Thread: finding parametric & symmetric equations of line

Have a question which is as above for line through points (0, 1/2, 1) and (2,1, -3)

They way I understand it this is what you do. Get a directional vector between the two points ie let (0, 1/2, 1) be point A and (2, 1, -3) be point B. Thus vector AtoB is < (2-0), (1-1/2), (-3-1)> giving < 2, 1/2, -4>

Now parametrics would be x= (0+2t) , y=( 1/2+1/2t) and z = (1+ (-4t))

giving symetric as being x/2 = (2y-1) = (z-1)/-4

But it looks like I'm wrong can someone please advise what I am doing wrong here.

Hello, Craka!

Find the equations for line through points

The way I understand it this is what you do.
Get a directional vector between the two points
Thus:

The parametrics would be: .

giving symmetric as: .

But it looks like I'm wrong . . . . WHO said you're wrong?

There is an unavoidable problem with the equations of a line.
. . I wish textbook authors and math teachers would be own up to it.

There are a zillion possible forms of the equations of a line.
. . There is no "one right answer."



You used point and . . . and your answers are correct.


Suppose you used
You would have: . . . also correct.



, we can use a parallel vector: .

With point , we would have: .


You see, the variations are endless . . .

Thankyou. That was causing me alot of anguish, as all the different sources I looked at seemed to have the method I was using, however the answer in the rear of my book was different to what I was getting.