How do I find the parametric equations for a plane?

[vg19]

Hi,

Given the plane 2x + y - z = 4, find its parametric equations.

This question seems simple, but the solution is not coming to me. I know the normal is (2,1,-1). But how do I find out the direction vectors and the points for the parametric equations?

 

[Parth Dave]

I would suggest finding three points on the plane and create two direction vectors and a point for your parametric equation. The easiest three points would be the intercepts. if y and z are 0, the co-ordinate becomes: (2,0,0). Using a similar process, you can find the two other intercepts, which are:
(2, 0, 0)
(0, 4, 0)
(0, 0, -4)

Thus, you can create two direction vectors and a third point. I think there is an easier way to approach this, but for now it eludes me.

 

[wais]

Hi there,

To find the parametric equations for a plane, we need to first find two direction vectors that lie on the plane. One way to do this is by finding the cross product of the normal vector with any other vector that is not parallel to it. So in this case, we can choose the vector (1,0,0) to find the first direction vector. The cross product of (2,1,-1) and (1,0,0) is (0,-1,-2).

Next, we can find a second direction vector by taking the cross product of the first direction vector and the normal vector. So the second direction vector would be (-2,2,-1).

Now, to find the points for the parametric equations, we can choose any point on the plane and use it as the origin. Let's say we choose the point (0,0,4). Then, our parametric equations would be:

x = 0 + at + bs
y = 0 - ct
z = 4 + dt

Where a,b,c,d are any real numbers and t and s are the parameters that will help us generate points on the plane.

I hope this helps! Let me know if you have any other questions.