Question

How would I convert cartesian equation to parametric?
For example:
2x-y+3z-6=0

Answer 1

This plane is 2x-y +3z =6
hence the normal vector n is \(\vec n=<2,-1,3>\),
and the direction vector d is \(\vec d=< 1,5,1>\), how, right? hehehe,,,,,
The logic is the normal vector is \(\vec n\perp\vec d\), Let \(\vec d=<a,b,c>\) , then inner product of n and d =0, that is \(<2,-1,3><a,b,c>=0\)
or 2a -1b +3c =0 --> b = 2a +3c. Hence if a =1, c =1, b = 5,

that is how I got d =<1,5,1>

Answer 2

Now, just find out only 1 point on the plane, very easy
2x-y+3z=6, let x =2, y =1, then z = 1 you have the point (2,1,1) on the plane, right?

Answer 3

combine all,
\(\vec d =<1,5,1>\), the point \((2,1,1)\)
hence the parametric equation is
\(x = 2+t\)
\(y =1+5t\)
\(z = 1+t\)