# When determining the equation of a plane given 3 points, why the election of the points to compute the normal vector matter?

Let's say I have three points $$P_1 = (1, 2, -1)$$, $$P_2 = (-1, 1, 4)$$ and $$P_3 = (1, 3, -2)$$ and I want to find the equation of the plane determined by it.

My reasoning is:

Given three points of a plane, there are three lines parallel to this plane determined by the pairs of points ($$P_1, P_2$$), ($$P_2, P_3$$) and ($$P_1$$, $$P_3$$), so I need to find a normal vector $$\vec{N}$$ such that it's perpendicular to two of those lines (and since all three lines are in the same plane, then if a $$\vec{N}$$ is perpendicular to two lines, then it must be perpendicular as well to the third line).

The standard way to compute $$\vec{N}$$ is by solving the equations system yielded by $$\vec{N} \cdot (\vec{P_3} - \vec{P_1}) = \vec{N} \cdot (\vec{P_2} - \vec{P_1}) = 0$$, but if I happen to choose $$\vec{N} \cdot (\vec{P_3} - \vec{P_1}) = \vec{N} \cdot (\vec{P_3} - \vec{P_2}) = 0$$ instead, although I get a $$\vec{N} = (4, -3, 1)$$ that is perpendicular to $$\overline{P_{1}P_{3}}$$ and $$\overline{P_{2}P_{3}}$$, it's not perpendicular to $$\overline{P_{1}P_{2}}$$ as I expected it to be. So at least the highlighted part of my reasoning above is false. But why? Why I seem to be forced to arbitrarily solve $$\vec{N} \cdot (\vec{P_3} - \vec{P_1}) = \vec{N} \cdot (\vec{P_2} - \vec{P_1}) = 0$$ if the line determined by $$P_2$$ and $$P_3$$ is in the same plane as well (and thus its normal vector should be parallel to the normal vectors of the other two lines, which doesn't happen)?

Solving $$\vec{N_1} \cdot (\vec{P_3} - \vec{P_1}) = \vec{N_1} \cdot (\vec{P_2} - \vec{P_1}) = 0$$ yields $$\vec{N_1} = (2, 1, 1)$$, which is the correct $$\vec{N}$$ of the plane.

Solving $$\vec{N_2} \cdot (\vec{P_3} - \vec{P_1}) = \vec{N_2} \cdot (\vec{P_3} - \vec{P_2}) = 0$$ yields $$\vec{N_2} = (4, -3, 1)$$.

But $$\vec{N_1}$$ and $$\vec{N_2}$$ are not parallel to each other, which is geometrically counterintuitive to me. What is going on?

• Your reasoning is correct but calculation is wrong. $\vec N(4,-3,1)$ is not perpendicular to $\vec{P_1P_3}=(0,1,-1)$ nor to $\vec{P_2P_3}=(2,2,-6)$. Also note that the two normal vectors should have been parallel for both your answers to be correct, which is not the case. Commented Oct 30, 2020 at 19:14
• @ShubhamJohri I know. And (2, 1, 1) is not perpendicular to $\vec{P_2, P_3} = (2, 2, -2)$, so what? I'm asking why none of $(4, -3, 1)$ and $(2, 1, 1)$ are perpendicular to all $P_1$, $P_2$ and $P_3$.
– Fran
Commented Oct 30, 2020 at 19:19
• $\vec{P_2P_3}=(2,2,-6)$. $(2,1,1)$ is the correct normal vector. Commented Oct 30, 2020 at 19:20
– Fran
Commented Oct 30, 2020 at 19:21
• Die Die Die! ðŸ”ª Commented Oct 30, 2020 at 19:22

There is no flaw in the reasoning. The calculation is wrong.

1. The two normal vectors are not parallel.
2. $$\vec N_2(4,-3,1)$$ is not orthogonal to $$\vec{P_1P_3}=(0,1,-1)$$ and $$\vec{P_2P_3}=(2,2,-6)$$ but it is orthogonal to $$\vec{P_1P_2}=(-2,-1,5)$$. Note that $$\vec N_1(2,1,1)$$ is orthogonal to all of them.

$$\vec{N_1}(2,1,1)$$ is the correct normal and you have run into an error while calculating $$\vec{N_2}$$.

$$\vec{N_2}\cdot\vec{P_1P_3}=0\implies y-z=0\\\vec{N_2}\cdot\vec{P_2P_3}=0\implies x+y-3z=0\\$$This gives $$\vec{N_2}=z(2,1,1)$$ parallel to $$\vec{N_1}$$.

Note that an alternative way to calculate the normal vector is to find $$\vec{P_1P_3}\times\vec{P_2P_3}$$ (or you can take the cross product of any two sides of the triangle $$P_1P_2P_3$$).

• Thanks, this makes sense. My error was computing $\vec{P_2, P_3}$, which equals to $(2, 2, -6)$ and in my calculations I set it to $(2, 2, -2)$. I have been the whole day squeezing my brain about why my reasoning could be wrong and I didn't care to re-check the numbers in the first place... :(
– Fran
Commented Oct 30, 2020 at 19:32
• Never mind that. You are welcome! Commented Oct 30, 2020 at 19:34