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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: $$x(t)=3-2t$$



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 $t.$

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The correct equation does not have a term $t<-2,1,1>$, but rather $t$ multiplied by the displacement vector $P-Q$. – David Mitra Dec 8 '12 at 20:49
up vote 5 down vote accepted

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.

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How is the symmetric equation correct? There is no value "t" that equates to (-2, 1, 1). – Craig Lafferty Sep 18 '14 at 19:32

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