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Thread: 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: $\displaystyle P_x(a,0,0),P_y(0,b,0),P_z(0,0,z)$ 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) $\displaystyle n\cdot(r-r_o)=0$ where r may be either vector $\displaystyle 0P_2$ or $\displaystyle 0P_3$ and $\displaystyle r_o$ is the vector with terminal point $\displaystyle P_o$ (Note that this notation corresponds to the diagram.)

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


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

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


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

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

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


    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: $\displaystyle P_x(a,0,0),P_y(0,b,0),P_z(0,0,z)$ 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) $\displaystyle n\cdot(r-r_o)=0$ where r may be either vector $\displaystyle 0P_2$ or $\displaystyle 0P_3$ and $\displaystyle r_o$ is the vector with terminal point $\displaystyle P_o$ (Note that this notation corresponds to the diagram.)

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


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

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


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

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

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


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

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

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

    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 $\displaystyle <bc,ac,ab>\cdot <x-a,0,0>=0$ for some reason and didn't notice it. I invented a rule for subtracting $\displaystyle <x,y,z>-<a,0,0>$ where $\displaystyle y-0=0,z-0=0$ as if it where multiplication $\displaystyle y0=0,z0=0$.

    Thanks
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