# Not coplanar points in $\mathbb{R}^3$ if and only if its determinant is nonzero [closed]

How to prove that points $P_1(x_1,y_1,z_1)$, $P_2(x_2,y_2,z_2)$, $P_3(x_3,y_3,z_3)$, $P_4(x_4,y_4,z_4)$ are not coplanar if and only if the determinant $$\left|\begin{array}{cccc} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ x_4 & y_4 & z_4 &1 \\ \end{array} \right| \neq 0$$

## closed as off-topic by Travis, graydad, Empty, Claude Leibovici, user223391 Oct 17 '15 at 7:49

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If the points are coplanar, then you can find the equation of a plane $P \equiv ax+by+cz+d=0$ such that your points all verify the equation with $(a,b,c,d) \neq (0,0,0,0)$. Hence the system $$\begin{cases} x_1 a + y_1 b + z_1 c + 1 d = 0 \\ x_2 a + y_2 b + z_2 c + 1 d = 0 \\ x_3 a + y_3 b + z_3 c + 1 d = 0 \\ x_4 a + y_4 b + z_4 c + 1 d = 0 \end{cases}$$ has a non trivial solution and the determinant vanishes (if not $(0,0,0,0)$ would be the only solution).

Conversely, if the determinant vanishes, the system has at least a non zero solution $(a,b,c,d)$ and this forms the equation of a plane satisfied by the points which are coplanar.

• Even more, this argument shows if the rank of this matrix is two then there multiple distinct planes containing the points and hence they must lie on a line. In this case the rank can't be 1, but in higher dimensions you can push this further. – Nate Oct 16 '15 at 20:12

Claim:

The absolute value of this determinant
$$\left|\begin{array}{cccc} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ x_4 & y_4 & z_4 &1 \\ \end{array} \right|$$
is the volume of a (skew) parallelepiped with vertices $(x_1,y_1,z_1),(x_2,y_2,z_2),(x_3,y_3,z_3),(x_4,y_4,z_4)$.

Proof: We know that the volume is the absolute value of the determinat that has as columns the vectors that represents the sides of the parallelepiped: $$d= \left|\begin{array}{cccc} x_1-x_4 & y_1-y_4 & z_1-z_4 \\ x_2-x_4 & y_2-y_4 & z_2-z_4 \\ x_3-x_4 & y_3-y_4 & z_3-z_4\\ \end{array} \right|$$ Now, by co-factor decomposition of a determinant we have: $$d= \left|\begin{array}{cccc} x_1-x_4 & y_1-y_4 & z_1-z_4&1 \\ x_2-x_4 & y_2-y_4 & z_2-z_4&1 \\ x_3-x_4 & y_3-y_4 & z_3-z_4&1\\ 0&0&0&1 \end{array} \right|$$ and, multiply the last columns bi $x_4$ and adding to the frst column we find $$d= \left|\begin{array}{cccc} x_1 & y_1-y_4 & z_1-z_4&1 \\ x_2 & y_2-y_4 & z_2-z_4&1 \\ x_3 & y_3-y_4 & z_3-z_4&1\\ x_4&0&0&1 \end{array} \right|$$ Doing the same,i.e. multiplying by $y_4$ and $z_4$ and adding to the second and third columns we arrive at: $$d= \left|\begin{array}{cccc} x_1 & y_1 & z_1&1 \\ x_2 & y_2 & z_2&1 \\ x_3 & y_3 & z_3&1\\ x_4&y_4&z_4&1 \end{array} \right|$$

That prove the claim.

Now the answer is obvious since the volume is null iff the four points are coplanar.