2.2 Vectors in Three Dimensions

Where is the lamp?

A floor plan puts a lamp 2 m along one wall and 3 m along the other (p111). That spot could be on the floor or on the ceiling.

(2, 3, 4)

The height pins it down. A map is a shadow of space: it keeps two coordinates and drops the third.

From the corner: $\sqrt{4+9+16}\approx5.39$ m, not the floor plan's $\sqrt{13}\approx3.61$ .

Grey: what the floor plan records. Blue: the lamp.

Axes, planes and octants

Three perpendicular axes, right-handed. Fingers along $+x$ , curl toward $+y$ : the thumb is $+z$ (p110).
A point is a walk. $(1,-2,3)$ : 1 along $x$ , 2 back along $y$ , 3 up (Example 2.11).
Three coordinate planes, eight octants. $z=0$ , $y=0$ , $x=0$ cut space by sign (p111). An octant is a quadrant with a sign for height.

coordinate planes octant signs the point (1, −2, 3)

Teal

z=0

, blue

y=0

, amber

x=0

. Top shows the floor plan.

One equation, three dimensions

$x=a$ on a line, in the plane and in space (Figure 2.32).

a in x = a3

$\mathbb R$ : a point.

$\mathbb R^2$ : a line;

y

is free.

$\mathbb R^3$ : a plane;

y,z

are free.

An equation keeps the points that pass. Every coordinate it leaves free stretches the set along that axis.

One picture

Exercise 61: across the floor

\sqrt{13}

, up 5, straight

\sqrt{38}\approx6.16

Space is the plane with one more square under the root.

Distance, planes, spheres, vectors

Distance · Theorem 2.2 · p112

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}

Pythagoras twice: across, then up.

Sphere · p116

(x-a)^2+(y-b)^2+(z-c)^2=r^2

Every point at distance $r$ from $(a,b,c)$ .

Planes parallel to coordinate planes · p115

x=a,\qquad y=b,\qquad z=c

Fix one coordinate; two stay free.

Vectors · p120–p122

\overrightarrow{PQ}=\langle x_2-x_1,\ y_2-y_1,\ z_2-z_1\rangle=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k

$k\mathbf v$ , $\mathbf v\pm\mathbf w$ component by component; $\|\mathbf v\|=\sqrt{x^2+y^2+z^2}$ .

Every operation runs one axis at a time; only the length mixes the components, through Pythagoras.

Examples 2.11–2.20

Example	given	result
2.11 · p110	sketch $(1,-2,3)$	octant $(+,-,+)$
2.12 · p112	$(3,-1,5)$ , $(2,1,-1)$	$d=\sqrt{41}\approx6.40$
2.13 · p116	through $(3,11,7)$ parallel to $yz$ ; points sharing $y=-2$	$x=3$ ; $y=-2$
2.14 · p116	centre $(10,7,4)$ through $(-1,3,-2)$	$(x-10)^2+(y-7)^2+(z-4)^2=173$
2.15 · p117	diameter $(-5,2,3)$ , $(3,4,-1)$	$(x+1)^2+(y-3)^2+(z-1)^2=21$
2.16 · p118	$(x-4)(z-2)=0$	the planes $x=4$ and $z=2$
2.17 · p119	$(x-2)^2+(y-1)^2=4$	cylinder, radius 2, axis $x=2,\ y=1$
2.18 · p121	$P(3,12,6)$ , $Q(-4,-3,2)$	$\overrightarrow{PQ}=\langle -7,-15,-4\rangle$
2.19 · p123	$\mathbf v=\langle -2,9,5\rangle$ , $\mathbf w=\langle 1,-1,0\rangle$	$3\mathbf v-2\mathbf w=\langle -8,29,15\rangle$ ; $\\|5\mathbf w\\|=5\sqrt2$ ; $\tfrac{1}{\sqrt{110}}\langle -2,9,5\rangle$
2.20 · p124	pass 20 yd down, 15 yd left, 60 mph at 30°	$\langle 24\sqrt3,\ 18\sqrt3,\ 30\rangle\approx\langle 41.57,\ 31.18,\ 30\rangle$

draw

Example 2.20's move: take the direction from geometry, then set the length with a scalar.

Lab 1 · plot a point

x1 y−2 z3

presets

coordinate planes

Try: CP 2.11, then top — the height is gone. Each flat view drops one coordinate.

Lab 2 · distance, two steps

P₁ x3 P₁ y−1 P₁ z5 P₂ x2 P₂ y1 P₂ z−1

presets

Try: CP 2.12 gives $5\sqrt2$ . Set $z_1=z_2$ : the up leg vanishes and the plane's formula is left (Figure 2.28).

Lab 3 · planes, spheres, cylinders

plane x = a, y = b, z = c sphere cylinder

centre a0 centre b0 centre c0 radius r3

preset

Try: Ex 2.15 diameter — the centre is the midpoint of $P$ and $Q$ . Cylinder from top is Example 2.17's circle. A missing variable is a free variable.

Lab 4 · from two points to a vector

P x3 P y12 P z6 Q x−4 Q y−3 Q z2

presets

standard-position copy unit vector

Try: swap P ↔ Q — every sign flips, the length stays. A vector is the trip, not the doorstep.

Lab 5 · combine vectors

v₁−2 v₂9 v₃5 w₁1 w₂−1 w₃0 a3 b−2

presets

Try: v + w — the diagonal of the parallelogram, as in the plane. CP 2.19: $\langle 1,-5,8\rangle$ , length $3\sqrt{10}$ .

Lab 6 · the forward pass

Example 2.20: $x$ downfield, $y$ left, $z$ up; yards and mph.

downfield (yd)20 to the left (yd)15 angle up30° speed (mph)60

A straight aim; gravity comes in Chapter 3.

Try: change only the speed — w stays, v scales. Direction from the field, length from the arm.

Common mistakes

Mistake	Result	Fix
$x$ and $y$ swapped on the page	a left-handed frame; §2.4's cross products flip	right-hand rule first
$x=3$ read as a line	in space it is a plane	count the free coordinates
Radius of $\dots=173$ taken as 173	a sphere 13 times too big	$r=\sqrt{173}\approx13.15$
Centre of $(x+1)^2+(y-3)^2+(z-1)^2=21$ as $(1,-3,-1)$	every sign flipped	$(x+1)^2=(x-(-1))^2$ : $(-1,3,1)$
Diameter used as the radius	$\sqrt{84}\approx9.17$ in Example 2.15	half of it: $\sqrt{21}$
Initial minus terminal	$\langle 7,15,4\rangle$ in Example 2.18: backwards	terminal minus initial
$\\|\mathbf v+\mathbf w\\|=\\|\mathbf v\\|+\\|\mathbf w\\|$	lengths added as if in line	add components, then take the length
$-2^2=-4$ inside a length	$\sqrt{102}$ for $\\|\langle -2,9,5\rangle\\|$	$(-2)^2=4$ : $\sqrt{110}$
$\mathbf w$ used as the velocity in Example 2.20	length 28.87, not 60	$\mathbf v=\tfrac{60}{\\|\mathbf w\\|}\mathbf w$

Practice and answers

Exercises from p126; the book's key: p833–834, p838–839.

Checkpoints 2.11–2.20

#	task	answer
2.11	sketch $(-2,3,-1)$	octant $(-,+,-)$
2.12	distance $(1,-5,4)$ to $(4,-1,-1)$	$\sqrt{9+16+25}=5\sqrt2$
2.13	plane through $(1,-6,-4)$ parallel to $xy$	$z=-4$
2.14	centre $(-2,4,-5)$ , through $(4,4,-1)$	$(x+2)^2+(y-4)^2+(z+5)^2=52$
2.15	diameter $P(2,-1,-3)$ , $Q(-2,5,-1)$	$x^2+(y-2)^2+(z+2)^2=14$
2.16	$(y+2)(z-3)=0$	the planes $y=-2$ and $z=3$
2.17	$x^2+(z-2)^2=16$	cylinder, radius 4, axis $x=0,\ z=2$
2.18	$\overrightarrow{ST}$ , $S(3,8,2)$ , $T(2,-1,3)$	$\langle -1,-9,1\rangle=-\mathbf i-9\mathbf j+\mathbf k$
2.19	unit vector along $5\mathbf v+3\mathbf w$ , $\mathbf v=\langle -1,-1,1\rangle,\ \mathbf w=\langle 2,0,1\rangle$	$\tfrac{1}{3\sqrt{10}}\langle 1,-5,8\rangle$
2.20	the pass at 40 mph, 45°	$\langle 16\sqrt2,\ 12\sqrt2,\ 20\sqrt2\rangle$

Exercises

#	task	answer	key step
61	box $O$ to $A(2,3,5)$ : other vertices, $OA$ · p126	$(2,0,5),(2,0,0),(2,3,0),(0,3,0),(0,3,5),(0,0,5)$ ; $\sqrt{38}$	each coordinate 0 or A's
63	$(y-5)(z-6)=0$	planes $y=5$ and $z=6$	a factor is 0
67	through $(1,1,1)$ parallel to $xy$	$z=1$	fix $z$
71	centre $(-1,7,4)$ , radius 4	$(x+1)^2+(y-7)^2+(z-4)^2=16$	right side $r^2$
73	diameter $P(-1,5,7)$ , $Q(-5,2,9)$	$(x+3)^2+(y-3.5)^2+(z-8)^2=\tfrac{29}{4}$	midpoint; $r^2=\tfrac14(16+9+4)$
75	$x^2+y^2+z^2-4z+3=0$	centre $(0,0,2)$ , radius 1	complete the square
77	$\overrightarrow{PQ}$ , $P(3,0,2)$ , $Q(-1,-1,4)$ · p127	$\langle -4,-1,2\rangle$	terminal minus initial
79	$P(-2,5,-8)$ , midpoint $M(1,-7,4)$	$\langle 6,-24,24\rangle$	$\overrightarrow{PQ}=2\overrightarrow{PM}$
83	$\mathbf a=\langle -1,-2,4\rangle,\ \mathbf b=\langle -5,6,-7\rangle$ : $\mathbf a+\mathbf b,\ 4\mathbf a,\ -5\mathbf a+3\mathbf b$	$\langle -6,4,-3\rangle$ , $\langle -4,-8,16\rangle$ , $\langle -10,28,-41\rangle$	component by component
87	$\\|\mathbf u-\mathbf v\\|,\ \\|-2\mathbf u\\|$ , $\mathbf u=2\mathbf i+3\mathbf j+4\mathbf k,\ \mathbf v=-\mathbf i+5\mathbf j-\mathbf k$	$\sqrt{38}$ , $2\sqrt{29}$	$\\|-2\mathbf u\\|=2\\|\mathbf u\\|$
93	unit vector along $\overrightarrow{PQ}$ , $P(-2,3,1)$ , $Q(0,-4,4)$	$\tfrac{1}{\sqrt{62}}\langle 2,-7,3\rangle$	the book's key prints $+7$ ; $-4-3=-7$
99	$\\|\mathbf u\\|=10$ along $\langle 7,-1,3\rangle$	$\tfrac{10}{\sqrt{59}}\langle 7,-1,3\rangle$	unit vector, then scale
103	length 5 along $\overrightarrow{AB}$ , $A(2,1,5)$ , $B(3,4,-7)$ · p128	$\tfrac{5}{\sqrt{154}}\langle 1,3,-12\rangle$	the Example 2.20 move
111	50 N along $\overrightarrow{OP}$ , $P(3,4,0)$ ; angle with $+x$	$\langle 30,40,0\rangle$ ; $53^\circ$	$\cos\alpha=30/50$

Pythagoras twice

Let $R=(x_2,y_2,z_1)$ : below or above $P_2$ , at $P_1$ 's height (p112).

\|P_1R\|^2=(x_2-x_1)^2+(y_2-y_1)^2,\qquad d^2=\|P_1R\|^2+(z_2-z_1)^2 .

The floor leg and the vertical leg meet at a right angle at $R$ . Each new dimension adds one right angle, so one more square under the root.

A sphere is a distance test

\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}=r \iff (x-a)^2+(y-b)^2+(z-c)^2=r^2

Squaring loses nothing: both sides are $\ge0$ (p116). Less than $r^2$ is inside the ball, more is outside.

Drop $z$ : $(x-2)^2+(y-1)^2=4$ measures distance to a line, not a point — Example 2.17's cylinder.

Centre and radius from general form

x^2+y^2+z^2+Dx+Ey+Fz+G=0\iff\Bigl(x+\tfrac D2\Bigr)^2+\Bigl(y+\tfrac E2\Bigr)^2+\Bigl(z+\tfrac F2\Bigr)^2=\frac{D^2+E^2+F^2}{4}-G

Exercise 75

$x^2+y^2+(z-2)^2=1$ : centre $(0,0,2)$ , radius 1.

Exercise 76

$(x-3)^2+(y+4)^2+(z-5)^2=25$ : centre $(3,-4,5)$ , radius 5.

The right side decides: positive, a sphere; zero, a point; negative, nothing at all.

What space allows

Lines that miss without being parallel; circles linked without touching (p113–114).

skew lines linked circles

Skew lines

$L_1=\{(t,1,2)\}$ , $L_2=\{(0,s,-1)\}$ : directions $\mathbf i$ , $\mathbf j$ ; a common point needs $z=2$ and $z=-1$ . Their shadows cross at $(0,1,0)$ ; the lines pass 3 apart.

In space a closed curve has no inside, so another can pass through it.

Where it is used next

Where	What it needs from §2.2
§2.3 dot product	components and length in space
§2.4 cross product	a right-handed frame: $\mathbf i\times\mathbf j=\mathbf k$
§2.5 lines and planes	$x=a$ grows into $ax+by+cz=d$ ; skew lines get a distance
§2.6 quadric surfaces	spheres, cylinders, completing the square

Objective	You can now
2.2.1 describe space	right-handed axes, coordinate planes, octants
2.2.2 locate points	plot $(x,y,z)$ as the far corner of a box
2.2.3 distance	$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
2.2.4 planes, spheres	$x=a$ , $y=b$ , $z=c$ ; $(x-a)^2+(y-b)^2+(z-c)^2=r^2$
2.2.5 vector operations	$\langle x,y,z\rangle=x\mathbf i+y\mathbf j+z\mathbf k$ , slot by slot

Where is the lamp?

Axes, planes and octants

One equation, three dimensions

One picture

Distance, planes, spheres, vectors

Examples 2.11–2.20

Lab 1 · plot a point

Lab 2 · distance, two steps

Lab 3 · planes, spheres, cylinders

Lab 4 · from two points to a vector

Lab 5 · combine vectors

Lab 6 · the forward pass

Common mistakes

Practice and answers

Pythagoras twice

A sphere is a distance test

Centre and radius from general form

What space allows

Where it is used next

Vectors in Three Dimensions

A point in space needs three coordinates, one per perpendicular axis

Right-hand rule: curl xxx toward yyy, the thumb is zzz

To plot (x,y,z)(x,y,z)(x,y,z), walk xxx, then yyy, then zzz

Each pair of axes spans a coordinate plane

The three coordinate planes cut space into eight octants

Distance in space is Pythagoras twice

Subtract, square, add, take the root: d=41d=\sqrt{41}d=41​

Two right triangles prove the distance formula

In space, lines that are not parallel can still miss each other

The same equation is a point, a line or a plane, depending on the space

A plane parallel to a coordinate plane fixes one coordinate

Find the coordinate that stays fixed; that is the plane's equation

A sphere is every point at one distance from a centre

Complete the square to read the centre and the radius

Radius from the distance formula; the equation keeps r2r^2r2

A diameter gives the centre (its midpoint) and the radius (half its length)

A product equal to zero describes a union of surfaces

A missing variable is free: a circle becomes a cylinder

A vector in space has three components: ⟨x,y,z⟩=xi+yj+zk\langle x,y,z\rangle=x\mathbf i+y\mathbf j+z\mathbf k⟨x,y,z⟩=xi+yj+zk

Every vector operation works one coordinate at a time, as in the plane

Component form is terminal minus initial, in all three slots

Scale, then combine slot by slot; ∥5w∥=5∥w∥\|5\mathbf w\|=5\|\mathbf w\|∥5w∥=5∥w∥

Take the direction from geometry, then fix the length with a scalar

Space is the plane with one more coordinate

Right-hand rule: curl $x$ toward $y$ , the thumb is $z$

To plot $(x,y,z)$ , walk $x$ , then $y$ , then $z$

Subtract, square, add, take the root: $d=\sqrt{41}$

Radius from the distance formula; the equation keeps $r^2$

A vector in space has three components: $\langle x,y,z\rangle=x\mathbf i+y\mathbf j+z\mathbf k$

Scale, then combine slot by slot; $\|5\mathbf w\|=5\|\mathbf w\|$