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

$y(t)=-1+t$

$z(t)=1+t$

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 $t.$

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    The correct equation does not have a term t<-2,1,1>, but rather $t$ multiplied by the displacement vector $P-Q$.2012-12-08

2 Answers 2

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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).2014-09-18
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Give a vector parametric equation for the line through the point (−4,1,1)(−4,1,1) that is parallel to the line ⟨−4−2t,2,−1−t⟩⟨−4−2t,2,−1−t⟩

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    The OP has probably taken the exam he talked about (the question is almost 4 years old). Your time would be better spend learning some MathJax than answering old questions that have accepted answers.2016-11-11