A question about an equation of a plane Let $A=(1,3,1)$; $B=(1,1,1)$; $C=(2,0,1)$; $D=(1,-2,3)$.
Determine the equation of a plane that passes through $D$ and is parallel with $(ABC)$.
I know the fact that $\mbox{dir}(\text{plane})=\mbox{dir}(ABC)$ if plane is parallel to $(ABC)$, but I don't know how to continue from here.
 A: In my previous answer, I used vectors to solve the problem. This will solve the problem for any plane you're given.
However, I should've spotted earlier that the plane you were given is quite simple - notice that $z=1$ for $A$, $B$ and $C$.
Therefore, the plane is parallel to the $xy$-plane and is $z=1$.
So you can then spot that $z=3$ for $D$, and then conclude the equation is $$z=3$$
A: The generic equation for a plane that includes the origin is:
$$
ax+by+cz=0
$$
it must be parallel to $\vec{AB}=\vec{OB}-\vec{OA}$ and $\vec{AC}=\vec{OC}-\vec{OA}$, that means $\vec{OB}-\vec{OA}$ and $\vec{OC}-\vec{OA}$ must satisfy the equation and so you can solve the system (two equations and three unknowns variables $a$, $b$ and $c$) that has infinite solutions representing the same plane, you just pick one.
The generic plane parallel to $ax+by+cz=0$ has an equation
$$
ax+by+cz=d
$$
and so you can find $d$ for your plane just by substitution of the coordinates of $D$.
A: An answer using vectors:
First of all, determine the equation of the first plane which contains $A$, $B$ and $C$.
This can be done by considering the lines $AB$ and $AC$. These lines are contained within the plane, and so if we consider the cross product of vectors $\vec{AB}$ and $\vec{AC}$, then we will get a vector which is a normal to the plane (a vector which is at right angles to the plane).
Next, if $\textbf{n}$ is a normal vector to the plane, then the equation for the plane is:
$$\textbf{r}.\textbf{n}=k$$
where $\textbf{r}$ is a general vector, and $k$ is a constant to be determined.
$\textbf{n}$ is a normal vector to the plane containing $A$, $B$ and $C$. But it also is a normal vector to any plane parallel to this plane.
Therefore, you could use the equation $$\textbf{r}.\textbf{n}=k$$ and plug in $\vec{OD}$ as $\textbf{r}$ to find $k$ which will give you the equation of the plane desired.
In this example:
$$\textbf{n}=\vec{AB}\times\vec{AC}$$
$$\textbf{n}=\pmatrix{1-1 \\ 1-3 \\ 1-1}\times\pmatrix{2-1 \\ 0-3 \\ 1-1}$$
$$\textbf{n}=\pmatrix{0 \\ -2 \\ 0}\times\pmatrix{1 \\ -3 \\ 0}$$
$$\textbf{n}=\pmatrix{0 \\ 0 \\ 2}$$
So the equation of any plane with the normal vector $\textbf{n}$ will be:
$$\textbf{r}.\pmatrix{0 \\ 0 \\ 2}=k$$
Next, as $D$ is contained in the plane which we are trying to find, then:
$$\vec{OD}.\pmatrix{0 \\ 0 \\ 2}=k$$
$$\pmatrix{1 \\ -2 \\ 3}.\pmatrix{0 \\ 0 \\ 2}=k$$
$$k=6$$
Therefore the equation is:
$$\textbf{r}.\pmatrix{0 \\ 0 \\ 2}=6$$
$$\textbf{r}.\pmatrix{0 \\ 0 \\ 1}=3$$
or in Cartesian form:
$$z=3$$
