# Equation of a line/planes chapter question

1. Sep 19, 2009

### tnutty

1. The problem statement, all variables and given/known data

Find the parametric equation and symmetric equation for the line :

11) The line through (1,-1,1) and parallel to the line x + 2 = y/2 = z-3

2. Relevant equations

Parametric equation :

x = x0 - at
y = y0 - bt
z = z0 - ct

symmetric equations :

(x - xo) / a = (y - yo)/b = (z - zo) /c

3. The attempt at a solution

I just need to find a parallel vector to the line x + 2 = y/2 = z-3.
Then I could use that vector with the initial point to find the parametric equation and the
symmetric equation.

I am just not sure how to find the parallel vector. Looking at the book for answer,
I see that the parallel vector is <2,3,2>, but I can't figure out how to obtain that answer.

2. Sep 19, 2009

### HallsofIvy

Staff Emeritus
One way to convert from "symmetric form" (x- x0)/a= (y- y0)/b= (z- z0)/c to "parametric form" is to use that "common value" as your parameter. That is, (x- x0)/a= (y- y0)/b= (z- z0)/c= t so you have (x- x0)/a= t, (y- y0)/b= t, and (z-z0)/c= t. Solve those for x, y, and z to get parametric equations for the same line.

Of course, once you have parametric equations, x= At+ x0, y= Bt+ y0, z= Ct+ z0, a vector parallel to the line is <A, B, C>.

(No, <2, 3, 2> is NOT parallel to x+ 2= y/2= z- 3. for example, on point that satisfies those is (-1, 2, 4) (I took x+2= y/2= z- 3= 1) and another is (0, 4, 5) (x+2= y/2= z- 3= 2). The vector from one of those to the other is NOT parallel to <2, 3, 2>.)

3. Sep 19, 2009

### tnutty

I though that the equation given "x + 2 = y/2 = z-3" looked like the symmetric form,
with a = b = c = 1. So <A,B,C> = <1,1,1> ?

The from that I could find the displacement vector of <1,-1,1> , <1,1,1>

Then use that vector, which must be parallel to find the answer?