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 $t.$
Best Answer
Your parameterization is not correct.
A parameterization of a line has the form $r(t)= P + t D$ where $P$ is a vector "touching" the line and $D$ is a direction vector for the line.
Given two points on the line, $P$ and $Q$, the equation $r(t) = P+ t Q$ is not the correct parameterization. This is where you are in error. So, instead of using $Q$ as you did, use the displacement vector from $P$ to $Q$. In your case, this would be $$ Q-P = (-2,1,1) - (3,-1,1) =\bigl(-2-3 , 1-(-1), 1-1 \bigr)=(-5,2,0). $$ The parameterization is then $$ r(t)=(3,-1,1)+t(-5,2,0). $$ (And don't forget to give a range of values for $t$.)
Your procedure for finding the symmetric equations looks ok.