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Homework Help: Parametric and symmetric equations

  1. Feb 12, 2008 #1

    tony873004

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    Find the parametric and symmetric equations of the line of intersection of the planes x+y+z=1 and x+z=0.

    I got the normal vectors, <1,1,1> and <1,0,1> and their cross product <1,0,-1> or i-k.

    I set z to 0 and got x=0, y=1, z=0.

    How do I form parametric equation out of this?? I know it's x=t, y=1, z=-t because this problem is nearly identical to one from lecture. But how did he do that step?

    This would make the symmetric equations x/1=y-1/0=z/-1. But I can't divide by 0, can I?
     
  2. jcsd
  3. Feb 12, 2008 #2
    You set z = 0.

    [tex]\pi_1: x+y=1[/tex]
    [tex]\pi_2: x=0[/tex]

    [tex]P(0,1,0)[/tex]

    The equation of the line is given by: [tex]\overrightarrow{r}=\overrightarrow{r}_0+t\overrightarrow{v}[/tex]

    And we know that the cross product of the two normal vectors of the plane is parallel to the line of intersection.

    [tex]\overrightarrow{r}=<0,1,0>+t<1,0,-1>[/tex]

    [tex]x=-z;y=y_0[/tex]

    Since we don't write 0 under the denominator.

    If [tex]x=x_0, y=y_0[/tex] vertical plane and [tex]z=z_0[/tex] horizontal plane.
     
    Last edited: Feb 13, 2008
  4. Feb 12, 2008 #3

    tony873004

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    Thanks for the explanation.

    I don't get this:
    [tex]\pi_2: x=1[/tex]
    If x+z=0 and I set z=0, then x+0=0. x=0 and 0+y+0=1, so y=1, hence P(0,1,0)
     
  5. Feb 13, 2008 #4
    Oh my. Sorry, I'm blind! I fixed it though.
     
  6. Feb 13, 2008 #5

    tony873004

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    Thanks. You explanation makes sense now.
     
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