# Find the equation of the plane knowing that it passes through 3 points

I have to find the equation of the plane that passes through $(0, 0, 0), (4, 0, -2), (0, 8, -6)$.

I have done the following:

The equation of the plane is of the form $$ax+by+cz+d=0$$

Since the points $(0, 0, 0), (4, 0, -2), (0, 8, -6)$ are points o fthe plane, we have that $$d=0 \\ 4a-2c=0 \Rightarrow a=\frac{c}{2} \\ 8b-6c=0 \Rightarrow b=\frac{3c}{4}$$

So $$\frac{c}{2}x+\frac{3c}{4}y+cz=0 \Rightarrow \frac{1}{2}x+\frac{3}{4}y+z=0$$

Is it correct??

Is there also an other way to find the equation of the plane??

Maybe using the cross-product??

Hint: first you find the normal to the plane: $N = \begin{pmatrix} 4 \\ 0 \\ -2 \end{pmatrix} \times \begin{pmatrix} 0 \\ 8 \\-6 \end{pmatrix} = (a,b,c)$, then your plane is:

$a(x-0)+b(y-0)+c(z-0) = 0$

Since you mentioned the cross product, let's use it. If you have a plane $\Pi$, and $P,Q,R \in \Pi$, then the vectors $Q-P$ and $R-P$ lie in the plane. Hence $(Q-P)\times(R-P)$ is normal to the plane.

In your exercise, $(4,0,-2)$ and $(0,8,-6)$ lie in the plane. Then $(2,0,-1)$ and $(0,4,-3)$ lie in the plane too. So: $$\begin{vmatrix} {\bf i} & {\bf j} & {\bf k} \\ 2 & 0 &-1 \\ 0 & 4 & -3\end{vmatrix} = (4,6,8)$$is normal to the plane. Let's use $(2,3,4)$ instead. We have that the plane is $\Pi: 2x+3y+4z = 0$. (I'm already using that ${\bf 0}\in \Pi \implies d = 0$).

A plane is uniquely determined given a point on the plane and a vector perpendicular to the plane. Such a vector is said to be normal to the plane.

Given a point $P_0 = (x_0, y_0, z_0)$ and a normal $\vec n = (a, b, c)$ to a plane, a point $P = (x, y, z)$ will be on the plane if $\vec{P_0P}$ is perpendicular to $\vec n$ that is,

$$\vec n \bullet \vec{P_0P}=0$$

If instead of being given a point and the normal, taken three non-colinear points $P_1(x_1,y_1,z_1), P_2(x_2,y_2,z_2)$ and $P_3(x_3,y_3,z_3)$, we form the vectors $\vec{P_1P_2}$ and $\vec{P_1P_3}$.

The cross product $\vec{P1P2}\times\vec{P1P3}$ is a vector perpendicular to both $\vec{P_1P_2}$ and $\vec{P_1P_3}$ and therefore perpendicular to the plane. Use one of the point, the vector obtained from the cross product in the above equation to derive the equation of the plane.

We get, $$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1\\ x_2-x_1 & y_2-y_1 & z_2-z_1\\ x_3-x_1 & y_3-y_1 & z_3-z_1\\ \end{vmatrix} =0$$

is the equation of plane , three points are given. $2x+4z+3y = 0$

We can do with some geometry and linear algebra.

We know area of of a triangle with three vertices is proportional to determinant :

\begin{vmatrix} x_1 & y_1 & z_1\\ x_2 & y_2 & z_2\\ x_3 & y_3 & z_3\\ \end{vmatrix}

The determinant will vanish if the points are collinear.

It is derived by setting cross product to zero. Now let us pull in our general point.

Let one point be the general point P (x,y). I leave it to you to draw simple difference vector diagram of triangle where vector P is subtracted and other modified. When the area is zero, vector lines are in the same direction leaving zero enclosed area, so it is a situation that the general straight must now pass through plane determined by three points.

$$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1\\ x_2-x_1 & y_2-y_1 & z_2-z_1\\ x_3-x_1 & y_3-y_1 & z_3-z_1\\ \end{vmatrix} =0$$