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Finding vector perpendicular to plane

  1. May 9, 2004 #1
    If you have a 3d vector, how do you find out the perpendicular vector to this (the normal plane and stuff)? I know that the scalar product has to be 0, but surely that leaves hundreds of ones that would do that, as 2 of the 3 numbers can be chosen and the last one changes the value to 0.
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  3. May 9, 2004 #2


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    Could you be more clear please, what exactly do you have and what do you want?

    For example if you have 2 vectors on a plane all you have to do is cross multiply them to get a vector perpendicular to the plane. If you have a plane with the equation r.n = a.n then n is a vector perpendicular to the plane.
  4. May 9, 2004 #3
    The book I have to learn this from says that given 2 points (A and R), you the normal (n) is given by:

    AR x n = 0.

    So if I know the vector AR, how exactly do I come up with the n that is perpendicular to the plane?
  5. May 9, 2004 #4


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    Solve the vector equation. Set each of the three components of the cross product equal to zero; this gives you three linear equations in three unknowns (the three components of n). By assumption the vectors AR and n are not parallel, so the determinant of the equations is not zero. Solve the equations.
  6. May 9, 2004 #5


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    You need be more clear about just what you're asking; precisely what are you given, and what do you want?

    Also, did you mean to write [itex]AR \cdot n = 0[/itex] or [itex]AR \times n = 0[/itex]? There's a big difference between the two; the former is a dot product, the latter is a cross product. (The usual ascii way to write a dot product would be AR.n or to use the alternative notation <AR, n>)
  7. May 9, 2004 #6
    Never heard of the terms dot product and cross product. I know what the scalar product is (mutiply out the rows and add them up), and I think that is what the book means.

    OK, let's see if I can explain this better.

    If I have the equation of a plane in the form ax+by+cz=d the direction perpendicular is just ai+bj+ck, right? Does this follow through to if you have the plane in vector form where you have three vectors with a lambda and mu in there too for the directions?
  8. May 9, 2004 #7


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    You'd better look up "cross product". The "scalar product" is the same as the dot product (although "multiply out the rows and add them up" makes no sense! I assume you mean "multiply corresponding components of the vectors and add the products")- and it is true that if A and R are two points and AR is the vector "from A to R", then AR . n= 0 (dot product) for any vector that is perpendicular to AR. However, if you are given two vectors, say u and v, the u x v (the cross product) perpendicular to both.

    That's one of the reasons everyone is asking for you to be clearer: given any vector AR there exist an infinite number of vectors perpendicular to it. In two senses: once you have found a vector perpendicular to AR, any
    multiple of that vector (same direction, different length) is also perpendicular to AR but more importantly, there exist an infinite number of directions perpendicular to AR. Given a vector or simply a line in a given direction, any vector (or line) in the plane perependicular to AR is itself perpendicular to AR.

    To answer your last questions: yes, if you have a plane written in the form
    Ax+ By+ Cz= D, then the vector Ai+ Bj+ Ck is perpendicular to it (as is any multiple). That is because if we let (x0,y0,0) be a point in the plane, it must satisfy Ax0+ By0+ Cz0= D so, for any (x,y,z) in the plane (which satisfies Ax+ By+ Cz= D), subtracting the first equation from the second gives A(x-x0)+ B(y-y0)+ C(z-z0)= D- D=0 or (Ai+ Bj+ Ck) dot ((x-x0)i+ (y-y0)j+ (z-z0))= 0- exactly the condition that Ai+ Bj+ Ck and (x-x0)i+ (y-y0)j+ (z-z0) be perpendicular.

    Now, suppose we have a plane given by r= (au+ bv+ c)i+ (du+ev+f)j+ (gu+hv+l)k- a plane determined by the two parameters u and v (your lambda and mu). (NOTE: this vector points from (0,0,0) TO a point in the plane. It is NOT in the plane itself unless (0,0,0) happens to be in the plane.)

    If we let u=0 and v=0, we see that r= ci+fj+lk gives a point in the plane: that is, the point P= (c,f,l) is in the plane.
    If we let u= 1 and v= 0, we see that (a+c)i+ (d+f)j+ (g+l)k gives a point in the plane: that is, the point Q= (a+c,d+f,g+l) is in the plane and so the vector PQ= ai+ dj+ gk is in the plane.
    If we let u= 0 and v= 1, we see that (b+c)i+ (e+f)j+ (h+l)k is gives a point in the plane: that is, the point R= (b+c, e+f, h+l) is in the plane and so the vector PR= bi+ ej+ hk is in the plane.

    That means that the vector PQ x PR (the cross product), which happens to be (ah-eg)i+ (bg-ah)j+ (ae-bd)k, is perpendicular to the plane.
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