# Get any $3$ points in a plane from normal and one point

I know how to get the normal of a plane from any 3 points, how can I do the inverse?

Suppose any $3$ points on the plane: $p_1$, $p_2$ ,$p_3$

We can get the vectors: $v_1 = p_2-p_1$,$v_2 =p_3-p_1$

And the plane normal n as: $n = v_1\times v_2$? If I got only $n$ and $p_1$, how can I get ANY other $2$ points: $p_2$ and $p_3$ ou two vectors on plane?

Looking for some vector form so I can easily write in matlab.

• As a hint: If you find $p_2$ on $n$ such that $v_1 = p_2-p_1$, $v_1 \cdot n = 0$ you know that $v_1$ must be on the plane – J. Bush Mar 18 '16 at 18:22
• Sorry, do you want to find the plane? Because the possible $p_2,p_3$ points are infinite, while the plane isn't – Nicolò Mar 18 '16 at 18:23
• I know, ANY p2 and p3, just different from p1. – Pedro77 Mar 18 '16 at 18:24
• Take the three axis-aligned unit vectors $e_1,e_2,e_3$. Then $n\times e_1$, etc. are orthogonal to $n$ and at least two of them are nonzero. So $p+(n\times e_1)$ etc. are at least two other points on the plane. – Rahul Mar 18 '16 at 18:43

Expanding on @EmilioNovati 's answer in case you need more help. If we take vectors $\vec x, \vec n, \vec p$ where $\vec x = \left[ x_1,x_2\right],\vec n = \left[ n_1,n_2\right],\vec p = \left[ p_1,p_2\right]$ we know $$\vec n \cdot (\vec x - \vec p) = 0 \\ \vec n \cdot \left[x_1-p_1,x_2 - p_2\right] = 0 \\ n_1x_1-n_1p_1 + n_2x_2 - n_2p_2 = 0 \\ x_2 = \frac{n_1p_1+n_2p_2-n_1x_1}{n_2}$$ You know $\vec p$ and $\vec n$. You can substitute any $x_1$ to find the corresponding $x_2$
If you know the normal $\mathbf{n}$ and a point on the plane $\mathbf{p_1}$ than you know the equation of the plane: $$\mathbf{n}\cdot(\mathbf{x}-\mathbf{p_1})=0$$