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Verify that the equation of a plane that passes through three points $(a_{1},a_{2},a_{3}),(b_{1},b_{2},b_{3}),(c_{1},c_{2},c_{13})$ of the space is:

$$\begin{vmatrix} 1 & x_{1} & x_{2} & x_{3} \\ 1 & a_{1} & a_{2} & a_{3} \\ 1 & b_{1} & b_{2} & b_{3} \\ 1 & c_{1} & c_{2} & c_{3} \\ \end{vmatrix}= 0$$

I have tried to reduce the determinant by making zeros, but it continues being so enormous to solve by Sarrus: $$\begin{vmatrix} 1 & x_{1} & x_{2} & x_{3} \\ 0 & a_{1}-x_{1} & a_{2}-x_{2} & a_{3}-x_{3} \\ 0 & b_{1}-x_{1} & b_{2}-x_{2} & b_{3}-x_{3} \\ 0 & c_{1}-x_{1} & c_{2}-x_{2} & c_{3}-x_{3} \\ \end{vmatrix}= 0$$

Is there a simple way to prove that?

Thanks in advance.

user9867
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    I would simply note that each of the three given points obviously satisfies the equation since then we have two identical lines in the determinant. Add to that the fact that the equation is linear and you are done- three points determine a plane. – user247327 Jan 15 '20 at 12:35
  • The determinant 1/6 of the volume of the tetrahedron with the points as vertices. The volume is zero iff the points are coplanar. – lhf Jan 15 '20 at 12:40
  • See https://math.stackexchange.com/a/2684551/265466. In that answer, the equation of a circle is found, but the underlying principle is the same: the points on the plane are exactly those for which the first row is a linear combination of the others. – amd Jan 15 '20 at 18:47

2 Answers2

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There are several ways you could go about proving this that have been mentioned in other answers and comments. Here’s one that hasn’t come up yet directly.

If you take the general Cartesian equation of a plane $Ax+By+Cz+D=0$ and plug in the coordinates of the three known points, you get a system of linear equations in the unknown coefficients: \begin{align}Aa_1+Ba_1+Ca_1+D &= 0\\Ab_1+Bb_1+Cb_1+D &= 0\\Ac_1+Bc_1+Cc_1+D &= 0\end{align} which can be written in matrix form as $$\begin{bmatrix}a_1&a_2&a_3&1\\b_1&b_2&b_3&1\\c_1&c_2&c_3&1\end{bmatrix} \begin{bmatrix}A\\B\\C\\D\end{bmatrix} = 0.$$ For a unique plane to be defined by the three points, then must not be colinear, in which case the coefficient matrix on the left has full rank and its null space is one-dimensional. This corresponds to the fact that multiplying both sides of the equation of the plane by a nonzero scalar produces an equivalent equation for the same plane.

For any other point on the plane, we can generate another linear equation in the unknown coefficients and add it to the system: $$\begin{bmatrix}x&y&z&1\\a_1&a_2&a_3&1\\b_1&b_2&b_3&1\\c_1&c_2&c_3&1\end{bmatrix} \begin{bmatrix}A\\B\\C\\D\end{bmatrix} = 0.$$ This system has a nontrivial solution iff the matrix on the left is singular, i.e., $$\begin{vmatrix}x&y&z&1\\a_1&a_2&a_3&1\\b_1&b_2&b_3&1\\c_1&c_2&c_3&1\end{vmatrix} = 0.$$ If you’re familiar with homogeneous coordinates, you can interpret the above equation as saying that every point on the plane is a linear combination of the three fixed noncolinear points, i.e., it is the join of those points.

The above idea can be applied to many other equations. For example, the determinant form of equation of a circle through three noncolinear points or a conic through five points in general position can also be understood in this way.

amd
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One approach is as follows. Let $a$ denote the vector $(a_1,a_2,a_3)$ and so-forth. It suffices to show that $x$ lies in the plane through $a,b,c$ if and only if the system of equations $$ \begin{cases} k_1 a + k_2 b + k_3 c = x\\ k_1 + k_2 + k_3 = 1 \end{cases} $$ has a solution for $k_1,k_2,k_3 \in \Bbb R$. Indeed, the $x$ of the form $x = k_1 a + k_2 b + k_3 c$ for some $k_i$ with $k_1+k_2+k_3 = 1$ form the affine space generated by $a,b,c$.

Another approach: $x$ lies in the plane through $a,b,c$ if and only if the vectors $x-c,a-c,b-c$ are linearly dependent.

Ben Grossmann
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