Question 345483
<pre>

We find a vector parallel to the line by the method of
"subtracting the coordinates" of (3,-2, 1) and (1, 1,-3).

(1-3)<b><i>i</b></i> + (1-(-2))<b><i>j</b></i> + (-3-1)<b><i>k</b></i> = -2<b><i>i</b></i> + 3<b><i>j</b></i> - 4<b><i>k</b></i> = < -2, 3, -4 >

A parametric form of the line is then

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

Since the direction vector < -2, 3, -4 > has no 0 coordinates,

we can write an equation in symmetric form:

{{{(x-3)/(-2)}}}{{{""=""}}}{{{(y-(-2))/3}}}{{{""=""}}}{{{(z-1)/(-4)}}}

Notice that I said "a" parametric or symmetric form, not "the" parametric 
of symmetric form, because neither form is unique.

Edwin</pre>