# Parametric equations of plane problem

• Jul 15th 2007, 05:52 AM
clockingly
Parametric equations of plane problem
This problem was confusing me so any help would be appreciated!

Write both the parametric and symmetric equations of the line of intersection of the planes with the equations 2x - y + z = 5 and x + y - z = 1.
• Jul 15th 2007, 06:42 AM
ThePerfectHacker
Quote:

Originally Posted by clockingly
This problem was confusing me so any help would be appreciated!

Write both the parametric and symmetric equations of the line of intersection of the planes with the equations 2x - y + z = 5 and x + y - z = 1.

If $z=0$ then:
$2x-y=5$
$x+y=1$
Thus, $x=2$ and $y=-1$

If $z=1$ then:
$2x-y=4$
$x+y=2$
Thus, $x=2$ and $y=0$

Thus, we have that this line must contains the points $(2,-1,0)\mbox{ and }(2,0,1)$

Thus, the Symettric Equation is:
$\frac{x-2}{0} = \frac{y+1}{-1} = \frac{z-0}{-1}$

Note, this is a zero in the denominator of the first fraction, it is a useful notation I saw in a Russian Textbook* which simply means $x=2$ constantly, so it is just a shorthand notation.

*)If you are interested the Textbook was written by Perelman! Not Perelman you think it is but his grandfather!
• Jul 15th 2007, 06:52 AM
Jhevon
Quote:

Originally Posted by ThePerfectHacker
Thus, the Symettric Equation is:
$\frac{x-2}{0} = \frac{y+1}{-1} = \frac{z-0}{-1}$

Note, this is a zero in the denominator of the first fraction, it is a useful notation I saw in a Russian Textbook* which simply means $x=2$ constantly, so it is just a shorthand notation.

*)If you are interested the Textbook was written by Perelman! Not Perelman you think it is but his grandfather!

Thanks for that TPH. i did this question the same as you did, but i had no idea what to do with that 0 in the denominator, it kind of freaked me out, so i decided not to post
• Jul 15th 2007, 06:57 AM
ThePerfectHacker
Quote:

Originally Posted by Jhevon
Thanks for that TPH. i did this question the same as you did, but i had no idea what to do with that 0 in the denominator, it kind of freaked me out, so i decided not to post

The way it is taught in American schools is if that every happens then the fuction does not have a "symettric form" to it. It only has parametric. But Grandpa's Perelman notation is really useful like I said.
• Jul 15th 2007, 07:52 AM
Soroban
Hello, clockingly!

Quote:

Write both the parametric and symmetric equations of the line of intersection
of the planes with the equations: . $\begin{array}{ccc}2x - y + z &= &5 \\ x + y - z &= &1\end{array}$

This is how I was taught to find the intersection of two planes.

We have: . $\begin{array}{cccc}2x - y + z & = & 5 & {\color{blue}[1]}\\ x + y -z & = & 1 & {\color{blue}[2]}\end{array}$

Add [1] and [2]: . $3x \,=\,6\quad\Rightarrow\quad x\,=\,2$

Substitute into [2]: . $2 + y - z \:=\:1\quad\Rightarrow\quad y \:=\:z-1$

We have $x,\,y,\,z$ as a function of $z\!:\;\;\begin{Bmatrix}x & = & 2 \\ y & = & z-1 \\ z & = & z\end{Bmatrix}$

On the right side, replace $z$ with a parameter $t$.

. . . . $\begin{Bmatrix}x & = & 2 \\ y & = & t - 1 \\ z & = & t\end{Bmatrix}$ . . . . There!

• Jul 15th 2007, 07:55 AM
Jhevon
Quote:

Originally Posted by Soroban
Hello, clockingly!

This is how I was taught to find the intersection of two planes.

We have: . $\begin{array}{cccc}2x - y + z & = & 5 & {\color{blue}[1]}\\ x + y -z & = & 1 & {\color{blue}[2]}\end{array}$

Add [1] and [2]: . $3x \,=\,6\quad\Rightarrow\quad x\,=\,2$

Substitute into [2]: . $2 + y - z \:=\:1\quad\Rightarrow\quad y \:=\:z-1$

We have $x,\,y,\,z$ as a function of $z\!:\;\;\begin{Bmatrix}x & = & 2 \\ y & = & z-1 \\ z & = & z\end{Bmatrix}$

On the right side, replace $z$ with a parameter $t$.

. . . . $\begin{Bmatrix}x & = & 2 \\ y & = & t - 1 \\ z & = & t\end{Bmatrix}$ . . . . There!

nice method
• Jul 15th 2007, 08:09 AM
Plato
The customary notation used in textbooks in North America for the symmetric form in which one direction number is zero is to use a semicolon: $\frac{{y + 1}}{{ - 1}} = \frac{z}{{ - 1}}\, ;\,x = 2$.