1.4: Equations of Planes in 3d

Specifying one point $(x_0,y_0,z_0)$ on a plane and a vector $\textbf{d}$ parallel to the plane does not uniquely determine the plane, because it is free to rotate about $\textbf{d}\text{.}$ On the other hand, giving one point

                

on the plane and one vector $\textbf{n}=\left \langle  n_x,n_y, n_z \right \rangle$ with direction perpendicular to that of the plane does uniquely determine the plane. If $(x,y,z)$ is any point on the plane then the vector $\left \langle  x-x_0,y-y_0,z-z_0 \right \rangle \text{,}$ whose tail is at $(x_0,y_0,z_0)$ and whose head is at $(x,y,z)\text{,}$ lies entirely inside the plane and so must be perpendicular to $\textbf{n}\text{.}$ That is,

Equation 1.4.1. The Equation of a Plane

$$\begin{gather*} \textbf{n}\cdot\left \langle  x-x_0,y-y_0,z-z_0 \right \rangle =0 \end{gather*}$$

Writing out in components

$$\begin{gather*} n_x(x-x_0)+n_y(y-y_0)+n_z(z-z_0)=0 \quad\text{or}\quad n_xx+n_yy+n_zz= d \end{gather*}$$

where $d=n_xx_0+n_yy_0+n_zz_0\text{.}$

Again, the coefficients $n_x,n_y,n_z$ of $x,\ y$ and $z$ in the equation of the plane are the components of a vector $\left \langle  n_x,n_y,n_z \right \rangle$ perpendicular to the plane. The vector $\textbf{n}$ is often called a normal vector for the plane. Any nonzero multiple of $\textbf{n}$ will also be perpendicular to the plane and is also called a normal vector.

Example 1.4.2

We have just seen that if we write the equation of a plane in the standard form

$$ax+by+cz=d \nonumber$$

then it is easy to read off a normal vector for the plane. It is just $\left \langle  a,b,c \right \rangle\text{.}$ So for example the planes

$$P:\ x+2y+3z=4 \qquad P':\ 3x+6y+9z=7 \nonumber$$

have normal vectors $\textbf{n}=\left \langle  1,2,3 \right \rangle$ and $\textbf{n}'=\left \langle  3,6,9 \right \rangle\text{,}$ respectively. Since $\textbf{n}'=3\textbf{n}\text{,}$ the two normal vectors $\textbf{n}$ and $\textbf{n}'$ are parallel to each other. This tells us that the planes $P$ and $P'$ are parallel to each other.

When the normal vectors of two planes are perpendicular to each other, we say that the planes are perpendicular to each other. For example the planes

$$P:\ x+2y+3z=4 \qquad P'':\ 2x-y=7 \nonumber$$

have normal vectors $\textbf{n}=\left \langle  1,2,3 \right \rangle$ and $\textbf{n}''=\left \langle  2,-1,0 \right \rangle\text{,}$ respectively. Since

$$\textbf{n}\cdot\textbf{n}'' = 1\times 2+2\times(-1)+3\times 0 = 0 \nonumber$$

the normal vectors $\textbf{n}$ and $\textbf{n}''$ are mutually perpendicular, so the corresponding planes $P$ and $P''$ are perpendicular to each other.

Here is an example that illustrates how one can sketch a plane, given the equation of the plane.

Example 1.4.3

In this example, we'll sketch the plane

$$P:\ 4x + 3y + 2z = 12 \nonumber$$

A good way to prepare for sketching a plane is to find the intersection points of the plane with the $x$-, $y$- and $z$-axes, just as you are used to doing when sketching lines in the $xy$-plane. For example, any point on the $x$ axis must be of the form $(x,0,0)\text{.}$ For $(x,0,0)$ to also be on $P$ we need $x=\frac{12}{4}=3\text{.}$ So $P$ intersects the $x$-axis at $(3,0,0)\text{.}$ Similarly, $P$ intersects the $y$-axis at $(0,4,0)$ and the $z$-axis at $(0,0,6)\text{.}$ Now plot the points $(3,0,0)\text{,}$ $(0,4,0)$ and $(0,0,6)\text{.}$ $P$ is the plane containing these three points. Often a visually effective way to sketch a surface in three dimensions is to

• only sketch the part of the surface in the first octant. That is, the part with $x\ge0\text{,}$ $y\ge 0$ and $z\ge 0\text{.}$
• To do so, sketch the curve of intersection of the surface with the part of the $xy$-plane in the first octant and,
• similarly, sketch the curve of intersection of the surface with the part of the $xz$-plane in the first octant and the curve of intersection of the surface with the part of the $yz$-plane in the first octant.

That's what we'll do. The intersection of the plane $P$ with the $xy$-plane is the straight line through the two points $(3,0,0)$ and $(0,4,0)\text{.}$ So the part of that intersection in the first octant is the line segment from $(3,0,0)$ to $(0,4,0)\text{.}$ Similarly the part of the intersection of $P$ with the $xz$-plane that is in the first octant is the line segment from $(3,0,0)$ to $(0,0,6)$ and the part of the intersection of $P$ with the $yz$-plane that is in the first octant is the line segment from $(0,4,0)$ to $(0,0,6)\text{.}$ So we just have to sketch the three line segments joining the three axis intercepts $(3,0,0)\text{,}$ $(0,4,0)$ and $(0,0,6)\text{.}$ That's it.

Here are two examples that illustrate how one can find the distance between a point and a plane.

Example 1.4.4

In this example, we'll compute the distance between the point

$$\textbf{x} = (1,-1,-3) \qquad\text{and the plane}\qquad P:\ x+2y+3z=18 \nonumber$$

By the “distance between $\textbf{x}$ and the plane $P$” we mean the shortest distance between $\textbf{x}$ and any point $\textbf{y}$ on $P\text{.}$ In fact, we'll evaluate the distance in two different ways. In the next Example 1.4.5, we'll use projection. In this example, our strategy for finding the distance will be to

• first observe that the vector $\textbf{n}=\left \langle 1,2,3 \right \rangle$ is normal to $P$ and then
• start walking¹ away from $\textbf{x}$ in the direction of the normal vector $\textbf{n}$ and
• keep walking until we hit $P\text{.}$ Call the point on $P$ where we hit, $\textbf{y}\text{.}$ Then the desired distance is the distance between $\textbf{x}$ and $\textbf{y}\text{.}$ From the figure below it does indeed look like distance between $\textbf{x}$ and $\textbf{y}$ is the shortest distance between $\textbf{x}$ and any point on $P\text{.}$ This is in fact true, though we won't prove it.

So imagine that we start walking, and that we start at time $t=0$ at $\textbf{x}$ and walk in the direction $\textbf{n}\text{.}$ Then at time $t$ we might be at

$$\textbf{x}+t\textbf{n} = (1,-1,-3) +t\,\left \langle  1,2,3 \right \rangle = (1+t, -1+2t, -3+3t) \nonumber$$

We hit the plane $P$ at exactly the time $t$ for which $(1+t, -1+2t, -3+3t)$ satisfies the equation for $P\text{,}$ which is $x+2y+3z=18\text{.}$ So we are on $P$ at the unique time $t$ obeying

$$\begin{align*} (1+t)+2(-1+2t)+3(-3+3t)=18 &\iff 14t = 28 \iff t=2 \end{align*}$$

So the point on $P$ which is closest to $\textbf{x}$ is

$$\begin{gather*} \textbf{y} = \big[\textbf{x}+t\textbf{n}\big]_{t=2} = (1+t, -1+2t, -3+3t)\big|_{t=2} = (3, 3, 3) \end{gather*}$$

and the distance from $\textbf{x}$ to $P$ is the distance from $\textbf{x}$ to $\textbf{y}\text{,}$ which is

$$\begin{gather*} |\textbf{y}-\textbf{x}| = 2|\textbf{n}| = 2\sqrt{1^2+2^2+3^2} =2\sqrt{14} \end{gather*}$$

Example 1.4.5. Example 1.4.4, revisited

We are again going to find the distance from the point

$$\textbf{x} = (1,-1,-3) \qquad\text{to the plane}\qquad P:\ x+2y+3z=18 \nonumber$$

But this time we will use the following strategy.

• We'll first find any point $\textbf{z}$ on $P$ and then
• we'll denote by $\textbf{y}$ the point on $P$ nearest $\textbf{x}\text{,}$ and we'll denote by $\textbf{v}$ the vector from $\textbf{x}$ to $\textbf{z}$ (see the figure below) and then
• we'll realize, by looking at the figure, that the vector from $\textbf{x}$ to $\textbf{y}$ is exactly the projection² of the vector $\textbf{v}$ on $\textbf{n}$ so that
• the distance from $\textbf{x}$ to $P\text{,}$ i.e. the length of the vector from $\textbf{x}$ to $\textbf{y}\text{,}$ is exactly $\left|\text{proj}_\textbf{n}\textbf{v} \right|\text{.}$

Now let's find a point on $P\text{.}$ The plane $P$ is given by a single equation, namely

$$x+2y+3z=18 \nonumber$$

in the three unknowns, $x\text{,}$ $y\text{,}$ $z\text{.}$ The easiest way to find one solution to this equation is to assign two of the unknowns the value zero and then solve for the third unknown. For example, if we set $x=y=0\text{,}$ then the equation reduces to $3z=18\text{.}$ So we may take $\textbf{z}=(0,0,6)\text{.}$

Then $\textbf{v}\text{,}$ the vector from $\textbf{x}=(1,-1,-3)$ to $\textbf{z}=(0,0,6)$ is $\left \langle  0-1\,,\,0-(-1)\,,\,6-(-3)  \right \rangle=\left \langle  -1,1,9 \right \rangle$ so that, by Equation 1.2.14,

$$\begin{align*} {\rm proj}_{\textbf{n}}\,\textbf{v}&=\frac{\textbf{v}\cdot\textbf{n}}{|\textbf{n}|^2}\,\textbf{n}\\ &= \frac{\left \langle  -1,1,9 \right \rangle\cdot\left \langle  1,2,3 \right \rangle}{|\left \langle  1,2,3 \right \rangle|^2}\, \left \langle  1,2,3 \right \rangle\\ &= \frac{28}{14} \left \langle  1,2,3 \right \rangle\\ &= 2 \left \langle  1,2,3 \right \rangle \end{align*}$$

and the distance from $\textbf{x}$ to $P$ is

$$\begin{gather*} \left|{\rm proj}_{\textbf{n}}\,\textbf{v}\right| = \big|2 \left \langle  1,2,3 \right \rangle\big| =2\sqrt{14} \end{gather*}$$

just as we found in Example 1.4.4.

In the next example, we find the distance between two planes.

Example 1.4.6

Now we'll increase the degree of difficulty a tiny bit, and compute the distance between the planes

$$P:\ x+2y+2z=1 \qquad\text{and}\qquad P':\ 2x+4y+4z=11 \nonumber$$

By the “distance between the planes $P$ and $P'$” we mean the shortest distance between any pair of points $\textbf{x}$ and $\textbf{x}'$ with $\textbf{x}$ in $P$ and $\textbf{x}'$ in $P'\text{.}$ First observe that the normal vectors

$$\textbf{n}=\left \langle  1,2,2 \right \rangle \qquad\text{and}\qquad \textbf{n}'=\left \langle  2,4,4 \right \rangle=2\textbf{n} \nonumber$$

to $P$ and $P'$ are parallel to each other. So the planes $P$ and $P'$ are parallel to each other. If they had not been parallel, they would have crossed and the distance between them would have been zero.

Our strategy for finding the distance will be to

• first find a point $\textbf{x}$ on $P$ and then, like we did in Example 1.4.4,
• start walking away from $P$ in the direction of the normal vector $\textbf{n}$ and
• keep walking until we hit $P'\text{.}$ Call the point on $P'$ that we hit $\textbf{x}'\text{.}$ Then the desired distance is the distance between $\textbf{x}$ and $\textbf{x}'\text{.}$ From the figure below it does indeed look like distance between $\textbf{x}$ and $\textbf{x}'$ is the shortest distance between any pair of points with one point on $P$ and one point on $P'\text{.}$ Again, this is in fact true, though we won't prove it.

We can find a point on $P$ just as we did on Example 1.4.5. The plane $P$ is given by the single equation

$$x+2y+2z=1 \nonumber$$

in the three unknowns, $x\text{,}$ $y\text{,}$ $z\text{.}$ We can find one solution to this equation by assigning two of the unknowns the value zero and then solving for the third unknown. For example, if we set $y=z=0\text{,}$ then the equation reduces to $x=1\text{.}$ So we may take $\textbf{x}=(1,0,0)\text{.}$

Now imagine that we start walking, and that we start at time $t=0$ at $\textbf{x}$ and walk in the direction $\textbf{n}\text{.}$ Then at time $t$ we might be at

$$\textbf{x}+t\textbf{n} = (1,0,0) +t\left \langle  1,2,2 \right \rangle = (1+t, 2t, 2t) \nonumber$$

We hit the second plane $P'$ at exactly the time $t$ for which $(1+t, 2t, 2t)$ satisfies the equation for $P'\text{,}$ which is $2x+4y+4z=11\text{.}$ So we are on $P'$ at the unique time $t$ obeying

$$\begin{align*} 2(1+t)+4(2t)+4(2t)=11 &\iff 18t = 9 \iff t=\frac{1}{2} \end{align*}$$

So the point on $P'$ which is closest to $\textbf{x}$ is

$$\begin{gather*} \textbf{x}' = \big[\textbf{x}+t\textbf{n}\big]_{t=\frac{1}{2}} = (1+t, 2t, 2t)\big|_{t=\frac{1}{2}} = (\frac{3}{2}, 1, 1) \end{gather*}$$

and the distance from $P$ to $P'$ is the distance from $\textbf{x}$ to $\textbf{x}'$ which is

$$\begin{gather*} \sqrt{(1-\frac{3}{2})^2+(0-1)^2+(0-1)^2} =\sqrt{\frac{9}{4}} =\frac{3}{2} \end{gather*}$$

Now we'll find the angle between two intersecting planes.

Example 1.4.7

The orientation (i.e. direction) of a plane is determined by its normal vector. So, by definition, the angle between two planes is the angle between their normal vectors. For example, the normal vectors of the two planes

$$\begin{alignat*}{2} P_1&:\quad & 2x+y-z&=3\\ P_2&: & x+y+z&=4 \end{alignat*}$$

are

$$\begin{align*} \textbf{n}_1&=\left \langle  2,1,-1 \right \rangle\\ \textbf{n}_2&=\left \langle  1,1,1 \right \rangle \end{align*}$$

If we use $\theta$ to denote the angle between $\textbf{n}_1$ and $\textbf{n}_2\text{,}$ then

$$\begin{align*} \cos\theta &=\frac{\textbf{n}_1\cdot\textbf{n}_2}{|\textbf{n}_1|\,|\textbf{n}_2|}\\ &=\frac{\left \langle  2,1,-1 \right \rangle\cdot\left \langle  1,1,1 \right \rangle} {|\left \langle  2,1,-1 \right \rangle|\,|\left \langle  1,1,1 \right \rangle|}\\ &=\frac{2}{\sqrt{6}\,\sqrt{3}} \end{align*}$$

so that

$$\begin{gather*} \theta =\arccos\frac{2}{\sqrt{18}} =1.0799 \end{gather*}$$

to four decimal places. That's in radians. In degrees, it is $1.0799\frac{180}{\pi}=61.87^\circ$ to two decimal places.

Exercises

Stage 1

1

The vector $\hat{\textbf{k}}$ is a normal vector (i.e. is perpendicular) to the plane $z=0\text{.}$ Find another nonzero vector that is normal to $z=0\text{.}$

2

Consider the plane $P$ with equation $3x+\frac{1}{2}y+z=4\text{.}$

1. Find the intersection of $P$ with the $y$-axis.
2. Find the intersection of $P$ with the $z$-axis.
3. Sketch the part of the intersection of $P$ with the $yz$-plane that is in the first octant. (That is, with $x,y,z\ge 0\text{.}$)

3

1. Find the equation of the plane that passes through the origin and has normal vector $\left \langle  1,2,3 \right \rangle\text{.}$
2. Find the equation of the plane that passes through the point $(0,0,1)$ and has normal vector $\left \langle  1,1,3 \right \rangle\text{.}$
3. Find, if possible, the equation of a plane that passes through both $(1,2,3)$ and $(1,0,0)$ and has normal vector $\left \langle  4,5,6 \right \rangle\text{.}$
4. Find, if possible, the equation of a plane that passes through both $(1,2,3)$ and $(0,3,4)$ and has normal vector $\left \langle  2,1,1 \right \rangle\text{.}$

4. ✳

Find the equation of the plane that contains $(1,0,0)\text{,}$ $(0,1,0)$ and $(0,0,1)\text{.}$

5

1. Find the equation of the plane containing the points $(1,0,1)\text{,}$ $(1,1,0)$ and $(0,1,1)\text{.}$
2. Is the point $(1,1,1)$ on the plane?
3. Is the origin on the plane?
4. Is the point $(4,-1,-1)$ on the plane?

6

What's wrong with the following exercise? “Find the equation of the plane containing $(1,2,3)\text{,}$ $(2,3,4)$ and $(3,4,5)\text{.}$”

Stage 2

7

Find the plane containing the given three points.

1. $\displaystyle (1,0,1),\ (2,4,6),\ (1,2,-1)$
2. $\displaystyle (1,-2,-3),\ (4,-4,4),\ (3,2,-3)$
3. $\displaystyle (1,-2,-3),\ (5,2,1),\ (-1,-4,-5)$

8

Find the distance from the given point to the given plane.

1. point $(-1,2,3)\text{,}$ plane $x+y+z=7$
2. point $(1,-4,3)\text{,}$ plane $x-2y+z=5$

9. ✳

A plane $\Pi$ passes through the points $A = (1, 1, 3)\text{,}$ $B = (2, 0, 2)$ and $C = (2, 1, 0)$ in $\mathbb{R}^3\text{.}$

1. Find an equation for the plane $\Pi\text{.}$
2. Find the point $E$ in the plane $\Pi$ such that the line $L$ through $D = (6, 1, 2)$ and $E$ is perpendicular to $\Pi\text{.}$

10. ✳

Let $A = (2, 3, 4)$ and let $L$ be the line given by the equations $x + y = 1$ and $x + 2y + z = 3\text{.}$

1. Write an equation for the plane containing $A$ and perpendicular to $L\text{.}$
2. Write an equation for the plane containing $A$ and $L\text{.}$

11. ✳

Consider the plane $4x + 2y - 4z = 3\text{.}$ Find all parallel planes that are distance $2$ from the above plane. Your answers should be in the following form: $4x + 2y - 4z = C\text{.}$

12. ✳

Find the distance from the point $(1,2,3)$ to the plane that passes through the points $(0,1,1)\text{,}$ $(1,-1,3)$ and $(2,0,-1)\text{.}$

Stage 3

13. ✳

Consider two planes $W_1\text{,}$ $W_2\text{,}$ and a line $M$ defined by:

$$\begin{align*} W_1\ &:\ -2x + y + z = 7\\ W_2\ &:\ -x + 3y + 3z = 6\\ M\ &:\ \frac{x}{2} = \frac{2y-4}{4} = z + 5 \end{align*}$$

1. Find a parametric equation of the line of intersection $L$ of $W_1$ and $W_2\text{.}$
2. Find the distance from L to M .

14

Find the equation of the sphere which has the two planes $x+y+z=3,\ x+y+z=9$ as tangent planes if the center of the sphere is on the planes $2x-y=0,\ 3x-z=0\text{.}$

15

Find the equation of the plane that passes through the point $(-2,0,1)$ and through the line of intersection of $2x+3y-z=0,\ x-4y+2z=-5\text{.}$

16

Find the distance from the point $\textbf{p}$ to the plane $\textbf{n}\cdot \textbf{x}= c\text{.}$

17

Describe the set of points equidistant from $(1,2,3)$ and $(5,2,7)\text{.}$

18

Describe the set of points equidistant from $\textbf{a}$ and $\textbf{b}\text{.}$

19. ✳

Consider a point $P(5,-10,2)$ and the triangle with vertices $A(0,1,1)\text{,}$ $B(1,0,1)$ and $C(1,3,0)\text{.}$

1. Compute the area of the triangle $ABC\text{.}$
2. Find the distance from the point $P$ to the plane containing the triangle.

20. ✳

Consider the sphere given by

$$(x-1)^2+(y-2)^2+(z+1)^2=2 \nonumber$$

Suppose that you are at the point $(2,2,0)$ on $S\text{,}$ and you plan to follow the shortest path on $S$ to $(2,1,-1)\text{.}$ Express your initial direction as a cross product.

1. To see why heading in the normal direction gives the shortest walk, revisit Example 1.3.5
2. Now might be a good time to review the Definition 1.2.13 of projection.