# Equation of a plane given point and line in parametric form

1. Sep 1, 2010

### musicmar

1. The problem statement, all variables and given/known data
Find the equation of the plane that contains P=(-1,0,1) and r(t)=<3t,t,8>

2. Relevant equations

3. The attempt at a solution

n * <r-r0>=0
n * <t+2, 2t, 3t> = 0

I distributed the n, adding the terms and obtained:

1/t = -2n(n+2n+3n)

Clearly, I've done something wrong. If someone could point me in the right direction with even how to start this problem correctly, that would be great.

Thanks!

2. Sep 1, 2010

### musicmar

Ok, I set t=1, found a point P0 (2,2,2) and found two vectors OP and OP0 and took the cross product.
I then took the dot product of the normal vector I found above and the given point P.

So, my equation of the plane is 2x + 2x = -1

Can someone tell me if this is valid/and/or correct?

3. Sep 1, 2010

### Staff: Mentor

You won't be able to get a unique plane if all you know is a point in the plane and a vector that lies in it.
I'm not following what you're doing above. I get that n is a normal to the plane, but where did <t + 2, 2t, 3t> come from?
???

4. Sep 2, 2010

### HallsofIvy

Staff Emeritus
2x+ 2x= -1?? The plane x= -1/4 is the plane parallel to the yz-plane at x= -1/4.
Did you mean 2x+ 2y= -1 or perhaps 2x+ 2z= -1?

musicmar, a line and a point not on the line determine a plane. But a "vector" is not a "line". It gives the direction but not specific points. Just given a "vector" we might have a line in the direction of the vector through the given point- and that will not determine a plane.

Anyway, to check if 2x + 2x= 4x = -1 is a solution, see if it meets the conditions. Does the given point (-1, 0, 1) lie in it? No, it doesn't; $4(-1)\ne -1$. Nor does it lie in 2x+ 2y= -1 or 2x+ 2z= -1. $2(-1)+ 2(0)= -2\ne -1$ and $2(-1)+ 2(1)= 0\ne -1$.