I am trying to find both the parametric and symmetric equations of a line passing through two points. This is for a study exam, so exact answers are not as helpful as detailed solutions.
$P(3,-1,1);$ $Q(-2,1,1)$
I found the points vectors then put them with respect to t: $r(t) = (3,-1,1)+t(-2,1,1)$
Then set each x,y,z to t, which should be the parametric equation: $x(t)=3-2t$
$y(t)=-1+t$
$z(t)=1+t$
Then to find the symmetric equation I set the points equal to giving me this: $\frac{(3-x)}{2}=1+y=z-1$
I am having trouble finding if I went about this the wrong way, primarily when creating the vectors and putting them in the equation for a line with respect to $t.$