Planes

The position of a given plane $E$ in space is completely determined by a point $A$ on this plane, and any vector $\vec n$ that forms a right angle with the plane. This vector is called a normal vector of $E$ .

Note:

Even though we draw borders to indicate a plane, it has no border and extends in all directions to infinity.
You can think of a plane as a set of infinitely many points. The statement " $A$ is on plane $E$ " can therefore be rewritten as $A \in E$ .

Make sure you understand the following examples.

Example 1

The plane containing point $A(0\vert 0\vert 1)$ and with normal vector $\vec{n}=\left(\begin{array}{r} 0\\ 0\\ 1 \end{array}\right)$ is parallel to the $xy$ -plane.
The plane containing point $A(0\vert 0\vert 1)$ and with normal vector $\vec{n}=\left(\begin{array}{r} 0\\ 0\\ -10 \end{array}\right)$ is the same plane as above.
The $yz$ -plane has the normal vector $\vec{n}=\left(\begin{array}{r} 1\\ 0\\ 0 \end{array}\right)$ or $\vec{n}=\left(\begin{array}{r} -0.1\\ 0\\ 0 \end{array}\right)$ or $\vec{n}=\left(\begin{array}{r} 140.321\\ 0\\ 0 \end{array}\right)$ .

Consider a plane $E$ containing point $A$ and with normal vector $\vec n$ . How can we decide if some other point $P$ in also on the plane?

From the figure below we see that

\boxed{P \in E\,\text{ if }\, \overrightarrow{AP} \perp \vec{n}}

Note that $\overrightarrow{AP} \perp \vec{n}$ if

\begin{array}{rll} \overrightarrow{AP} \bullet \vec{n} &=& (\vec{P}-\vec{A})\bullet \vec n=0\\ &=& \left(\begin{array}{r} P_x-A_x\\ P_y-A_y\\ P_z-A_z \end{array}\right)\bullet \left(\begin{array}{r} n_x\\ n_y\\ n_z \end{array}\right)=0\\ &=& (P_x-A_x)n_x+(P_y-A_y)n_y+(P_z-A_z)n_z\\ &=& P_x n_x + P_y n_y + P_z n_z -(A_x n_x+A_y n_y+A_z n_z)=0 \end{array}

So $P$ is in the plane $E$ if, and only if

\boxed{n_x P_x + n_y P_y + n_z P_z=\underbrace{n_x A_x + n_y A_y + n_z A_z}_{=d}}

This is called the normal equation of the plane. Let us summarise:

Theorem 1

A plane $E$ with normal vector $\vec n$ contains point $A$ . Assume they are given, that is, we know their coordinates and components. A point $P(x\vert y\vert z)$ is on $E$ if, and only if, its coordinates fulfil the normal equation

n_x x+ n_y y+ n_z z = d

where $d$ is calculated as follows:

d=n_x A_x + n_y A_y + n_z A_z

Note that, as $A$ and $\vec{n}$ are known, so is the number $d$ .

Example 2

A plane $E$ passes through the point $A(1\vert 0\vert 1)$ and has normal vector $\vec n=\left(\begin{array}{r} 7\\ -4\\ 5 \end{array}\right)$ . Is point $P(3\vert 1\vert 2)$ in $E$ ?

Solution

Method 1: No, because

\overrightarrow{AP} \bullet \vec{n} = \left(\begin{array}{r} 3-1\\ 1-0\\ 2-1 \end{array}\right) \bullet \left(\begin{array}{r} 7\\ -4\\ 5 \end{array}\right) = 14-4+5=15\neq 0

thus $\overrightarrow{AP} \not\perp \vec{n}$ .

Method 2: All points $(x\vert y \vert z)$ on $E$ fulfil the normal equation

n_x x +n_y y+ n_z z= d

where $n_x=7, n_y=-4$ and $n_z=5$ and $d$ is calculated as follows:

\begin{array}{lll} d &=& A_x n_x + A_y n_y + A_z n_z\\ &=& 7\cdot 1+ (-4)\cdot 0+ 5\cdot 1 = 12 \end{array}

Thus, the normal equation is

7 x -4 y+ 5 z= 12

Inserting the coordinates of $P$ into the equation, we get

7\cdot 3+ (-4)\cdot 1+ 5\cdot 2 = 27 \neq 12

so $P$ is not in $E$ .

Example 3

A plane $E$ contains the point $A(2\vert 3\vert -1)$ and has normal vector

\vec n=\left(\begin{array}{r} 2\\ -3\\ 8 \end{array}\right)

Write down the normal equation of $E$ . Use the normal equation to find out if point $U(1.5\vert 4\vert -0.5)$ in $E$ .

Solution

$d=2\cdot 2+(-3)\cdot 3+ 8\cdot (-1)=-13$ . Thus a point $(x \vert y\vert z)$ is in $E$ if, and only if

2x -3y + 8z = -13

This last equation is the normal equation of $E$ . To check if $U$ is in $E$ , we can insert its coordinates into the normal equation:

2\cdot 1.5 - 3\cdot 4+8\cdot (-0.5)=-13

So yes, $U$ is in $E$ .

Example 4

A plane $E$ has the normal equation

5x-2y+6z=2

Find a normal vector of $E$ . Also, one point in $E$ .

Solution

A normal vector is

\vec n=\left(\begin{array}{r} 5\\ -2\\ 6 \end{array}\right)

Any point $P(\vert x \vert y \vert z)$ with

5x-2y+6z=2

is in $E$ . So for example, let's use $x=0$ and $y=1$ , so we have to find $z$ with

0-2+6z=2

and it follows $z=\frac{2}{3}$ . Thus, one point in $E$ is $P(0\vert 1\vert \frac{2}{3})$ .

Exercise 1

Q1

A plane $E$ passes through the point $A(1\vert 4 \vert 1)$ and has normal vector $\left(\begin{array}{r} 8\\ 4\\ 5 \end{array}\right)$ .

Write down the normal equation of $E$ .
Find another point $Q$ in $E$ .

Q2

A plane $F$ contains the point $A(3\vert 0\vert 0)$ and has normal vector $\vec{n}=\left(\begin{array}{r} 6\\ 3\\ 2 \end{array}\right)$ .

Write down the normal equation of $F$ .
Is the point $P(1\vert 2\vert 3)$ in $F$ ?
Is the point $Q(2\vert 4\vert 6)$ in $F$ ?
Where does $F$ intersect the y-axis?

Q3

Find a normal vector of $E$ , and also a point in $E$ :

$-2x+4y-z=1$
$3x+3y+3z=3$
$x+y+z=0$
$x+y=3$
$z=1$

Q4

A plane $E$ contains the point $C(2\vert 4\vert -1)$ and has normal vector $\vec{n}=\left(\begin{array}{r} 10\\ -5\\ 20 \end{array}\right)$ .

Find the normal equation of $E$ .
Decide if point $P(1\vert 3\vert -2)$ is in $E$ .

Q5

The normal equations of two planes are shown below. Are the planes parallel?

3x-2y-4z=6

-6x+4y+8z=5

Solution

A1

The normal equation is $8x+4y+5z=d$ , where $d=8+16+5=29$ . Thus, we have
$8x+4y+5z=29$
Every point $(x\vert y\vert z)$ which fulfils this equation is in $E$ .
Choose, for example, $x=0, y=0$ , and thus $5z=29$ and therefore $z=5.8$ . One point in $E$ is therefore $Q(0\vert 0\vert 5.8)$ .

A2

The normal equation is $6x+3y+2z=d$ , where $d=3\cdot 6=18$ . Thus, we have
$6x+3y+2z=18$
Insert the coordinates of $P(1\vert 2\vert 3)$ into the normal equation, and check if the result is $d=18$ :
$6\cdot 1+3\cdot 2+2\cdot 3=18$
Indeed, so $P \in F$ .
Insert the coordinates of $Q(2\vert 4\vert 6)$ into the normal equation, and check if the result is $d=18$ :
$6\cdot 2+3\cdot 4+2\cdot 6 \neq 18$
So $Q \not\in F$ .
Let us denote the intersection point by $S$ . As $S$ is on the $y$ -axis, it must have the coordinates $S(0\vert S_y\vert 0)$ . As $S$ is in $F$ , it must fulfil the normal equation of $F$ :
$0 +3\cdot S_y+0 =18$
It follows that $S_y=6$ . So $F$ intersects the $y$ -axis at $y=\underline{6}$ .

A3

$-2x+4y-z=1$ , thus $\vec n = \left(\begin{array}{c} -2\\ 4\\ -1\end{array}\right)$ , e.g. $P(1\vert 1\vert 1)\in E$ .
$3x+3y+3z=3$ , thus $\vec n = \left(\begin{array}{c} 3\\ 3\\ 3\end{array}\right)$ , e.g. $P(-1\vert 0\vert 2)\in E$ .
$1 x+1 y+1 z=0$ , thus $\vec n = \left(\begin{array}{c} 1\\ 1\\ 1\end{array}\right)$ , e.g. $P(0\vert 0\vert 0)\in E$ .
$1 x+1 y+0 z=3$ , thus $\vec n = \left(\begin{array}{c} 1\\ 1\\ 0\end{array}\right)$ , e.g. $P(2\vert 1\vert 100)\in E$ .
$0x+0y+1z=1$ , thus $\vec n = \left(\begin{array}{c} 0\\ 0\\ 1\end{array}\right)$ , e.g. $P(10.4\vert 1000\vert 1)\in E$ .

A4

Normal equation: $10x-5y+20x=-20$ .
Insert $P(1\vert 3\vert -2)$ into the equation and check if result is $-20$ :
$10\cdot 1-5\cdot 3+20\cdot(-2)\neq -20$
So $P \not\in E$ .

A5

Two planes are parallel if their normal vectors are collinear. Normal vector of first plane is $\vec n = \left(\begin{array}{c} 3\\ -2\\ -4\end{array}\right)$ , normal vector of second plane is $\vec m = \left(\begin{array}{c} -6\\ 4\\ 8\end{array}\right)$ . They are collinear, so the planes are parallel.