# Thread: Equation of a plane given point and norm.

1. ## Equation of a plane given point and norm.

Hi guys I am stuck on something,

Find the equation of the plane p containing the point P(0;2;3) and normal to the line with parametric
form x = 1 + 2t; y = 0; z = t.

So norm is (1,0,0) + t(2,0,1)?

Next do I have to define another point say, R on plane so-

vector PR will be perpendicular to norm direction vector and-

PR x n = 0 (cross product of 2 perp. vectors)..??

If this is the right logic how do I use the norm in the form I have? I think I'd be able to do this if I had the norm as a 3 number direction vector, not (1,0,0) + t(2,0,1)..

Am I right in saying that as a plane is Ax + By + Cz = D, norm (x,y,z) provides the coefficients in this formula and point P(x,y,z) gives the rest of the information to algebraically solve?

2. ## Re: Equation of a plane given point and norm.

Originally Posted by camjerlams
Find the equation of the plane p containing the point P(0;2;3) and normal to the line with parametric
form x = 1 + 2t; y = 0; z = t.
So norm is (1,0,0) + t(2,0,1)?
Next do I have to define another point say, R on plane so-
vector PR will be perpendicular to norm direction vector and-
Am I right in saying that as a plane is Ax + By + Cz = D, norm (x,y,z) provides the coefficients in this formula and point P(x,y,z
The norm is simply a vector. In this case, $\displaystyle N=<2,0,1>$

A point on the plane is $\displaystyle R=(x,y,z)$. Then $\displaystyle \overrightarrow {PR} = \left\langle {x - 0,y-2,z - 3} \right\rangle$ is a vector "in" the plane.
As you say that $\displaystyle \overrightarrow {PR}$ must be perpendicular to the normal $\displaystyle N$.

Thus $\displaystyle N\cdot\overrightarrow {PR}=0$ or $\displaystyle 2(x)+1(z-3)=0$