Points, Straight Lines and Shapes in 3D

Lessons

Senior 5 · Mathematics

Points, Straight Lines and Shapes in 3D

5 lessons

Lesson 4 of 5

Planes: equations, positions, angles and distances

45 min

A floor, a roof slope, a solar panel

A solar panel on a roof in Nyagatare must face the sun at a good angle; a ramp must not be too steep; a satellite dish must point along a given direction. Each surface is part of a plane, and the question is always about its tilt: the direction perpendicular to it. A plane is fixed by one point and that perpendicular direction, its normal. From the normal come the plane's equation, its angle with a line or another plane, and the distance from any point to it.

Definition

The plane through the point $A (x_{1}, y_{1}, z_{1})$ with two non-parallel direction vectors $u$ and $v$ has

vector equation $r = a + \lambda u + \mu v$ , $\lambda, \mu \in \mathbb{R}$ , and parametric equations from its components;
normal vector $n = u \times v$ , perpendicular to every vector in the plane;
Cartesian equation $n \cdot (r - a) = 0$ , that is $a^{'} x + b^{'} y + c^{'} z + d^{'} = 0$ where $n = (a^{'}, b^{'}, c^{'})$ .

Conversely, $a x + b y + c z + d = 0$ (with $a, b, c$ not all 0) is a plane with normal $(a, b, c)$ . Four points $A, B, C, D$ are coplanar exactly when $[\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}] = 0$ .

Worked example 1: the plane through three points

Find the Cartesian equation of the plane through $A (1, 0, 2)$ , $B (3, 1, 1)$ , $C (0, 2, 3)$ . For which $k$ does $D (2, k, 2)$ lie in it?

Solution. From unit 7, $\overrightarrow{AB} \times \overrightarrow{AC} = (2, 1, -1) \times (-1, 2, 1) = (3, -1, 5)$ . So

3(x - 1) - (y - 0) + 5(z - 2) = 0 \iff 3x - y + 5z - 13 = 0 .

Check $B$ : $9 - 1 + 5 - 13 = 0$ ✓; $C$ : $0 - 2 + 15 - 13 = 0$ ✓. $D$ lies in the plane when $6 - k + 10 - 13 = 0$ , so $k = 3$ : then $A, B, C, D$ are coplanar.

Important

A line and a plane. Let $L: r = p + t\,d$ and $\Pi: n \cdot r + d' = 0$ .

If $d \cdot n \ne 0$ , $L$ meets $Π \Pi$ in exactly one point: substitute the parametric equations and solve for $t$ . If moreover $d \parallel n$ , $L$ is perpendicular to $Π \Pi$ .
If $d \cdot n = 0$ , $L$ is parallel to $Π \Pi$ : it lies in $Π \Pi$ if $p$ satisfies the equation, and has no common point otherwise.

The angle $\varphi$ between $L$ and $Π \Pi$ is the angle between $L$ and its projection on $Π \Pi$ . It is $90^\circ$ minus the angle between $d$ and $n$ , so

\sin\varphi = \frac{|d \cdot n|}{|d|\,|n|}, \qquad 0^\circ \le \varphi \le 90^\circ .

Distance from a point to a plane. For $P (x_{0}, y_{0}, z_{0})$ and $\Pi: ax + by + cz + d = 0$ ,

d(P, \Pi) = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}} .

Reason. For any point $A$ of $Π \Pi$ , the distance is the length of the projection of $\overrightarrow{AP}$ on $n$ : $\frac{|\overrightarrow{AP} \cdot n|}{|n|}$ , and $\overrightarrow{AP} \cdot n = ax_0 + by_0 + cz_0 + d$ because $A$ satisfies the equation.

Worked example 2: a line meeting a plane, and a parallel line

Let $\Pi: 2x + y - 2z = 4$ , $L : r = (1, 1, 1) + t (1, 1, 1)$ and $M : r = (1, 0, 1) + t (1, 2, 2)$ . Find where $L$ meets $Π \Pi$ and the angle between them; show that $M$ is parallel to $Π \Pi$ and find its distance from $Π \Pi$ .

Solution. For $L$ : $2 (1 + t) + (1 + t) - 2 (1 + t) = 4$ , so $1 + t = 4$ , $t = 3$ , and $L$ meets $Π \Pi$ at $(4, 4, 4)$ . With $n = (2, 1, - 2)$ , $∣ n ∣ = 3$ : $\sin\varphi = \frac{|2 + 1 - 2|}{\sqrt{3} \times 3} = \frac{1}{3\sqrt{3}}$ , so $\varphi \approx 11.1^\circ$ .

For $M$ : $(1, 2, 2) \cdot (2, 1, -2) = 2 + 2 - 4 = 0$ , so $M \parallel \Pi$ . Its point $(1, 0, 1)$ gives $2 + 0 - 2 = 0 \ne 4$ , so $M$ is not in $Π \Pi$ . Every point of $M$ is at the same distance: $\frac{|2 + 0 - 2 - 4|}{3} = \frac{4}{3}$ .

Worked example 3: a height from a distance

Find the distance from $D (2, 3, 4)$ to the plane $3 x - y + 5 z - 13 = 0$ of worked example 1, and use it to recover the volume of the tetrahedron $A B C D$ .

Solution. $d = \frac{|6 - 3 + 20 - 13|}{\sqrt{9 + 1 + 25}} = \frac{10}{\sqrt{35}} \approx 1.69$ . The base $A B C$ has area $\frac{\sqrt{35}}{2}$ (unit 7), so the volume is $\frac{1}{3} \times \frac{\sqrt{35}}{2} \times \frac{10}{\sqrt{35}} = \frac{5}{3}$ , the same as with the mixed product. ✓

Important

Two planes $\Pi_1$ and $\Pi_2$ with normals $n_{1}$ and $n_{2}$ :

if $n_1 \parallel n_2$ , the planes are parallel; they coincide when their equations are proportional;
otherwise they meet in a line with direction $n_1 \times n_2$ ; a point of it is found by fixing one coordinate and solving the two equations.

The angle between the planes is the acute angle between their normals: $\cos\theta = \frac{|n_1 \cdot n_2|}{|n_1|\,|n_2|}$ . The planes are perpendicular when $n_1 \cdot n_2 = 0$ .

Worked example 4: the line where two planes meet

Find the line of intersection of $x + y + z = 6$ and $2 x - y + z = 3$ , and the angle between the planes.

Solution. $n_1 \times n_2 = (1, 1, 1) \times (2, -1, 1) = (2, 1, -3)$ . Put $z = 0$ : $x + y = 6$ and $2 x - y = 3$ give $x = 3$ , $y = 3$ . So the line is $r = (3, 3, 0) + t (2, 1, - 3)$ . Check $t = 1$ : $(5, 4, - 3)$ gives $5 + 4 - 3 = 6$ ✓ and $10 - 4 - 3 = 3$ ✓.

$\cos\theta = \frac{|2 - 1 + 1|}{\sqrt{3}\sqrt{6}} = \frac{2}{\sqrt{18}} \approx 0.471$ , so $\theta \approx 61.9^\circ$ .

Warning

Line and plane use sine, two lines or two planes use cosine. The normal is perpendicular to the plane, which turns the cosine of the angle with $n$ into the sine of the angle with $Π \Pi$ .
Put the plane in the form $a x + b y + c z + d = 0$ before using the distance formula: for $2 x + y - 2 z = 4$ , $d = - 4$ .
$d \cdot n = 0$ means the line is parallel to the plane, not perpendicular to it. Perpendicular means $d \parallel n$ .

Tip

The coefficients of $x, y, z$ in a plane's equation are its normal: read $n = (3, - 1, 5)$ straight off $3 x - y + 5 z - 13 = 0$ .

MIT OpenCourseWare: Equations of planes, a recitation from 18.02SC Multivariable Calculus (about 14 min). Watch a plane's equation built from a point and a normal vector found with the cross product.

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Practice

Questions 1 to 3 are routine; 4 to 7 need more care; 8 asks for a reason.

Find the Cartesian equation of the plane through $(2, - 1, 3)$ with normal $(1, 4, - 2)$ .
Find the equation of the plane through $A (1, 1, 0)$ , $B (2, 0, 1)$ , $C (0, 2, 2)$ .
Are $A (1, 0, 0)$ , $B (0, 1, 0)$ , $C (0, 0, 1)$ , $D (1, 1, - 1)$ coplanar?
Find the distance from $(1, 2, 3)$ to the plane $2 x - y + 2 z + 4 = 0$ .
Find where the line $\frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z}{-1}$ meets the plane $x + y + z = 3$ , and the angle between them.
Find the angle between the planes $x + 2 y - 2 z = 5$ and $2 x - y + 2 z = 1$ .
Find the line of intersection of $x - y + z = 1$ and $2 x + y - z = 5$ .
Explain why $2 x - 4 y + 6 z = 1$ and $x - 2 y + 3 z = 7$ never meet, and find the distance between them.

Show answersHide answers

$x + 4 y - 2 z + 8 = 0$ .
$\overrightarrow{AB} \times \overrightarrow{AC} = (1, -1, 1) \times (-1, 1, 2) = (-3, -3, 0)$ , so $n = (1, 1, 0)$ and the plane is $x + y = 2$ .
Yes: $A, B, C$ lie in $x + y + z = 1$ , and so does $D$ ( $1 + 1 - 1 = 1$ ).
$\frac{|2 - 2 + 6 + 4|}{3} = \frac{10}{3}$ .
$(1 + 2 t) + (- 2 + 3 t) + (- t) = 3$ gives $t = 1$ : the point $(3, 1, - 1)$ . $\sin\varphi = \frac{|2 + 3 - 1|}{\sqrt{14}\sqrt{3}} = \frac{4}{\sqrt{42}}$ , $\varphi \approx 38.1^\circ$ .
$\cos\theta = \frac{|2 - 2 - 4|}{3 \times 3} = \frac{4}{9}$ , $\theta \approx 63.6^\circ$ .
Adding the equations gives $3 x = 6$ , $x = 2$ ; with $z = 0$ , $y = 1$ . Direction $(1, -1, 1) \times (2, 1, -1) = (0, 3, 3)$ : $r = (2, 1, 0) + t (0, 1, 1)$ .
The normals $(2, - 4, 6)$ and $(1, - 2, 3)$ are parallel but the equations are not proportional ( $2 \times 7 \ne 1$ ), so the planes are parallel and distinct. Using $(7, 0, 0)$ on the second plane: $\frac{|14 - 1|}{\sqrt{56}} \approx 1.74$ .

Sources and credits

Written by Ganzaa: the explanations, the diagram (drawn from computed coordinates), the worked examples, the practice items and the quiz questions. Every equation, intersection, angle and distance was checked by computer.
Video: Equations of planes, MIT 18.02SC Multivariable Calculus, Fall 2010, MIT OpenCourseWare, embedded from YouTube.

1A floor, a roof slope, a solar panel

A floor, a roof slope, a solar panel