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Math Help - Planes in space (again)

  1. #1
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    Planes in space (again)

    Ok, I'm bad at space geometry and this is a tricky one.

    I almost forgot the diagram, it's in the reply.

    Find an equation of the plane with x-intercept a, y-intercept b, and z-intercept c.

    To find the equation of a plane I need a normal vector and a point. I have the points: P_x(a,0,0),P_y(0,b,0),P_z(0,0,z) which can give me some vectors. I attached a picture so you can see my reasoning, however it isn't suppose to represent this particular plane. In general:

    (1) n\cdot(r-r_o)=0 where r may be either vector 0P_2 or 0P_3 and r_o is the vector with terminal point P_o (Note that this notation corresponds to the diagram.)

    If I arbitrarily take P_x(a,0,0) as the terminal point of r_o, I have r_o=<a,0,0>. The vector along P_xP_y is r_{xy}=<-a,b,0>, along P_xP_z, we have r_{xz}=<-a,o,c>. Using equation (1) I have:


    I find the normal vector n=r_{xy}\times r_{xz} using the determinant. I obtain:

    (2) n=(bc)i+(ac)j+(ab)k


    (3) <bc,ac,ab>\cdot <x-x_o,y-y_o,z-z_o>=0

    If I plug in any point P_x(a,0,0),P_y(0,b,0),P_z(0,0,z) into (3), I don't get the correct answer which is:

    (x/a)+(y/b)+(z/c)=1


    Where am I going wrong here?
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  2. #2
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    Here is the diagram
    Attached Thumbnails Attached Thumbnails Planes in space (again)-plane-diagram.jpg  
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  3. #3
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    Quote Originally Posted by adkinsjr View Post
    ...

    Find an equation of the plane with x-intercept a, y-intercept b, and z-intercept c.

    To find the equation of a plane I need a normal vector and a point. I have the points: P_x(a,0,0),P_y(0,b,0),P_z(0,0,z) which can give me some vectors. I attached a picture so you can see my reasoning, however it isn't suppose to represent this particular plane. In general:

    (1) n\cdot(r-r_o)=0 where r may be either vector 0P_2 or 0P_3 and r_o is the vector with terminal point P_o (Note that this notation corresponds to the diagram.)

    If I arbitrarily take P_x(a,0,0) as the terminal point of r_o, I have \bold{\color{red}r_o=<a,0,0>}. The vector along P_xP_y is r_{xy}=<-a,b,0>, along P_xP_z, we have r_{xz}=<-a,o,c>. Using equation (1) I have:


    I find the normal vector n=r_{xy}\times r_{xz} using the determinant. I obtain:

    (2) n=(bc)i+(ac)j+(ab)k <<<<<< OK


    (3) <bc,ac,ab>\cdot <x-x_o,y-y_o,z-z_o>=0

    If I plug in any point P_x(a,0,0),P_y(0,b,0),P_z(0,0,z) into (3), I don't get the correct answer which is:

    (x/a)+(y/b)+(z/c)=1


    Where am I going wrong here?
    Using the vector \overrightarrow{r_0}=\langle a,0,0 \rangle your equation [3] becomes:

    \langle bc, ac, ab\rangle \cdot \langle x-a, y, z\rangle = 0

    bcx -bca +acy+abz=0~\implies~bcx  +acy+abz= bca

    Now divide the quation by the term at the RHS and you'll get the given result.
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  4. #4
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    lol, I see what I did now. I wrote <bc,ac,ab>\cdot <x-a,0,0>=0 for some reason and didn't notice it. I invented a rule for subtracting <x,y,z>-<a,0,0> where y-0=0,z-0=0 as if it where multiplication y0=0,z0=0.

    Thanks
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