Answer
$x=-8+11t,y=1-3t,z=4$
and
$\frac{x+8}{11}=\frac{y-1}{-3},z=4$
Work Step by Step
The direction vector for a line through the $(-8,1,4)$ and the point $(3,-2,4)$ is $\lt 11,-3,0 \gt$ .
Parametric equations defined by:
$x=x_0+at$, $y=y_0=bt$ and $z=z_0+ct$
Thus, the parametric equations are:
$x=-8+11t,y=1-3t,z=4-(0)t=4$
The symmetric equations are defined by:
$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$
Hence, the symmetric equations are:
$\frac{x+8}{11}=\frac{y-1}{-3},z=4$
