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How do you find a normal of a direction vector in 3 space?

  1. Mar 2, 2005 #1
    Question #1. How do you find a normal of a direction vector in 3 space?
    Question #2.:

    What is the scalar equation of the plane that contains the x-axis and the point (4,-2,1)?
     
  2. jcsd
  3. Mar 2, 2005 #2

    HallsofIvy

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    There is no single normal to a vector in 3-space. There exist an entire plane normal to any vector and any vector in that plane is normal to the original plane.

    If a plane includes the entire x-axis, then it also includes the vector from (0,0,0) to (1, 0, 0): i. Since this plane includes the point (4, -2, 1) as well as (0,0,0) it includes the vector 4i- 2j+ k and so the cross product of those two vectors is perpendicular to the plane: -j- 2k is perpendicular to the entire plane. If (x, y, z) is any point in the plane, then the vector from (0,0,0) to (x,y,z), xi+ yj+ zk, is perpendicular to that vector: (xi+ yj+ zk).(-j- 2k)= 0 or -y- z= 0. The desired plane is z= -y.
     
  4. Mar 2, 2005 #3
    hallsofivy.....that is not the answer.......I think you are wrong.......can you please make it a bit more clear on the steps made to get the answer.

    for my first question ok sorry maybe I worded it wrong....how do i find a normal plane of a direction vector in 3 space?
     
  5. Mar 2, 2005 #4
    or maybe ur right .....thats the plane......but not the answer

    I'm not sure how to find the normal.....(1,0,0)cross what? to get a normal.....then after that put the (x,y,z) values in the equation Ax+By+Cz+D=0
    Then find D by inserting the point (4,-2,1) and you've got the equation?

    Is this process correct
     
    Last edited: Mar 2, 2005
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