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?

  • 1
    $\begingroup$ 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. $\endgroup$ Commented Oct 30, 2020 at 19:14
  • $\begingroup$ @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 $P_3$. $\endgroup$
    – Fran
    Commented Oct 30, 2020 at 19:19
  • $\begingroup$ $\vec{P_2P_3}=(2,2,-6)$. $(2,1,1)$ is the correct normal vector. $\endgroup$ Commented Oct 30, 2020 at 19:20
  • $\begingroup$ @ShubhamJohri Kill me please. $\endgroup$
    – Fran
    Commented Oct 30, 2020 at 19:21
  • $\begingroup$ Die Die Die! 🔪 $\endgroup$ Commented Oct 30, 2020 at 19:22

1 Answer 1


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 $P_1P_2P_3$).

  • $\begingroup$ 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... :( $\endgroup$
    – Fran
    Commented Oct 30, 2020 at 19:32
  • $\begingroup$ Never mind that. You are welcome! $\endgroup$ Commented Oct 30, 2020 at 19:34

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