A question about an equation of a plane

Question

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.

Do you know vector algebra? If so, that could help you out here. Btw, it's "plane", not "plan". — Shuri2060, Mar 20 '16 at 16:21
I know some,but don't know how to continue that problem.Please,could you answer me? — Adrian Costea, Mar 20 '16 at 16:33
If you understand the cross and dot product, then you should be able to use my method to find the equation. — Shuri2060, Mar 20 '16 at 16:34
I've just spotted a much easier solution to this particular problem because the first plane is parallel to the $xy$-plane. I've added this second solution. — Shuri2060, Mar 20 '16 at 16:46

score 3 · Answer 1 · answered Mar 20 '16 at 16:45

3

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 $$z=3$$

answered Mar 20 '16 at 16:45

Shuri2060

4,353

Why,if z=1,the plane is parallel to the xy-plane? – Adrian Costea Mar 20 '16 at 16:47
Do you have a 3D-graph plotter program? This might help you see why. But to visualise it, since $z=1$, if you let $z$ be the "height" of a point, then you will see all points must have a height of 1. If this doesn't help, then think of the analogy in 2D. If $y=1$, then the line must be parallel to the $x$-axis. – Shuri2060 Mar 20 '16 at 16:51

Marco Disce · Accepted Answer · 2016-03-21T07:41:56.717

2

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 $D$.

edited Mar 21 '16 at 07:41

answered Mar 20 '16 at 16:56

Marco Disce

1,960

The plane through $A$ , $B$ and $C$ doesn't go through the origin? – Shuri2060 Mar 20 '16 at 17:07
No,it goes through the point D. – Adrian Costea Mar 20 '16 at 17:12
@QuestionAsker right, in fact there is no point passing through $A$, $B$ and $C$ here – Marco Disce Mar 20 '16 at 17:15
@QuestionAsker thank you for your edit – Marco Disce Mar 20 '16 at 17:16
No problem - I think I understand your method here now. I thought you initially assumed $A$, $B$, $C$ to go through the origin, but you're just simply finding the plane parallel to the first through the origin before translating it to go through $D$ which makes sense. – Shuri2060 Mar 20 '16 at 17:17
How can I find D only by substitution of the coordinates of D? – Adrian Costea Mar 20 '16 at 17:27
d is your only unknown variable at that point, you can find with one linear equation – Marco Disce Mar 20 '16 at 17:42

Shuri2060 · Answer 3 · 2016-03-20T16:52:28.470

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:

$$z=3$$

A question about an equation of a plane

3 Answers3