Find the parametic equation of the line given by x+y-z=2,2x+3y+z=4
Attempt to solution:
Is the line given by this equation the intersection of these two planes ?
Dear nyasha,
I don't know what row reduced echeloen form is. But you could easily find the line in the following method.
$\displaystyle x+y-z=2$------(1)
$\displaystyle 2x+3y+z=4$---------(2)
$\displaystyle (1)+(2)\Rightarrow{3x+4y=6}$
This is the equation of the line. Since you need to find the parametric form,
$\displaystyle x=\frac{6-4y}{3}=t$
Therefore, $\displaystyle x=t$
$\displaystyle y=\frac{6-3t}{4}$
Hope this helps.
Hello, nyasha!
Here's my approach . . .
Find the parametic equation of the line given by: .$\displaystyle \begin{array}{cccc} x+y-z&=&2 & [1] \\ 2x+3y+z&=&4 & [2] \end{array}$
Add [1] and [2]: .$\displaystyle 3x + 4y \:=\:6$
. . Solve for $\displaystyle x\!:\;\;x \:=\:\frac{6-4y}{3}$
Substitute into [1]: .$\displaystyle \frac{6-4y}{3} + y - z \:=\:2$
. . Solve for $\displaystyle z\!:\;\;z \:=\:-\frac{y}{3}$
We have: .$\displaystyle \begin{Bmatrix}x &=& 2 - \frac{4}{3}y \\ \\[-4mm] y &=& y \\ \\[-3mm] z &=& -\frac{1}{3}y \end{Bmatrix}$
On the right, replace $\displaystyle y$ with $\displaystyle 3t$
. . and we have: .$\displaystyle \begin{Bmatrix}x &=& 2-4t \\ y &=& 3t \\ z &=& -t \end{Bmatrix}$