Calculus Volume 3 · Chapter 2 · Section 2.1

Vectors in the Plane

An amount with a direction. Draw it as an arrow, write it as two numbers, add it tip to tail.

Idea 1 · length and directionAn arrow you may slide anywhere.Same length, same direction: same vector. Idea 2 · tip to tailAdding is walking one arrow, then the next.The sum runs from start to finish. Idea 3 · two numbers⟨x, y⟩ = how far across, how far up.Then every operation is arithmetic.

Colour key v w u unit vector result key idea · round handles can be dragged · page links open the textbook

L1Is 300 lb + 150 lb = 450 lb?
L1An arrow you can slide
L1Stretch, add, subtract
L1One picture
L2Components and magnitude
L2The rules
L2Angles, unit vectors, i and j
L2Examples 2.1–2.10
L2Lab 1 · drag v and w
L2Lab 2 · length and angle
L2Lab 3 · a stuck car and a crosswind
L2Common mistakes
L2Practice and answers
L3Why the rules hold
L3The triangle inequality
L3Length and normalizing
L3The quadrant trap
L3Every vector from two others
L3The same rules in space

Level 1 · see itWhat a vector isbefore any formula

Is 300 lb + 150 lb = 450 lb?

A truck pulls a stuck car with 300 lb. Two people push with 150 lb, aimed 15° off the truck's line (Example 2.9).

446.6 lb

at about 5° — not 450. Arrows add in the plane; their sizes add only when they point the same way.

A vector has size and direction (p90); a scalar has size only.

To scale: truck, push, sum.

An arrow you can slide

An arrow from an initial to a terminal point; length $\|\mathbf v\|$ , direction by the head (p90).
Only length and direction count. A vector says "go 7 right, 4 up" — not where to start. (p90)
Names: $\mathbf v$ , $\vec v$ , $\overrightarrow{PQ}$ ; $\mathbf 0$ has no direction.

Drag $P$ : the arrow slides, $\langle 7,4\rangle$ stays. Drag $Q$ : the vector changes.

Example 2.1 (p91). Dashed: the same vector from the origin.

Stretch, add, subtract

Example 2.2 (p93) on a grid: $\mathbf v=\langle 2,4\rangle$ , $\mathbf w=\langle 3,-2\rangle$ .

Both addition methods land on the same point: v then w, or w then v, are two halves of one parallelogram.

Scalar multiple · p91

$k\mathbf v$ : $|k|$ times as long, flipped if $k<0$ . It never turns an arrow.

Sum · p92

Tail of $\mathbf w$ on tip of $\mathbf v$ . $\mathbf v+\mathbf w=\mathbf w+\mathbf v$ .

Difference · p92

$\mathbf v-\mathbf w$ runs from the tip of $\mathbf w$ to the tip of $\mathbf v$ .

One picture

Two numbers, or a length and an angle — the same vector, joined by a right triangle.

A vector is a right triangle you can move.

Level 2 · compute itComponents, rules, unit vectorsand three labs

Components and magnitude

Component form · p96

\mathbf v=\langle x_t-x_i,\ y_t-y_i\rangle

Terminal minus initial. $\langle 4,-2\rangle$ is a vector, $(4,-2)$ a point.

Magnitude · p97

\|\langle x,y\rangle\|=\sqrt{x^2+y^2}

Pythagoras. Zero only for $\mathbf 0$ .

Operations · p98

k\mathbf v=\langle kx_1,\,ky_1\rangle,\qquad \mathbf v\pm\mathbf w=\langle x_1\pm x_2,\ y_1\pm y_2\rangle

Across adds to across, up adds to up.

Triangle inequality · p93

\|\mathbf v+\mathbf w\|\le\|\mathbf v\|+\|\mathbf w\|

Equal only when $\mathbf v$ , $\mathbf w$ point the same way.

The rules

Theorem 2.1 (p99).

rule	name	means
$\mathbf u+\mathbf v=\mathbf v+\mathbf u$	commutative	walk either arrow first
$(\mathbf u+\mathbf v)+\mathbf w=\mathbf u+(\mathbf v+\mathbf w)$	associative	no brackets needed
$\mathbf u+\mathbf 0=\mathbf u,\quad \mathbf u+(-\mathbf u)=\mathbf 0$	identity, inverse	there and back
$r(s\mathbf u)=(rs)\mathbf u$	scaling	stretch twice = once by the product
$(r+s)\mathbf u=r\mathbf u+s\mathbf u,\quad r(\mathbf u+\mathbf v)=r\mathbf u+r\mathbf v$	distributive	scale a triangle, scale every side
$1\mathbf u=\mathbf u,\quad 0\mathbf u=\mathbf 0$	identity, zero	—

Number rules survive because a vector is two numbers side by side. Multiplying two vectors needs new ideas: §2.3 and §2.4.

Angles, unit vectors, i and j

From length and angle · p99

\mathbf v=\langle\|\mathbf v\|\cos\theta,\ \|\mathbf v\|\sin\theta\rangle

Unit vector · p100

\mathbf u=\frac{\mathbf v}{\|\mathbf v\|}

Length 1, same direction.

i and j · p101

\langle x,y\rangle=x\,\mathbf i+y\,\mathbf j

A unit vector is a pure direction — a point on the unit circle. Multiply by a length to get any vector.

Examples 2.1–2.10

Example	given	result
2.1 · p91	$P(1,1)$ , $Q(8,5)$	$\overrightarrow{PQ}=\langle 7,4\rangle$ , length $\sqrt{65}$
2.2 · p93	$3\mathbf w$ , $\mathbf v+\mathbf w$ , $2\mathbf v-\mathbf w$	on the grid: $\langle 9,-6\rangle,\ \langle 5,2\rangle,\ \langle 1,10\rangle$
2.3 · p94	equivalent?	(a) no · (b) yes
2.4 · p96	$(-3,4)\to(1,2)$	$\langle 4,-2\rangle$
2.5 · p98	$\mathbf v=\langle 6,8\rangle$ , $\mathbf w=\langle -2,4\rangle$	$\\|\mathbf v\\|=10$ ; $\mathbf v+\mathbf w=\langle 4,12\rangle$ ; $3\mathbf v=\langle 18,24\rangle$ ; $\mathbf v-2\mathbf w=\langle 10,0\rangle$
2.6 · p100	length 4 at $-45^\circ$	$\langle 2\sqrt2,-2\sqrt2\rangle$
2.7 · p100	along $\langle 1,2\rangle$ ; length 7	$\tfrac1{\sqrt5}\langle 1,2\rangle$ ; $\tfrac7{\sqrt5}\langle 1,2\rangle$
2.8 · p101	$\langle 3,-4\rangle$ ; unit vector at 60°	$3\mathbf i-4\mathbf j$ ; $\tfrac12\mathbf i+\tfrac{\sqrt3}2\mathbf j$
2.9 · p102	300 lb + 150 lb at 15°	446.6 lb at 5°
2.10 · p103	425 mph west, wind 40 mph from the NE	454.17 mph, 3.57° south of west

Put one force on the $x$ -axis first — it makes every force problem shorter.

Lab 1 · drag v and w

show

slide w to the tip of v100% scalar k1.5

presets

Try: same direction — the only case where the lengths add exactly. k v through 0 — the arrow shrinks and flips.

Lab 2 · length and angle

length ‖v‖4 angle θ315°

presets

Try: play the angle — amber $x$ , blue $y$ . Length moves the tip along a ray, angle moves it round a circle; the unit vector only sees the angle.

Lab 3 · a stuck car and a crosswind

scenario

truck ‖F₁‖ (lb)300 push ‖F₂‖ (lb)150 angle between15°

Try: play the angle — 450 lb at 0°, 335.4 lb at 90°, 150 lb at 180°. A 40 mph wind barely slows a 425 mph plane, but it pushes it off its line.

Common mistakes

Mistake	Result	Fix
Initial minus terminal	$\langle -4,2\rangle$ in Example 2.4: backwards	terminal minus initial
$(4,-2)$ for a vector	points and vectors mixed up	angle brackets (p96)
Adding magnitudes	450 lb instead of 446.6 lb	add components, then take the length
$\\|k\mathbf v\\|=k\\|\mathbf v\\|$ for $k<0$	a negative length	$\|k\|\,\\|\mathbf v\\|$
$\tan^{-1}(y/x)$ without the signs	30° for $-\sqrt3\,\mathbf i-\mathbf j$ ; true 210°	sketch; add 180° when $x<0$
Radian mode	about 210 lb in Example 2.9	degree mode
"Wind from the NE" drawn toward the NE	the plane drifts the wrong way	add 180°: it blows toward 225°
Dividing by $\|x\|+\|y\|$ to normalize	$\langle 1,2\rangle/3$ has length $\sqrt5/3$	divide by $\sqrt{x^2+y^2}$
$2\mathbf v-\mathbf w$ read as $2(\mathbf v-\mathbf w)$	$\langle -2,12\rangle$ instead of $\langle 1,10\rangle$	the scalar multiplies one vector

Practice and answers

Exercises from p104; the book's key: p832–833, p837–838.

Checkpoints 2.1–2.10

#	task	answer
2.1	$\overrightarrow{ST}$ , $S(3,-1)$ , $T(-2,3)$	$\langle -5,4\rangle$
2.2	sketch $2\mathbf w-\mathbf v$	on the grid $\langle 4,-8\rangle$
2.3	which are equivalent? (p95)	a, b, e
2.4	$(-4,-5)\to(-1,2)$	$\langle 3,7\rangle$
2.5	$\mathbf a=\langle 7,1\rangle$ , $\mathbf b$ : $(3,2)\to(-1,-1)$	$\\|\mathbf a\\|=5\sqrt2$ ; $\mathbf b=\langle -4,-3\rangle$ ; $3\mathbf a-4\mathbf b=\langle 37,15\rangle$
2.6	prove $\mathbf u+(-\mathbf u)=\mathbf 0$	$\langle x-x,\ y-y\rangle=\langle 0,0\rangle$
2.7	length 10 at 120°	$\langle -5,\ 5\sqrt3\rangle$
2.8	length 5, opposite to $\langle 9,2\rangle$	$-\tfrac5{\sqrt{85}}\langle 9,2\rangle$
2.9	$\langle 16,-11\rangle$ ; unit vector at 225°	$16\mathbf i-11\mathbf j$ ; $-\tfrac{\sqrt2}2\mathbf i-\tfrac{\sqrt2}2\mathbf j$
2.10	550 mph north, wind 50 mph from the NW	about 516 mph

Exercises

1–9 use $P(-1,3)$ , $Q(1,5)$ , $R(-3,7)$ .

#	task	answer	key step
1	$\overrightarrow{PQ}$	$\langle 2,2\rangle=2\mathbf i+2\mathbf j$	terminal minus initial
5	$\overrightarrow{PQ}+\overrightarrow{PR}$	$\langle 0,6\rangle$	add componentwise
7	$2\overrightarrow{PQ}-2\overrightarrow{PR}$	$\langle 8,-4\rangle$	scale, then subtract
11	unit vector, $(-1,-3)\to(2,1)$	$\langle\tfrac35,\tfrac45\rangle$	$\langle 3,4\rangle$ has length 5
13	$Q$ on the $y$ -axis, $\\|\mathbf v\\|=\sqrt5$	$Q(0,2)$	$\sqrt{1+y^2}=\sqrt5$
15	$\mathbf a=2\mathbf i+\mathbf j$ , $\mathbf b=\mathbf i+3\mathbf j$	$\langle 3,4\rangle$ ; $\langle 1,-2\rangle$ ; $\langle 4,2\rangle$ ; $\langle -1,-3\rangle$ ; $\langle 3,-1\rangle$	$5\le\sqrt5+\sqrt{10}$
17	$\\|-3\mathbf a+\mathbf b-4\mathbf i+\mathbf j\\|$	15	sum is $\langle 0,15\rangle$
27	length 7 along $\langle 3,-5\rangle$ (p105)	$\approx\langle 3.60,-6.00\rangle$	normalize, scale by 7
29	length 2 at 30°	$\langle\sqrt3,1\rangle$	$\langle 2\cos30^\circ,2\sin30^\circ\rangle$
33	length 10 at $5\pi/6$	$\langle -5\sqrt3,5\rangle$	quadrant II
35	angle of $5\sqrt2\,\mathbf i-5\sqrt2\,\mathbf j$	$7\pi/4$	$-45^\circ+360^\circ$
43	$D$ for parallelogram $ABCD$ (p106)	$D(6,1)$	$\overrightarrow{AD}=\overrightarrow{BC}$
49	45 lb and 52 lb at 25° (p107)	94.71 lb at 13.42°	Lab 3
53	550 mph at N43°E, wind 25 mph toward N15°E (p108)	572.19 mph, N41.82°E	bearing → angle $90^\circ-b$

Level 3 · why it worksWhere the rules come fromshort arguments

Why the rules hold

Each proof: write components, use the number rule in each slot (p99).

\mathbf u+\mathbf v=\langle x_1+x_2,\ y_1+y_2\rangle=\langle x_2+x_1,\ y_2+y_1\rangle=\mathbf v+\mathbf u

r(\mathbf u+\mathbf v)=\langle rx_1+rx_2,\ ry_1+ry_2\rangle=r\mathbf u+r\mathbf v

Vector algebra is two copies of number algebra running side by side.

The triangle inequality

\bigl(\|\mathbf v\|+\|\mathbf w\|\bigr)^2-\|\mathbf v+\mathbf w\|^2=2\bigl(\|\mathbf v\|\,\|\mathbf w\|-(v_1w_1+v_2w_2)\bigr)

\|\mathbf v\|^2\|\mathbf w\|^2-(v_1w_1+v_2w_2)^2=(v_1w_2-v_2w_1)^2\ \ge 0

So the gap is never negative; it is zero only for parallel vectors pointing the same way. The gap measures how much two arrows disagree in direction.

Length and normalizing

\|k\mathbf v\|=\sqrt{k^2}\,\sqrt{x^2+y^2}=|k|\,\|\mathbf v\|,\qquad \Bigl\|\frac{\mathbf v}{\|\mathbf v\|}\Bigr\|=1

Every nonzero vector is "how long" times "which way": $\mathbf v=\|\mathbf v\|\,\mathbf u$ .

The quadrant trap

$\tan\theta=y/x$ has two solutions 180° apart (Figure 2.20).

Exercise 35

$5\sqrt2\,\mathbf i-5\sqrt2\,\mathbf j$ : $-45^\circ$ , quadrant IV → $315^\circ$ .

Exercise 36

$-\sqrt3\,\mathbf i-\mathbf j$ : $30^\circ$ , both negative → $210^\circ$ .

A ratio cannot tell $\langle 1,1\rangle$ from $\langle -1,-1\rangle$ ; the signs carry the rest of the direction.

Every vector from two others

If $a_1b_2-a_2b_1\neq0$ , then $\mathbf c=\alpha\mathbf a+\beta\mathbf b$ with (Exercise 37)

\alpha=\frac{c_1b_2-c_2b_1}{a_1b_2-a_2b_1},\qquad \beta=\frac{a_1c_2-a_2c_1}{a_1b_2-a_2b_1} .

$\mathbf a=\langle 2,1\rangle$ , $\mathbf b=\langle 1,3\rangle$ , $\mathbf c=\langle 7,11\rangle$ : $\alpha=2$ , $\beta=3$ .

Two non-parallel arrows are a coordinate system; $\mathbf i,\mathbf j$ are just the tidiest pair.

The same rules in space

§2.2 adds a third component: $\langle x,y,z\rangle$ , $\|\mathbf v\|=\sqrt{x^2+y^2+z^2}$ , $\mathbf i,\mathbf j,\mathbf k$ . Nothing else changes.

Tip to tail and components are the whole chapter, in more dimensions.

\langle 1,3,2\rangle+\langle 3,-1,1\rangle=\langle 4,2,3\rangle

. Drag to turn.

	Property	Rule	As arrows
i	commutative	$\mathbf u+\mathbf v=\mathbf v+\mathbf u$	parallelogram
ii	associative	$(\mathbf u+\mathbf v)+\mathbf w=\mathbf u+(\mathbf v+\mathbf w)$	three legs, one trip
iii	additive identity	$\mathbf u+\mathbf 0=\mathbf u$	a walk of length 0
iv	additive inverse	$\mathbf u+(-\mathbf u)=\mathbf 0$	there and back
v	associativity of scalar multiplication	$r(s\mathbf u)=(rs)\mathbf u$	stretch twice
vi	distributive	$(r+s)\mathbf u=r\mathbf u+s\mathbf u$	stretches add along a line
vii	distributive	$r(\mathbf u+\mathbf v)=r\mathbf u+r\mathbf v$	scaled triangle
viii	identity and zero	$1\mathbf u=\mathbf u,\ \ 0\mathbf u=\mathbf 0$	—

Objective	You can now
2.1.1 plane vector, notation	draw $\overrightarrow{PQ}$ ; equivalent = same length and direction
2.1.2 operations	$k\mathbf v$ , $\mathbf v\pm\mathbf w$ : as arrows and in components
2.1.3 component form	$\langle x_t-x_i,\ y_t-y_i\rangle$
2.1.4 magnitude	$\sqrt{x^2+y^2}$ : Pythagoras = distance formula
2.1.5 unit vectors	$\mathbf v/\\|\mathbf v\\|$ ; $\ x\,\mathbf i+y\,\mathbf j$
2.1.6 vector quantities	force (Example 2.9), velocity (Example 2.10)

Is 300 lb + 150 lb = 450 lb?

An arrow you can slide

Stretch, add, subtract

One picture

Components and magnitude

The rules

Angles, unit vectors, i and j

Examples 2.1–2.10

Lab 1 · drag v and w

Lab 2 · length and angle

Lab 3 · a stuck car and a crosswind

Common mistakes

Practice and answers

Why the rules hold

The triangle inequality

Length and normalizing

The quadrant trap

Every vector from two others

The same rules in space

Vectors in the Plane

Speed alone is not enough: a vector also carries a direction

Draw a vector as an arrow: length = magnitude, head = direction

Where an arrow starts is not part of the vector

To sketch PQ→\overrightarrow{PQ}PQ​, start at PPP and put the head at QQQ

A scalar stretches or flips an arrow, never turns it

Add tip to tail: v+w\mathbf v+\mathbf wv+w runs from the tail of v\mathbf vv to the head of w\mathbf ww

The parallelogram gives the same sum, so v+w=w+v\mathbf v+\mathbf w=\mathbf w+\mathbf vv+w=w+v

Subtracting w\mathbf ww means adding −w-\mathbf w−w

Magnitudes do not simply add: ∥v+w∥≤∥v∥+∥w∥\|\mathbf v+\mathbf w\|\le\|\mathbf v\|+\|\mathbf w\|∥v+w∥≤∥v∥+∥w∥

For a combination like 2v−w2\mathbf v-\mathbf w2v−w: scale first, then add tip to tail

On a grid, equal vectors make the same moves across and up

To test equivalence, count each arrow's moves

A vector from the origin is named by its head: ⟨x,y⟩\langle x,y\rangle⟨x,y⟩

Component form = terminal point minus initial point

The magnitude is the hypotenuse: ∥⟨x,y⟩∥=x2+y2\|\langle x,y\rangle\|=\sqrt{x^2+y^2}∥⟨x,y⟩∥=x2+y2​

Tip to tail on a grid: across adds to across, up adds to up

In components, vector arithmetic is number arithmetic in each slot

Vectors obey the same algebra rules as numbers

Each property holds because it holds for numbers in every slot

The triangle inequality comes down to a perfect square

Magnitude and angle give the components: the legs of a right triangle

Length 4 at −45∘-45^\circ−45∘ is ⟨22, −22⟩\langle 2\sqrt2,\ -2\sqrt2\rangle⟨22​, −22​⟩

Divide a vector by its length to get its direction

For a chosen length in a given direction, scale the unit vector

Every vector is xxx steps of i\mathbf ii plus yyy steps of j\mathbf jj

A unit vector at angle θ\thetaθ is cos⁡θ i+sin⁡θ j\cos\theta\,\mathbf i+\sin\theta\,\mathbf jcosθi+sinθj

One vector, four descriptions: all the same right triangle

Forces add as arrows: 300 lb + 150 lb gives less than 450 lb

Put one force on the xxx-axis, add components, then find length and angle

Velocities add too: a crosswind changes both speed and heading

Wind toward 225°: add components to get 454.17 mph, 3.57° south of west

A vector is a right triangle you can slide

To sketch $\overrightarrow{PQ}$ , start at $P$ and put the head at $Q$

Add tip to tail: $\mathbf v+\mathbf w$ runs from the tail of $\mathbf v$ to the head of $\mathbf w$

The parallelogram gives the same sum, so $\mathbf v+\mathbf w=\mathbf w+\mathbf v$

Subtracting $\mathbf w$ means adding $-\mathbf w$

Magnitudes do not simply add: $\|\mathbf v+\mathbf w\|\le\|\mathbf v\|+\|\mathbf w\|$

For a combination like $2\mathbf v-\mathbf w$ : scale first, then add tip to tail

A vector from the origin is named by its head: $\langle x,y\rangle$

The magnitude is the hypotenuse: $\|\langle x,y\rangle\|=\sqrt{x^2+y^2}$

Length 4 at $-45^\circ$ is $\langle 2\sqrt2,\ -2\sqrt2\rangle$

Every vector is $x$ steps of $\mathbf i$ plus $y$ steps of $\mathbf j$

A unit vector at angle $\theta$ is $\cos\theta\,\mathbf i+\sin\theta\,\mathbf j$

Put one force on the $x$ -axis, add components, then find length and angle