# Equation of a line from Parametric equations

1. Sep 16, 2004

### Spectre32

How does one find the equation of a line from parametric equations?

In spefiic i'm looking at this: x(t) = 1+2t , y(t) = -1 + 3t , z(t) = 4+t.... I think i gotta use something liek x-1/a = y-1/b=z-1/c or something like that. If what i just said is true, then i'm lost on what to do next.

Thanks

2. Sep 16, 2004

### Tide

t = (x-1)/2 = (y+1)/3 = z-4

3. Sep 16, 2004

### Spectre32

or wait.... do i uses this:

<x_0,y_0,z_0>+ t<a,b,c> ...

now i'm really confused.

4. Sep 16, 2004

### Spectre32

Yeah tide, i did that, but then isn't there something else i gotta do? I dont know i'm kinda confused about this stuff. Also allow me to add the rest of the question. IT's find the equation of a plane containing that line, and the point (1,-1,5). Sooo i need the N =<a,b,c> vector and a point ot get my equation of a plane.. So could i use use the second half of that equation i posted aboce ( t<a,b,c>) to get what i need for my N vector?

Last edited: Sep 16, 2004