# Find equation of plane given conditions

1. Mar 1, 2017

### Mr Davis 97

1. The problem statement, all variables and given/known data
Find an equation of the plane that passes through the point (1,2,-2) and that contains the line
x=2t,y=3-t,z=1+3t.

2. Relevant equations

3. The attempt at a solution

I know that a plane is determined by a base vector and a normal vector, and the equation of the plane is $\vec{n} \cdot (\vec{r} - \vec{r_0})$, where n is the normal vector, and r0 is the base vector.

So we know the base vector, and now must determine what the normal vector is to the plane. We need to find two vectors in the plane, and take the cross product of those vectors. I am not sure where to find those two vectors... Would any two non-parallel vectors whose endpoints are on the line suffice?

2. Mar 1, 2017

### Buzz Bloom

Hi Mr Davis:
I suggest thinking about the problem a bit differently.

I think it would help if you write down, in section 2, the general form of the "equation of the plane". The identity of three points in the plane should lead you to three simultaneous equations to solve.

Good luck.

Regards,
Buzz

3. Mar 1, 2017

### Staff: Mentor

Is the given point on the line? If so, there is no unique plane.

If the given point is not on the line, use the parametric equations of the line to find two points. then you'll have three points, from which you can get two vectors in the plane. From these vectors, you can get a normal to the plane, and from it and the given point, getting the equation of the plane is straightforward.