Parametric and symmetric equations

[tony873004]

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?

 

[rocomath]

You set z = 0.

$$\pi_1: x+y=1$$
$$\pi_2: x=0$$

$$P(0,1,0)$$

The equation of the line is given by: $$\overrightarrow{r}=\overrightarrow{r}_0+t\overrightarrow{v}$$

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

$$\overrightarrow{r}=<0,1,0>+t<1,0,-1>$$

$$x=-z;y=y_0$$

Since we don't write 0 under the denominator.

If $$x=x_0, y=y_0$$ vertical plane and $$z=z_0$$ horizontal plane.

 

[tony873004]

Thanks for the explanation.

I don't get this:
$$\pi_2: x=1$$
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)

 

[rocomath]

Thanks for the explanation.

I don't get this:
$$\pi_2: x=1$$
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)

Oh my. Sorry, I'm blind! I fixed it though.

 

[tony873004]

Thanks. You explanation makes sense now.