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 $\displaystyle z=0$ 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!
Hello, clockingly!
This is how I was taught to find the intersection of two planes.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!