tags


/mathhelpforum @mathhelpforum