# Proof, equation of a plan, linear algebra

1. Mar 14, 2009

### sciencegirl1

1. The problem statement, all variables and given/known data
Equation of a plan that contains the point p0=x0,y0,z0 and the normalvector N=A,B,C is N*P*P0=0 where P is(x,y,z)
=>A(x-xo)+B(y-y0)+c(z-z0)=0

Is it possible to prove this?
Can you help me if it is?

2. Mar 14, 2009

### yyat

Maybe it should be N*(P-P0)=0 instead of N*P*P0?

If I understand correctly, you are supposed to prove that the plane that contains the point P0 and has normal N is given by the equation N*(P-P0)=0. This is true. Did you attempt to prove it?

3. Mar 14, 2009

### sciencegirl1

No I didn´t because I have no idea how to or if it´s possible.
Is it?
And if it is, can you tell me how to do it.

4. Mar 14, 2009

### Count Iblis

P - P0 is a vector that is parallel to the plane, so this must be orhogonal to the normal vector. Therefore:

N dot (P - P0) = 0

5. Mar 14, 2009

### lanedance

Hi sciencegirl1

drawing a picture always seems to help, but in line with yyat's comments, but the geometric reasoning is as follows

first take a point on the plane
$$\textbf{p}_0 = (x_0,y_0,z_0)$$

next take any other point on the plan
$$\textbf{p} = (x,y,z)$$

the difference is a vector parallel to the plane
$$\textbf{p}-\textbf{p}_0 = (x- x_0,y-y_0,z-z_0)$$

so if we have a vector normal to the plane
$$\textbf{n}= (a,b,c)$$

then by defintion this is perpindciular to any vector parallel to the plane, so
$$\textbf{n}*(\textbf{p}-\textbf{p}_0)= 0$$

i'm not too sure what you want to do here, but if you want to prove this try writing a formula for a plane, find the normal vector, then take the dot product with the difference of any 2 points on the plane and see if you can show its 0

Last edited: Mar 14, 2009
6. Mar 14, 2009

### Count Iblis

If you have shown that if P is in the plane, then:

N dot (P-P0) = 0

Then what you need to show is that the reverse is also true. This is equivalent to saying that if P is not in the plane, you have

N dot (P - P0) not equal to zero.

This follows from the fact that if P - P0 is not in the plane, it must have a component along N.

So, you then have a necessary and sufficient condition for P to be in the plane: The equation

N dot (P - P0) = 0

This is therefore the equation of the plane.