1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Discrete Mathematics 12

  1. Jan 2, 2005 #1
    Find the scalar eq'n of a plane that is perpendicular to the plane with normal vector [3,1,2] and passes through points A(2,-6,-1) and B(1,2,-4).

    I think that the normal vector can be the direction vector of this new plane. But then, in order to find the scalar eq'n I need a normal vector of this new plane. How do I find this?
     
  2. jcsd
  3. Jan 2, 2005 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Strictly speaking a plane doesn't have a "direction vector". What is true is that the vector [3,1,2] is in the plane you want. You also know that the vector from A(2,6,-1) to B(1,2,-4) (which is, of course, [1-2,2-6,-4-(-1)]= [-1, -4, -3] is in the plane. Do you know how to find a vector that is perpendicular to both [3,1,2] and [1,4,-3]?
     
  4. Jan 2, 2005 #3
    I understand what you just found. But no, I do not know how to find a vector's that perpendicular to both those vectors.
     
  5. Jan 3, 2005 #4
    eme_girl,
    do u know the direction of a vector that is a cross product of two vectors?

    -- AI
     
  6. Jan 3, 2005 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    TenaliRaman's point: the cross product of two vectors is always perpendicular to both.

    The cross product of [a1,a2,a3] and [b1,b2,b3] is the vector [a2b3-a3b2,a3b1-a1b3,a1b2-a2b1].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Discrete Mathematics 12
  1. Kinematics 12 (Replies: 1)

Loading...