General equation of a plane through $3$ points.

Question

Find the general equation of the plane through the points $A(1,1,0),\, B (1,0,1),$ and $C(0,1,2)$.

How is this done?

Answer 1

The general equation for a plane is $ax+by+cz=k$

Plug in the values for points A,B, and C and you have a linear system of three equations which you can solve. One way to do this is to represent the system with a matrix and use Gaussian elimination.

Note that there are 4 unknowns, so there is no single solution for a,b, and c since you can just multiply the equation by some constant and it stays the same, so you just solve for ratio of a to b to c.

Also if there is no single solution, then there is a solution set (multiple planes can satisfy the conditions).

Answer 2

Equation of the plane if given $3$ points can be found by $B-A=(1,0,1)-(1,1,0)=(0,-1,1)$ and $C-A=(0,1,2)-(1,1,0)=(-1,0,2)$. You now find the cross-product of $$(B-A)\times(C-A)=\begin{bmatrix}i & j & k \\0 & -1 & 1 \\-1 & 0 & 2 \end{bmatrix}=(-2,-1,-1)= \vec{n}$$ $i$, $j$ and $k$ are unit vectors that are directed along the $x$, $y$ and $z$ axis respectively. $\vec{n}$ is known as the normal vector $\vec{n}$ and is perpendicular to the equation of the plane. Now use the general vector equation of the plane:$$\vec{r}\cdot\vec{n}=\vec{a}\cdot\vec{n}$$ where $\vec{r}$ is the position vector relative to the origin of any point $(x,y,z)$ in the plane and $\vec{a}$ is a known point that lies on the plane. So using $\vec{a}=A$ we get $$(x,y,z)\cdot(-2,-1,-1)=(1,1,0)\cdot(-2,-1,-1)=-2-1+0=-3$$ In Cartesian coordinates this is $$-2x-y-z=-3=\color{blue}{2x+y+z=3}$$ as the equation of the plane.

Answer 3

Let the general equation of the plane be $ax+by+cz+d = 0\;,$ Then the equation of the plane passes through the point $A(1,1,0)$, that is $$a(x-1)+b(y-1)+c(z-0) = 0\tag1$$

The plane also passes through $B(1,0,1)\;,$ so we put $x=1,y=0,z=1\;,$ and we get

$$a(1-1)+b(0-1)+c(1-0) = 0\tag2$$

Now the plane also passes through $C(0,1,2)\;,$ so we put $x=0,y=1,z=2\;\bf{in \; (1)\;,}$ and we also get

$$a(0-1)+b(1-1)+c(2-0) = 0\tag3$$

Now eliminating $a,b$ and $c$ from these $3$ equations we get an equation of the plane.

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We can also eliminate $a,b$ and $c$ as

$$\displaystyle \begin{vmatrix} x-1& y-1 & z\\ 0& -1& 1\\ -1& 0 & 2 \end{vmatrix}=0\Rightarrow -1\left(y-1+z\right)+2(-x+1) = 0\Rightarrow 2x+y+z = 3$$