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How to find a vector perpendicular to a plane.

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    If 2x+y-2z=5 is the equation of a plane, how would you find a normal to this plane. The answer is (2,1,-2)
    The proof I was given was that (x,y,z).(2,1,-2)=5

    This gives 2x+y-2z=5 which obviously doesn't help very much to understand why (2,1-2) is a normal to this plane.
    Am i just meant to say for any equation of a plane e.g 7x-4y+2z=11 a normal to this plane would be (7,-4,2)??

    Any help greatly appreciated.
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    (Original post by anonstudent1)
    If 2x+y-2z=5 is the equation of a plane, how would you find a normal to this plane. The answer is (2,1,-2)
    The proof I was given was that (x,y,z).(2,1,-2)=5

    This gives 2x+y-2z=5 which obviously doesn't help very much to understand why (2,1-2) is a normal to this plane.
    Am i just meant to say for any equation of a plane e.g 7x-4y+2z=11 a normal to this plane would be (7,-4,2)??

    Any help greatly appreciated.
    When you see an equation of the form ax+by+cz=k, the normal to the plane is always (a,b,c).

    Why? Well let \mathbf{r}=(x,y,z) and \mathbf{n}=(a,b,c). Choose numbers u,v,w such that au+bv+cw=k (that is, we can choose (u,v,w) to be any point in the plane). Then if we write \mathbf{u}=(u,v,w), we have

    ax+by+cz=au+bv+cw

    which is precisely the statement that \mathbf{r} \cdot \mathbf{n} = \mathbf{u} \cdot \mathbf{n}

    Equivalently, (\mathbf{r} - \mathbf{u}) \cdot \mathbf{n} = 0

    That is, \mathbf{n} is perpendicular to (x-u,y-v,z-w) for every choice of x,y,z.

    But any direction vector in the plane can be written in the form (x-u,y-v,z-w), since (u,v,w) is a fixed point in the plane and (x,y,z) is the coordinates (position vector) of any given point. So what the equation (\mathbf{r}-\mathbf{u}) \cdot \mathbf{n} = 0 tells us is that \mathbf{n} is perpendicular to all directions in the plane. That is, \mathbf{n} is normal to the plane.

    I hope I haven't obfuscated this too much with the heavy use of notation. Let me know if you want anything clarified.
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    (Original post by nuodai)
    When you see an equation of the form ax+by+cz=k, the normal to the plane is always (a,b,c).
    Thank you for your help
    I think for the level of exams i'm doing right now, the above will be sufficient. Wow the rest looks complicated, will give it a read through when i have more time.
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    (Original post by anonstudent1)
    Thank you for your help
    I think for the level of exams i'm doing right now, the above will be sufficient. Wow the rest looks complicated, will give it a read through when i have more time.
    The rest isn't something you need to know, as such, but it shows why it's true so it's worth trying to understand once even if you don't remember it. (But as you say, maybe it's too late for that :p:)
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    (Original post by anonstudent1)
    If 2x+y-2z=5 is the equation of a plane, how would you find a normal to this plane. The answer is (2,1,-2)
    The proof I was given was that (x,y,z).(2,1,-2)=5

    This gives 2x+y-2z=5 which obviously doesn't help very much to understand why (2,1-2) is a normal to this plane.
    Am i just meant to say for any equation of a plane e.g 7x-4y+2z=11 a normal to this plane would be (7,-4,2)??

    Any help greatly appreciated.
    Yes, the normal wolud be (7,-4,2) in your example.
    And why?
    For the equation of a plane we need a point on the plane
    P_0(x_0,y_0,z_0) and a vector being perpendicular to
    the plane.This is the normal vector, and let \vec n=A\vec i+B\vec j+C\vec k
    THe position vector pointing to P0 is \vec r_0=x_0\cdot \vec i+y_0\cdot \vec j+z_0\cdot \veck.
    Let a point in the plane P(x,y,z), sor the position vector
    pointing this point is \vec r=x\cdot \vec i+y\cdot \vec j+z\cdot \vec k.
    It is clear that the vector \vec r-\vec r_0 is in the plane, so
    perpendicular to the normal vector, that their dot product is zero.
    This fact gives the vector equation of the plane:
    \left (\vec r -\vec r_0\right )\cdot \vec n=0
    Calculating the dot product
    A(x-x_0)+B(y-y_0)+C(z-z_0)=0 gives the scalar equation.
    Arranging
    Ax+By+Cz=Ax_0+By_0+Cz_0=D
    whre D is a constant value.
    From this you can read down the coordinates of the normal vector (A,B,C)
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    (Original post by nuodai)
    When you see an equation of the form ax+by+cz=k, the normal to the plane is always (a,b,c).

    Why? Well let \mathbf{r}=(x,y,z) and \mathbf{n}=(a,b,c). Choose numbers u,v,w such that au+bv+cw=k (that is, we can choose (u,v,w) to be any point in the plane). Then if we write \mathbf{u}=(u,v,w), we have

    ax+by+cz=au+bv+cw

    which is precisely the statement that \mathbf{r} \cdot \mathbf{n} = \mathbf{u} \cdot \mathbf{n}

    Equivalently, (\mathbf{r} - \mathbf{u}) \cdot \mathbf{n} = 0

    That is, \mathbf{n} is perpendicular to (x-u,y-v,z-w) for every choice of x,y,z.

    But any direction vector in the plane can be written in the form (x-u,y-v,z-w), since (u,v,w) is a fixed point in the plane and (x,y,z) is the coordinates (position vector) of any given point. So what the equation (\mathbf{r}-\mathbf{u}) \cdot \mathbf{n} = 0 tells us is that \mathbf{n} is perpendicular to all directions in the plane. That is, \mathbf{n} is normal to the plane.

    I hope I haven't obfuscated this too much with the heavy use of notation. Let me know if you want anything clarified.
    Are you a wizard?
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    n=2i +1j -2k is already the normal vector.

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Updated: April 26, 2014
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