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Given a vector normal to a plane, how do I get a vector perpendicular to that plane?

  1. Jan 30, 2012 #1
    If I have a vector nhat = [nx, ny, nz] which is normal to some plane, how can I write the vectors (I assume there are infinitely many) which are perpendicular to that plane?
     
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  3. Jan 30, 2012 #2

    Char. Limit

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    Re: Given a vector normal to a plane, how do I get a vector perpendicular to that pla

    Well, normal is perpendicular. So the set of vectors perpendicular to that plane would be any non-zero multiple of [itex]\hat{n}[/itex].
     
  4. Jan 30, 2012 #3
    Re: Given a vector normal to a plane, how do I get a vector perpendicular to that pla

    Sorry, not thinking very clearly. I meant to type the set of unit vectors which are PARALLEL to that same plane
     
  5. Jan 30, 2012 #4

    Char. Limit

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    Re: Given a vector normal to a plane, how do I get a vector perpendicular to that pla

    Ah, well that is different. We need to find the set of all vectors normal to our normal vector. Now, one requirement of normality is that the dot-product be zero, i.e.

    [tex]\vec{T} \cdot \hat{n} = 0[/tex]

    I use T here because any vector that's normal to the normal will be tangent to the plane. Now, with that...


    hm. This will require a bit more thought. I'll be right back. In the meantime, hopefully someone who has already figured out the answer will stop by!
     
  6. Jan 30, 2012 #5

    Char. Limit

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    Re: Given a vector normal to a plane, how do I get a vector perpendicular to that pla

    All right, I'm back and I've figured it out. Start with T dot N = 0, which we know to be true. Expand this out to get:

    [tex]T_1 n_x + T_2 n_y + T_3 n_z = 0[/tex]

    [tex]T_1 = - T_2 \frac{n_y}{n_x} - T_3 \frac{n_z}{n_x}[/tex]

    And from there, it should be trivial to express T as a linear combination of vectors with coefficients (n_y)/(n_x) and (n_z)/(n_x), respectively. That'll give you your plane. :)

    That was fun!

    Note: We assume n_x is not zero. If it is, this problem becomes a lot more trivial.
     
  7. Jan 30, 2012 #6
    Re: Given a vector normal to a plane, how do I get a vector perpendicular to that pla

    Thank you so much for the help!
     
  8. Jan 30, 2012 #7

    HallsofIvy

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    Re: Given a vector normal to a plane, how do I get a vector perpendicular to that pla

    Of course, any vector in the plane is a vector parallel to the plane. You don't need a normal vector to find that.

    If the plane is given by z= ax+ by+ c and we take x= y= 0, z= c so (0, 0, c) is a point in the plane. And for any numbers, X and Y, (X, Y, aX+ bY+ c) is also a point in the plane. The vector from the first to the second is [itex]X\vec{i}+ Y\vec{j}+ (aX+ bY)\vec{k}[/itex] is a vector in (parallel to) the plane. That can be written as [itex]X(\vec{i}+ a\vec{k})+ Y(\vec{j}+ b\vec{k})[/itex] indicating that [itex]\vec{i}+ a\vec{k}[/itex] and [itex]\vec{j}+ b\vec{k}[/itex] form a basis for the vector space of all vectors parallel to the plane z= ax+ by+ c.
     
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