Originally Posted by
Soroban Hello, statman101!
The cross-product approach is the best way.
Here's another (primitive and longer) approach.
(Use it when you're desperate.)
$\displaystyle \begin{array}{cccccccc}
\text{Multiply [2] by -4:} & \text{-}16x - 4y - 4z &=& \text{-}60 \\
\text{Add [1]:} & 3x + 4y - 5x &=& 12 \end{array}$
And we have: .$\displaystyle -13x - 9z \:=\:-48 \quad\Rightarrow\quad x \:=\:\dfrac{48-9z}{13}$
Substitute into [2]: .$\displaystyle 4\left(\frac{48-9z}{13}\right) + y + z \:=\:15 \quad\Rightarrow\quad y \:=\:\dfrac{3+23z}{13} $
So we have: .$\displaystyle \begin{Bmatrix}x &=& \frac{48-9z}{13} &=& \frac{48}{13} - \frac{9}{13}z\\ \\[-3mm]
y &=& \frac{3 + 23z}{13} &=& \frac{3}{13} + \frac{23}{13}z \\ \\[-3mm] z &=& z \end{Bmatrix}$
On the right side, replace $\displaystyle \,z$ with a parameter $\displaystyle \,t.$
. . . . . $\displaystyle \begin{Bmatrix} x &=& \frac{48}{13} - \frac{9}{13}t \\ \\[-3mm]
y &=& \frac{3}{13} + \frac{23}{13}t \\ \\[-3mm]
z &=& 0 + t \end{Bmatrix}$
These are the parametric equations of the line through $\displaystyle \left(\frac{48}{13},\:\frac{3}{13},\:0\right) $
. . with direction vector: .$\displaystyle \vec v \:=\:\langle \text{-}\frac{9}{13},\:\frac{23}{13},\:1\rangle \;=\;\langle \text{-}9,\:23,\:13\rangle $
The symmetric equation is: .$\displaystyle \displaystyle \frac{x-\frac{48}{13}}{-9} \;=\;\frac{y-\frac{3}{13}}{23} \;=\;\frac{z-0}{1} $
Note: There are fifteen brizillion different forms of the symmetric equation.
. . . . .This is just one of them.