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.

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- Jul 15th 2007, 05:52 AMclockinglyParametric 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 AMThePerfectHacker
If $\displaystyle z=0$ then:

$\displaystyle 2x-y=5$

$\displaystyle x+y=1$

Thus, $\displaystyle x=2$ and $\displaystyle y=-1$

If $\displaystyle z=1$ then:

$\displaystyle 2x-y=4$

$\displaystyle x+y=2$

Thus, $\displaystyle x=2$ and $\displaystyle y=0$

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

Thus, the Symettric Equation is:

$\displaystyle \frac{x-2}{0} = \frac{y+1}{-1} = \frac{z-0}{-1}$

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 $\displaystyle x=2$ 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 AMJhevon
- Jul 15th 2007, 06:57 AMThePerfectHacker
- Jul 15th 2007, 07:52 AMSoroban
Hello, clockingly!

Quote:

Write both the parametric and symmetric equations of the line of intersection

of the planes with the equations: .$\displaystyle \begin{array}{ccc}2x - y + z &= &5 \\ x + y - z &= &1\end{array}$

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

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

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

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

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

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

- Jul 15th 2007, 07:55 AMJhevon
- Jul 15th 2007, 08:09 AMPlato
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: $\displaystyle \frac{{y + 1}}{{ - 1}} = \frac{z}{{ - 1}}\, ;\,x = 2$.