Find Linear Equation of a plane

1. Sep 22, 2013

PsychonautQQ

1. The problem estatement, all variables and given/known data
Find the Linear equation of a plane passing through the origin and points (6,2,2) and (5,-1,1)

2. Relevant equations
Linear equation of a plane takes the form ax+by+cz+d = 0

3. The attempt at a solution
So I believe I can find the direction numbers a,b,c by taking the first position minus the second position.

a = 6 - 5 = 1
b = 2 - -1 = 3
c = 2 - 1 = 2

I don't know what to do from here though. I could plug in one of the locations into the formula of format
a(x-xI) + b(y-yi) + c(z-zi) = 0
say I plug in the first position (6,2,2)
1(x-6) + 3(y-2) + 2(z-2) = 0
to get x+3y+2z-16=0
but my i'm typing that into the answer box and it's saying i'm wrong

2. Sep 22, 2013

Staff: Mentor

It's obviously wrong, since the origin (x = 0, y = 0, z = 0) doesn't satisfy the equation.

If we call your two given points P and Q, the cross product of the vectors OP and OQ is perpendicular to both. That's a normal to your plane. Once you know a normal and a point on the plane, you can quickly get the plane equation.

BTW, it's good practice to check your work by making sure that the points that are supposed to be on the plane are actually solutions to the equation you came up with. If all of them are, you can be pretty sure that your equation is correct. If at least one is not a solution to the equation, that's a tipoff that something is wrong.

3. Sep 22, 2013

CAF123

The equation of the plane you got does not have (5,-1,1) or (0,0,0) as a solution, I.e it does not pass through that point. The eqn a(x-xI) + b(y-yI) + c(z-zI) = 0 can be recast as $\vec{n} \cdot \vec{r} = 0$ where $\vec{n}$ is a normal to the plane and $\vec{r}$ is a vector in the plane.

4. Sep 22, 2013

PsychonautQQ

I cross OP with OQ and got
0i+4j-16k and that gives me n.

n = (a,b,c) = (0,4,-16)
and then say I use the point (6,2,2)

Can i plug this values into the equation
a(x-xi) + b(y-yi) + c(z-zi) = 0
0 + 4(y-2) - 16(z-2) = 0
4y - 8 - 16z + 32 = 0
4y - 16z + 24 = 0
Web work is saying this is wrong but I don't know why it looks right?

And then you said n dot r = 0
so (0,4,-16) dot (xi+yj+zk) = 0 to get r which maybe are the real (a,b,c) values to the equation above..?
if I do this to solve for r there can be more than one correct answer.. the lowest one I guess would be x=0 y=4 and z=1 but I don't know if i'm on the right track here...

5. Sep 22, 2013

Staff: Mentor

That's not what I get.

When you calculate a cross product, there are lots of opportunities to make errors, especially sign errors. When you're done it's a good idea to check your work. Is <6, 2, 2> * <0, 4, -16> = 0? Is <5, -1, 1> * <0, 4, -16> = 0?

6. Sep 23, 2013

CAF123

Once you fix the normal, you want to find a vector in the plane between some arbritary point (x,y,z) in the plane and a known point. Draw some vector $\langle x,y,z\rangle$ with its tail at the origin and the tip at a point in the plane. This will allow you to find a vector in the plane which extends between some arbitrary (x,y,z) and a known point in the plane.