Calculus Volume 3 · Chapter 2 · Section 2.5

Equations of Lines and Planes in Space

A line is a point and a direction. A plane is a point and a normal. Every distance is a cross product or a projection.

Idea 1 · point + direction $\mathbf r=\mathbf r_0+t\mathbf v$ walks the line.Each

t

is one point. Idea 2 · point + normal $\mathbf n\cdot\overrightarrow{PQ}=0$ is the plane.Every arrow in it is square to

\mathbf n

. Idea 3 · distancesWalk along the normal until you hit.Lines: cross product. Planes: dot product.

Colour key line v direction n normal distance, result key idea · drag any 3D figure to turn it · page links open the textbook

L1Where does a straight path land?
L1A point and a direction
L1A point and a normal
L1Two lines, two planes
L1One picture
L2Equations of a line
L2Equations of a plane
L2Distances and angles
L2Examples 2.45–2.54
L2Lab 1 · walk a line
L2Lab 2 · parallel, crossing or skew
L2Lab 3 · a point and a plane
L2Lab 4 · two planes
L2Common mistakes
L2Practice and answers
L3Distance to a line is an area
L3Distance to a plane is a shadow
L3Why normals measure the angle
L3The gap between skew lines

Level 1 · see itWhat pins a line or a plane downbefore any formula

Where does a straight path land?

A path passes $(1,-2,3)$ heading along $\langle 1,2,3\rangle$ (Exercise 247). Where does it meet the floor $z=0$ ?

(0, −4, 0)

at $t=-1$ : one step backwards. One point and one direction fix every point of a line.

A point and a direction

Start at a point $P$ , position vector $\mathbf r_0$ .
Walk $t$ steps of the direction vector $\mathbf v$ (p170): $\mathbf r=\mathbf r_0+t\mathbf v$ .
Every real $t$ is one point; $0\le t\le 1$ is just the segment. The parameter is a clock on the line.

Example 2.45: through $(1,4,-2)$ and $(-3,5,0)$ , so $\mathbf v=\langle -4,1,2\rangle$

Points at

t=-1,0,1,2

: equal steps of

\mathbf v

. Drag to turn.

A point and a normal

Fix a point $P$ and a normal vector $\mathbf n$ (p177).
Collect every $Q$ with $\overrightarrow{PQ}\perp\mathbf n$ : $\mathbf n\cdot\overrightarrow{PQ}=0$ .
That set is a plane. A plane is everything you can reach without moving along n.

turn

Q

round

P

40°

Plane

x+y+2z=4

through

P(1,1,1)

\mathbf n=\langle 1,1,2\rangle

Two lines, two planes

Lines (Figure 2.68):

	share a point	no common point
directions parallel	equal	parallel
not parallel	intersecting	skew

Skew is new in space: not parallel, yet they never meet. Two distinct planes are simpler: parallel, or meeting in a line (p181).

show

One picture

Example 2.51: from

P(3,1,2)

walk along

\mathbf n=\langle 1,-2,1\rangle

. The line

P+t\mathbf n

hits

x-2y+z=5

t=\tfrac13

, so

d=\tfrac13\|\mathbf n\|=\tfrac{2}{\sqrt6}\approx 0.82

A line along the normal joins the two ideas: the shortest way to a plane is straight along n.

Level 2 · compute itEquations, distances, anglesand four labs

Equations of a line

Theorem 2.11 (p171): through $(x_0,y_0,z_0)$ , direction $\langle a,b,c\rangle$ .

Vector · (2.12)

\mathbf r=\mathbf r_0+t\mathbf v

Parametric · (2.13)

x=x_0+ta,\ \ y=y_0+tb,\ \ z=z_0+tc

Symmetric · (2.14)

\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}

Only when $a,b,c\neq0$ ; a zero gives a separate equation, e.g. $y=1$ .

Segment · (2.15) · p172

\mathbf r=(1-t)\,\mathbf p+t\,\mathbf q,\qquad 0\le t\le 1

Equations of a line are not unique: any point on it and any parallel vector will do.

Equations of a plane

Vector · (2.17) · p178

\mathbf n\cdot\overrightarrow{PQ}=0

Scalar · (2.18)

a(x-x_0)+b(y-y_0)+c(z-z_0)=0

General form

ax+by+cz+d=0

The coefficients are the normal: $\langle a,b,c\rangle$ .

Three points · Example 2.49

$\mathbf n=\overrightarrow{PQ}\times\overrightarrow{QR}$ , then any of the points.

A point and a line · Example 2.50

$\mathbf n=\mathbf v_1\times\mathbf v_2$ : the line's direction and an arrow from the line to the point.

To find a plane, find two arrows inside it; their cross product is the normal (§2.4).

Distances and angles

Point to line · Theorem 2.12 · p173

d=\frac{\|\overrightarrow{PM}\times\mathbf v\|}{\|\mathbf v\|}

Point to plane · Theorem 2.14 · p183

d=\frac{|ax_0+by_0+cz_0+k|}{\sqrt{a^2+b^2+c^2}}

Plane $ax+by+cz+k=0$ .

Angle between planes · p182

\cos\theta=\frac{|\mathbf n_1\cdot\mathbf n_2|}{\|\mathbf n_1\|\,\|\mathbf n_2\|}

Line of intersection · Example 2.52

\mathbf v=\mathbf n_1\times\mathbf n_2

plus one point that satisfies both equations.

Parallel normals: parallel planes. Perpendicular normals: perpendicular planes. Parallel planes are as far apart as any point of one is from the other.

Examples 2.45–2.54

Example	given	result
2.45 · p171	$(1,4,-2)$ , $(-3,5,0)$	$x=1-4t,\ y=4+t,\ z=-2+2t$ ; $\ \frac{x-1}{-4}=y-4=\frac{z+2}{2}$
2.46 · p172	segment $(2,1,4)$ to $(3,-1,3)$	$x=2+t,\ y=1-2t,\ z=4-t,\ 0\le t\le1$
2.47 · p173	$M(1,1,3)$ , $\frac{x-3}{4}=\frac{y+1}{2}=z-3$	$d=\frac{2\sqrt{38}}{\sqrt{21}}\approx 2.69$
2.48 · p175	three pairs of lines	(a) skew · (b) meet at $(1,-1,1)$ · (c) parallel
2.49 · p178	$(1,1,-2)$ , $(0,2,1)$ , $(-1,-1,0)$	$\mathbf n=\langle 8,-4,4\rangle$ ; $\ 8x-4y+4z+4=0$
2.50 · p178	$(1,4,3)$ and $x=\frac{y-1}{2}=z+1$	$5x-3y+z+4=0$
2.51 · p180	$(3,1,2)$ , $x-2y+z=5$	$d=\frac{2}{\sqrt6}\approx 0.82$
2.52 · p181	$x+y+z=0$ , $2x-y+z=0$	$x=2t,\ y=t,\ z=-3t$
2.53 · p182	three pairs of planes	(a) parallel · (b) orthogonal · (c) 1.27 rad ≈ 73°
2.54 · p184	$2x+y-z=2$ , $2x+y-z=8$	$d=\sqrt6\approx 2.45$

Lab 1 · walk a line

parameter t0.5

segment only,

0\le t\le1

presets

Try: Exercise 249 — the direction has no $z$ -part, so the line never reaches the floor and the symmetric form loses a fraction.

Lab 2 · parallel, crossing or skew

presets

slide

L_2

sideways0

Try: Ex 2.48a, then slide until they meet. A skew pair is an intersecting pair pushed apart along $\mathbf v_1\times\mathbf v_2$ .

Lab 3 · a point and a plane

presets

x_0

y_0

z_0

Try: move $P$ parallel to the plane — $d$ stays put; only the part of $\overrightarrow{QP}$ along $\mathbf n$ counts. Exercise 297: the foot $R$ is where a sphere centred at $P$ touches the plane.

Lab 4 · two planes

presets

a_2

b_2

−1

c_2

k_2

Plane 2 is $a_2x+b_2y+c_2z=k_2$ . Try: make $\langle a_2,b_2,c_2\rangle$ a multiple of $\mathbf n_1$ — the line disappears and a gap opens.

Common mistakes

Mistake	Result	Fix
A point used as the direction	$\langle 1,4,-2\rangle$ in Example 2.45: a different line	direction = $Q-P$
Dividing by a zero component	$\frac{y-1}{0}$ in Exercise 249	write $y=1$ separately
One letter $t$ for both lines	forces them to be at the same place at the same "time"	$t$ on one line, $s$ on the other
"Not intersecting, so parallel"	Example 2.48a called parallel	compare directions first; different and apart is skew
Sign of the constant in Theorem 2.14	$x-2y+z=5$ read as $k=+5$ : $8/\sqrt6$	move it over: $k=-5$ , $2/\sqrt6$
Dropping the absolute value	$-2/\sqrt6$ , a negative distance	$\|\,\cdot\,\|$ in the numerator
$\cos\theta$ without $\|\mathbf n_1\cdot\mathbf n_2\|$	1.70 rad for Checkpoint 2.50	take the acute angle, 1.44 rad

Practice and answers

Exercises from p186; the book's key: p835, p842–844.

Checkpoints 2.43–2.51

#	task	answer
2.43	line through $(1,-3,2)$ , $(5,-2,8)$	$x=1+4t,\ y=-3+t,\ z=2+6t$ ; $\ \frac{x-1}{4}=y+3=\frac{z-2}{6}$
2.44	segment $(-1,3,6)$ to $(-8,2,4)$	$x=-1-7t,\ y=3-t,\ z=6-2t,\ 0\le t\le1$
2.45	$(0,3,6)$ to $x=1-t,\ y=1+2t,\ z=5+3t$	$\sqrt{10/7}\approx 1.20$
2.46	$x=1-4t,\ y=3+t,\ z=8-6t$ ; $x=2+3s,\ y=2s,\ z=-1-3s$	skew
2.47	plane containing $x=-y=z$ and $\frac{x-3}{2}=y=z-2$	$-2x+y+3z=0$
2.48	$(5,-1,0)$ to $4x+2y-z=3$	$15/\sqrt{21}\approx 3.27$
2.49	$x+y-z=3$ meets $3x-y+3z=5$	$x=t,\ y=7-3t,\ z=4-2t$
2.50	angle between those planes	1.44 rad
2.51	$5x-2y+z=6$ and $5x-2y+z=-3$	$9/\sqrt{30}\approx 1.64$

Exercises

#	task	answer	key step
243	line through $(-3,5,9)$ , $(4,-7,2)$	$\frac{x+3}{7}=\frac{y-5}{-12}=\frac{z-9}{-7}$	$\mathbf v=\langle 7,-12,-7\rangle$
247	$(1,-2,3)$ , $\mathbf v=\langle 1,2,3\rangle$ : meets $z=0$ ?	$(0,-4,0)$	$3+3t=0$
249	$(3,1,5)$ , $\mathbf v=\langle 1,0,0\rangle$	$y=1,\ z=5$ ; never meets $z=0$	no $z$ -part
251	origin to $x=1+t,\ y=3+t,\ z=5+4t$	$\sqrt3$	$\\|\langle 7,1,-2\rangle\\|/\sqrt{18}$
253	$A(-3,1,1)$ to $x=-y=-z$	$2\sqrt2/\sqrt3$	Theorem 2.12
255	$L_1$ : $x=1+t,\ y=t,\ z=2+t$ ; $L_2$ : $x-3=y-1=z-3$	parallel, $\sqrt2/\sqrt3$ apart	both along $\langle 1,1,1\rangle$
259	$x=-2y=3z$ meets $x=-5-t,\ y=-1+t,\ z=t-11$	$(-12,6,-4)$	$r=-2,\ t=7$
261	$x=y-1=-z$ , $x-2=-y=\frac z2$	skew	$\mathbf w\cdot(\mathbf v_1\times\mathbf v_2)=5\neq0$
263	(p187)	equal	same direction, shared point $(1,4,7)$
269	$P(1,2,3)$ , $\mathbf n=\langle 1,2,3\rangle$	$x+2y+3z-14=0$	scalar form, expand
271	$4x+5y+10z-20=0$	$\mathbf n=4\mathbf i+5\mathbf j+10\mathbf k$ ; $(5,0,0),(0,4,0),(0,0,2)$	two coordinates zero
277	through $(-2,1,3)$ , $\perp\ 2x-3y+z=7$	$x=-2+2t,\ y=1-3t,\ z=3+t$	direction = normal
281	$(1,1,1),(2,4,3),(-1,-2,-1)$	$-2y+3z-1=0$	$\overrightarrow{PQ}\times\overrightarrow{PR}=\langle 0,-2,3\rangle$
287	$x=1+2t,\ y=-2t,\ z=2+t$ and $3x+4y+6z-7=0$ (p189)	$(-3,4,0)$	$8+4t=0$
289	$(1,5,-4)$ to $3x-y+2z-6=0$	$16/\sqrt{14}$	Theorem 2.14
291	$x+y+z=0$ , $2x-y+z-7=0$	neither; 62°	$\cos\theta=2/\sqrt{18}$
295	skew lines (p190)	$1/\sqrt6$	$\mathbf N=\langle -2,1,1\rangle$
297	sphere at $(-3,2,4)$ tangent to $2x+4y-3z=8$	$r=18/\sqrt{29}$ ; $\big(-\tfrac{51}{29},\tfrac{130}{29},\tfrac{62}{29}\big)$	Lab 3

Level 3 · why it worksWhere the formulas come fromshort arguments

Distance to a line is an area

$\overrightarrow{PM}$ and $\mathbf v$ span a parallelogram with base $\|\mathbf v\|$ and height $d$ (Figure 2.65):

\|\overrightarrow{PM}\times\mathbf v\|=\|\mathbf v\|\,d

Example 2.47: $\sqrt{152}=\sqrt{21}\,d$ , so $d\approx 2.69$ whichever point $P$ of the line you pick.

Sliding $P$ along the line shears the parallelogram but keeps its base and height.

Distance to a plane is a shadow

The shortest arrow to the plane is the part of any $\overrightarrow{QP}$ along $\mathbf n$ (Theorem 2.13, p180). With $Q=(x_1,y_1,z_1)$ in $ax+by+cz+k=0$ :

d=\frac{|\overrightarrow{QP}\cdot\mathbf n|}{\|\mathbf n\|}=\frac{|a(x_0-x_1)+b(y_0-y_1)+c(z_0-z_1)|}{\sqrt{a^2+b^2+c^2}}=\frac{|ax_0+by_0+cz_0+k|}{\sqrt{a^2+b^2+c^2}}

since $ax_1+by_1+cz_1=-k$ . Without the absolute value the sign says which side of the plane $P$ is on.

Why normals measure the angle

Turn the pair of planes together; each normal turns with its plane, so the angle between normals equals the angle between planes (Figure 2.72).

Two planes make two angles, $\theta$ and $180^\circ-\theta$ . $|\mathbf n_1\cdot\mathbf n_2|$ picks the one at most $90^\circ$ : Example 2.53c gives $73^\circ$ , not $107^\circ$ .

A normal is a plane's handle: whatever the plane does, the handle does too.

The gap between skew lines

$\mathbf N=\mathbf v_1\times\mathbf v_2$ is square to both lines. Project any arrow $\mathbf v_{12}$ from one line to the other onto it (student project):

d=\frac{|\mathbf v_{12}\cdot(\mathbf v_1\times\mathbf v_2)|}{\|\mathbf v_1\times\mathbf v_2\|}

$\frac{x-5}{2}=\frac{y-3}{4}=\frac{z-1}{3}$ , $\frac{x-6}{3}=\frac{y-1}{5}=\frac z7$ : $\mathbf N=\langle 13,-5,-2\rangle$ , $d=\frac{|-25|}{\sqrt{198}}\approx 1.78$ .

Zero numerator: a triple product of zero, so the lines are coplanar and meet (§2.4). Parallel lines make $\mathbf N=\mathbf 0$ ; use Theorem 2.12 instead. One formula sorts intersecting from skew; parallel lines need their own test.

Objective	You can now
2.5.1 equations of a line	$\mathbf r=\mathbf r_0+t\mathbf v$ ; parametric, symmetric; $\mathbf v=Q-P$ ; segment $0\le t\le 1$
2.5.2 point to line	$\\|\overrightarrow{PM}\times\mathbf v\\|\,/\,\\|\mathbf v\\|$
2.5.3 equations of a plane	$\mathbf n\cdot\overrightarrow{PQ}=0$ ; $\ a(x-x_0)+b(y-y_0)+c(z-z_0)=0$
2.5.4 point to plane	$\|\overrightarrow{QP}\cdot\mathbf n\|/\\|\mathbf n\\|=\|ax_0+by_0+cz_0+k\|/\\|\mathbf n\\|$
2.5.5 angle between planes	$\cos\theta=\|\mathbf n_1\cdot\mathbf n_2\|/(\\|\mathbf n_1\\|\\|\mathbf n_2\\|)$

Where does a straight path land?

A point and a direction

A point and a normal

Two lines, two planes

One picture

Equations of a line

Equations of a plane

Distances and angles

Examples 2.45–2.54

Lab 1 · walk a line

Lab 2 · parallel, crossing or skew

Lab 3 · a point and a plane

Lab 4 · two planes

Common mistakes

Practice and answers

Distance to a line is an area

Distance to a plane is a shadow

Why normals measure the angle

The gap between skew lines

Equations of Lines and Planes in Space

Parallel vectors are scalar multiples of each other

A line is one point plus every multiple of a direction: r=r0+tv\mathbf r=\mathbf r_0+t\mathbf vr=r0​+tv

Read off components for parametric equations; solve each for ttt for symmetric ones

Two points give the direction: v=PQ→\mathbf v=\overrightarrow{PQ}v=PQ​

A segment is the line with the clock stopped at t=0t=0t=0 and t=1t=1t=1

Distance to a line = parallelogram area ÷ base

Read PPP and v\mathbf vv off the symmetric form, cross, divide

The distance does not depend on which point of the line you pick

In space, lines with different directions need not meet: skew lines

Directions first; if not parallel, solve for a common point with two different parameters

A plane is every point QQQ with PQ→⊥n\overrightarrow{PQ}\perp\mathbf nPQ​⊥n

Expand n⋅PQ→=0\mathbf n\cdot\overrightarrow{PQ}=0n⋅PQ​=0: the coefficients of x,y,zx,y,zx,y,z are the normal

Three points: cross two arrows in the plane to get the normal

A point and a line: the line gives one arrow in the plane, the point the other

Distance to a plane = the shadow of QP→\overrightarrow{QP}QP​ on n\mathbf nn

Pick any point in the plane, project onto n\mathbf nn: d=2/6d=2/\sqrt6d=2/6​

Two distinct planes are either parallel or meet in a line

Line of intersection: solve the two equations together, one variable free

The angle between two planes is the angle between their normals

Classify two planes by their normals: multiples, zero dot product, or compute the angle

Normals measure the angle because each is square to its own plane

Plug the point into the plane's equation, then divide by ∥n∥\|\mathbf n\|∥n∥

Theorem 2.14 is the projection formula with the arithmetic done

The gap between parallel planes is a point-to-plane distance

The gap between skew lines is the shadow of a connecting arrow on v1×v2\mathbf v_1\times\mathbf v_2v1​×v2​

A line needs a point and a direction; a plane needs a point and a normal

A line is one point plus every multiple of a direction: $\mathbf r=\mathbf r_0+t\mathbf v$

Read off components for parametric equations; solve each for $t$ for symmetric ones

Two points give the direction: $\mathbf v=\overrightarrow{PQ}$

A segment is the line with the clock stopped at $t=0$ and $t=1$

Read $P$ and $\mathbf v$ off the symmetric form, cross, divide

A plane is every point $Q$ with $\overrightarrow{PQ}\perp\mathbf n$

Expand $\mathbf n\cdot\overrightarrow{PQ}=0$ : the coefficients of $x,y,z$ are the normal

Distance to a plane = the shadow of $\overrightarrow{QP}$ on $\mathbf n$

Pick any point in the plane, project onto $\mathbf n$ : $d=2/\sqrt6$

Plug the point into the plane's equation, then divide by $\|\mathbf n\|$

The gap between skew lines is the shadow of a connecting arrow on $\mathbf v_1\times\mathbf v_2$