# Math Help - finding the parametric equation of a line

1. ## finding the parametric equation of a line

question: Find the parametric equation of the line through the points (0,.5,1) and (2,1,-3)

work:r=r_0+tv the vector parallel to the line is <2-0,1-.5,-3-1> I thought I would take (0,.5,1) to be the r_0 but the text book didn't. How do I know which initial point to use?

Am I doing this right? I mean is it necessary to find the vector equation and then from that take the parametric equation or is there a faster way?

2. Originally Posted by superdude
question: Find the parametric equation of the line through the points (0,.5,1) and (2,1,-3)

work:r=r_0+tv the vector parallel to the line is <2-0,1-.5,-3-1> I thought I would take (0,.5,1) to be the r_0 but the text book didn't. How do I know which initial point to use?

Am I doing this right? I mean is it necessary to find the vector equation and then from that take the parametric equation or is there a faster way?

the equation of a line passing through two points $a , b$ is

$r = ta + (1-t)b$

$r = ( 2t , 5 - 4t , 1 - 4t )$

3. so your saying the problem can't be solved using the formula I was using?

4. Hi

Actually , we are talking the same method but my is the simplier expression :

We can take either $(0,5,1)$ or $(2,1,-3)$
to be the $r_0$

After we calculate the vector parallel to the required line $a - b = (0,5,1) - (2,1,-3) = ( -2,4,4)$ , we put the parameter [tex] t ATH] next to the vector and add up with $a$ or $b$ , the sum is what we are looking for

Don't believe ?

if we take $(0,5,1 )$ to be the initial point

the required equation is $(0,5,1) + t (-2,4,4) = -2t , 5 + 4t , 1 + 4t )$

now if we take $(2,1,-3)$ to be the initial point ,
the equation becomes $( 2 - 2t , 1 + 4t , -3 + 4t )$

Well , obviously , they are totally different but if we change the parameter t by (t+1) , the equation (2) becomes

$( 2(-t) , 1 + 4t +4 , -3 + 4 + 4t ) = ( - 2t , 5 + 4t , 1 + 4t )$

5. I don't think you found the direction vector right, it's point b-a not a-b