HL · Vectors 1 · Vectors: size, direction & arithmetic

Module 24 of 46Lesson 1 of 6 · HL · Vectors 1 · Vectors: size, direction & arithmetic

Vectors, lines, scalar & vector products · Essential understanding 1

Vectors: size, direction & arithmetic

IB AA AHL 3.12with Prof. Kitso

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

The big idea: A vector is a quantity with a size and a direction. We write it as a column of numbers, add vectors by adding the matching numbers, and find its length with Pythagoras.

By the end of this lesson you can:

tell a scalar from a vector and write a vector as a column or with i, j, k
add and subtract vectors and multiply a vector by a scalar
find the magnitude of a vector in 2D and 3D
find a unit vector and a vector of a given length in a given direction
decide whether two vectors are parallel and use this to find unknowns
find a resultant of two or more vectors

💡 AA Paper 1 has no calculator

In AA, Paper 1 is calculator-free, so the by-hand working is what earns the marks there. Use the TI-Nspire steps for Paper 2, and to check your hand answers.

🧑🏾 Imagine this…

A friend in Maun asks, “How far is the airport?” You say “5 km”. That is only half an answer. 5 km which way? If you say “5 km to the north-east”, now they can find it. “5 km” on its own is a scalar (just a size). “5 km to the north-east” is a vector (size and direction).

Scalar

a quantity with size only: mass 60 kg, time 8 s, speed 12 m/s.

Vector

a quantity with size AND direction: a velocity of 12 m/s east, a force of 30 N downwards, a move of 3 steps right and 2 steps up.

Column vector

the numbers written in a column, e.g.

\left(\begin{smallmatrix}3\\-2\end{smallmatrix}\right)

: “3 along the x-direction, 2 down the y-direction”. In 3D there is a third number for z.

Base vectors i, j, k

the unit steps along the axes:

\left(\begin{smallmatrix}1\\0\\0\end{smallmatrix}\right)

= i,

\left(\begin{smallmatrix}0\\1\\0\end{smallmatrix}\right)

= j,

\left(\begin{smallmatrix}0\\0\\1\end{smallmatrix}\right)

= k. So

\left(\begin{smallmatrix}3\\-2\\5\end{smallmatrix}\right)

= 3i − 2j + 5k.

Magnitude |v|

the length of the vector. For

\left(\begin{smallmatrix}a\\b\\c\end{smallmatrix}\right)

it is

\sqrt{a^2+b^2+c^2}

Unit vector

a vector of length exactly 1. It only tells you a direction.

Zero vector

0 =

\left(\begin{smallmatrix}0\\0\\0\end{smallmatrix}\right)

. It has no size and no direction.

Parallel vectors

two vectors pointing the same way or exactly opposite ways. One is a scalar multiple of the other: b = λa.

★ The rules

Add / subtract: do each position separately. $\begin{pmatrix}a\\b\end{pmatrix}+\begin{pmatrix}c\\d\end{pmatrix}=\begin{pmatrix}a+c\\b+d\end{pmatrix}$
Scalar multiple: multiply every number. $k\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}ka\\kb\end{pmatrix}$ A negative k turns the vector round (it points the opposite way). |kv| = |k||v|.
Magnitude: $|\mathbf{v}|=\sqrt{v_1^2+v_2^2+v_3^2}$
Unit vector in the direction of v: $\hat{\mathbf{v}}=\dfrac{1}{|\mathbf{v}|}\mathbf{v}$
A vector of length L in the direction of v: $L\,\hat{\mathbf{v}}=\dfrac{L}{|\mathbf{v}|}\mathbf{v}$
The resultant of several vectors is their sum: one single vector with the same total effect.
Parallel: b = λa for some number λ ≠ 0. So the numbers must scale by the same factor.

⚠️ Marks are lost here

|a + b| is not |a| + |b|. Add the vectors first, then find the length of the answer.
To make a unit vector you divide every number by the length |v| (not just one of them).
A negative scalar does not make the vector “shorter”: −3v is three times as long as v, pointing the opposite way.
Vectors are written in bold in print. By hand, write a curly underline or an arrow above the letter.

💡 Store vectors on your TI-Nspire

On a Calculator page, type a vector in square brackets (it is a one-row matrix), then press ctrl💻 Ctrl var💻 Ctrl+L for the store arrow and a letter: [2,-1,3]→p. After that p+q, 3p-2q and norm(p) all work. You can also find norm under menu💻 click menu on keypad ▸ 7: Matrix & Vector ▸ Vector, or simply type it. In IB exam mode the unit-vector, dot-product and cross-product commands are blocked, so in these lessons you divide by norm( ) and type dot and cross products from their formulas.

🧪 Try it yourselfAdd vectors by dragging

Drag the tips of a (blue) and b (orange). The green arrow is a + b (the resultant). Make a = (4, −3) and check |a| = 5. Then try to make b a multiple of a: when the two arrows line up, they are parallel.

Worked examples

Example 1.1Writing and measuring a vector

A drone leaves a lodge near Kasane and flies 4 km east and 3 km south. Let east be the x-direction and north the y-direction, and call the move a.

(a) Write a as a column vector and using i and j.
(b) Find the magnitude of a.
(c) Find the unit vector in the direction of a.

🧠 Think it through

East is positive x and south is negative y, so a = 4 along x and −3 along y.
Magnitude by Pythagoras: $|\mathbf{a}|=\sqrt{4^2+(-3)^2}$ .
Unit vector: divide each number by the magnitude.

✍️ By hand

\mathbf{a}=\begin{pmatrix}4\\-3\end{pmatrix}=4\mathbf{i}-3\mathbf{j}

|\mathbf{a}|=\sqrt{16+9}=\sqrt{25}=5

\hat{\mathbf{a}}=\dfrac{1}{5}\begin{pmatrix}4\\-3\end{pmatrix}=\begin{pmatrix}0.8\\-0.6\end{pmatrix}

🔢 On your TI-Nspire CX II

Type norm([4,-3]) and press enter ↵💻 Enter. Then divide the vector by it: [4,-3]/norm([4,-3]). (In exam mode there is no unit-vector key, so divide by norm( ) yourself.)
📍 Where to find it on the TI-Nspire
norm( is in menu💻 click menu on keypad▸7: Matrix & Vector▸7: Norms▸1: Norm
norm( ) is the magnitude; divide the vector by it

✅ Answer (a) a = (4, −3) = 4i − 3j. (b) |a| = 5 km. (c) (0.8, −0.6).

Check: the unit vector must have length 1. 0.8² + 0.6² = 0.64 + 0.36 = 1 ✓.

Example 1.2Adding, subtracting and scaling in 3D

Given

\mathbf{p}=\begin{pmatrix}2\\-1\\3\end{pmatrix}

and

\mathbf{q}=\begin{pmatrix}-4\\5\\1\end{pmatrix}

, find (a)

\mathbf{p}+\mathbf{q}

(b)

\mathbf{q}-\mathbf{p}

(c)

3\mathbf{p}-2\mathbf{q}

🧠 Think it through

Work position by position: top with top, middle with middle, bottom with bottom.
For (c) scale first (3p and 2q), then subtract.

✍️ By hand

\mathbf{p}+\mathbf{q}=\begin{pmatrix}2+(-4)\\-1+5\\3+1\end{pmatrix}=\begin{pmatrix}-2\\4\\4\end{pmatrix}

\mathbf{q}-\mathbf{p}=\begin{pmatrix}-4-2\\5-(-1)\\1-3\end{pmatrix}=\begin{pmatrix}-6\\6\\-2\end{pmatrix}

3\mathbf{p}-2\mathbf{q}=\begin{pmatrix}6\\-3\\9\end{pmatrix}-\begin{pmatrix}-8\\10\\2\end{pmatrix}=\begin{pmatrix}14\\-13\\7\end{pmatrix}

🔢 On your TI-Nspire CX II

Store the vectors: [2,-1,3]→p and [-4,5,1]→q. Then type p+q, q-p and 3p-2q.
Vectors behave like one-row matrices

✅ Answer (a) (−2, 4, 4). (b) (−6, 6, −2). (c) (14, −13, 7).

Example 1.3Magnitude and unit vectors in 3D

The vector

\mathbf{v}=\begin{pmatrix}2\\3\\-6\end{pmatrix}

is given.

(a) Find |v|.
(b) Write down the unit vector in the direction of v.
(c) Find a vector of magnitude 14 in the opposite direction to v.

🧠 Think it through

Magnitude is the 3D Pythagoras: square the three numbers, add, square root.
Divide by the magnitude to get the unit vector.
Magnitude 14 is twice 7. Opposite direction means a negative scalar.

✍️ By hand

|\mathbf{v}|=\sqrt{4+9+36}=\sqrt{49}=7

\hat{\mathbf{v}}=\dfrac{1}{7}\begin{pmatrix}2\\3\\-6\end{pmatrix}=\begin{pmatrix}\frac{2}{7}\\\frac{3}{7}\\-\frac{6}{7}\end{pmatrix}

-2\mathbf{v}=\begin{pmatrix}-4\\-6\\12\end{pmatrix}

, and

|-2\mathbf{v}|=2\times 7=14

✓

🔢 On your TI-Nspire CX II

Type norm([2,3,-6]) and then [2,3,-6]/norm([2,3,-6]). The GDC gives the exact fractions $\dfrac{2}{7}$ , $\dfrac{3}{7}$ , − $\dfrac{6}{7}$ , so copy them into your answer.
📍 Where to find it on the TI-Nspire
norm( is in menu💻 click menu on keypad▸7: Matrix & Vector▸7: Norms▸1: Norm

✅ Answer (a) 7. (b) (

\dfrac{2}{7}

\dfrac{3}{7}

, −

\dfrac{6}{7}

). (c) (−4, −6, 12).

Spot the pattern: 2, 3, 6 and 7 is a favourite “nice” 3D length (2² + 3² + 6² = 7²). So are (1, 2, 2) → 3, and (2, 6, 9) → 11, and (4, 4, 7) → 9.

Example 1.4Parallel vectors and unknowns

(a) The vectors

\mathbf{a}=\begin{pmatrix}2\\-3\\1\end{pmatrix}

and

\mathbf{b}=\begin{pmatrix}k\\12\\-4\end{pmatrix}

are parallel. Find k.
(b) The vectors

\mathbf{p}=\begin{pmatrix}k\\4\end{pmatrix}

and

\mathbf{q}=\begin{pmatrix}9\\k\end{pmatrix}

are parallel. Find the possible values of k.

🧠 Think it through

Parallel means b = λa. Use the position where both numbers are known to find λ.
In (b) the factor is the same for both rows: $\dfrac{k}{9} = \dfrac{4}{k}$ . Cross-multiply.

✍️ By hand

(a)

12=\lambda(-3)\Rightarrow\lambda=-4

. Check the last row:

1\times(-4)=-4

✓. So

k=2\times(-4)=-8

(b) Write

\mathbf{q}=\lambda\mathbf{p}

9=\lambda k

and

k=4\lambda

. Substitute:

9=4\lambda^2

, so

\lambda=\pm 1.5

and

k=4\lambda=\pm 6

. (Same result from

\dfrac{k}{9}=\dfrac{4}{k}\Rightarrow k^2=36

Check

k=6

\mathbf{p}=\begin{pmatrix}6\\4\end{pmatrix}

\mathbf{q}=\begin{pmatrix}9\\6\end{pmatrix}=1.5\mathbf{p}

✓. Check

k=-6

\mathbf{q}=\begin{pmatrix}9\\-6\end{pmatrix}=-1.5\begin{pmatrix}-6\\4\end{pmatrix}

✓.

✅ Answer (a) k = −8. (b) k = 6 or k = −6.

In (b), k = −6 gives vectors pointing in opposite directions. They are still parallel. Do not throw away a negative scalar!

Example 1.5Resultant of two velocities

A mokoro on the Okavango is paddled at (1.2, 0.9) m/s relative to the water, where i is east and j is north. The current flows at (0.8, −0.3) m/s.

(a) Find the resultant velocity of the mokoro.
(b) Find its speed, to 3 s.f.
(c) Find the angle between the resultant velocity and the east direction.
(d) How long, to the nearest second, does it take to travel 400 m in this direction?

🧠 Think it through

Resultant velocity = paddling velocity + current (add the vectors).
Speed is the magnitude of the velocity.
The angle above east: $\tan\theta=\dfrac{\text{north part}}{\text{east part}}$ .
$\text{Time} = \dfrac{\text{distance}}{\text{speed}}$ .

✍️ By hand

\mathbf{r}=\begin{pmatrix}1.2\\0.9\end{pmatrix}+\begin{pmatrix}0.8\\-0.3\end{pmatrix}=\begin{pmatrix}2.0\\0.6\end{pmatrix}

m/s

|\mathbf{r}|=\sqrt{2^2+0.6^2}=\sqrt{4.36}=2.088\ldots\approx 2.09

m/s

\theta=\tan^{-1}\!\left(\dfrac{0.6}{2}\right)=16.7^\circ

north of east

t=\dfrac{400}{2.0881}=191.6\ldots\approx 192

🔢 On your TI-Nspire CX II

Set the angle mode to Degree (doc ▾💻 File menu ▸ Settings). Type norm([2,0.6]), tan⁻¹(0.6/2) and 400/norm([2,0.6]).
📍 Where to find it on the TI-Nspire
norm( is in menu💻 click menu on keypad▸7: Matrix & Vector▸7: Norms▸1: Norm

✅ Answer (a) (2.0, 0.6) m/s. (b) 2.09 m/s. (c) 16.7° north of east. (d) 192 s.

Keep the unrounded speed (2.08806) in your calculator for part (d). Do not use the rounded 2.09.

Challenge questions

⭐ Challenge 1.6Finding an unknown from a lengthPaper 1 style

The vector

\mathbf{w}=\begin{pmatrix}k\\2k\\-2k\end{pmatrix}

has magnitude 12, where

k>0

(a) Find the value of k.
(b) Write down the unit vector in the direction of w.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

|w| = $\sqrt{k^2+4k^2+4k^2}=\sqrt{9k^2}=3k$ (since k > 0).
Set 3k = 12.
Divide w by 12 (its length).

✍️ By hand

3k=12\Rightarrow k=4

, so

\mathbf{w}=\begin{pmatrix}4\\8\\-8\end{pmatrix}

\hat{\mathbf{w}}=\dfrac{1}{12}\begin{pmatrix}4\\8\\-8\end{pmatrix}=\begin{pmatrix}\frac{1}{3}\\\frac{2}{3}\\-\frac{2}{3}\end{pmatrix}

✅ Answer (a) k = 4. (b) (

\dfrac{1}{3}

\dfrac{2}{3}

, −

\dfrac{2}{3}

Always say k > 0 was used. Without it, k = −4 also gives length 12.

⭐ Challenge 1.7Forces on a gate postPaper 2 style

Three forces (in newtons) act on a gate post. F₁ = (12, −5, 0), F₂ = (−4, 9, 6) and F₃ is unknown. The z-direction is vertical.

(a) Find the resultant of F₁ and F₂.
(b) The post is in equilibrium, so all three forces add up to the zero vector. Find F₃.
(c) Find the magnitude of F₃ to 3 s.f.
(d) Find the unit vector in the direction of F₁.
(e) A fourth force is m times F₂. Find m so that F₁ + mF₂ has no i-component, and write down this resultant.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

Resultant = add. Equilibrium means F₁ + F₂ + F₃ = 0, so F₃ = −(F₁ + F₂).
Magnitude = 3D Pythagoras.
|F₁| = √(144 + 25) = 13.
No i-component means the top number of F₁ + mF₂ is 0: 12 − 4m = 0.

✍️ By hand
(a) 
(b) 
(c)  N
(d) 
(e) ; 

🔢 On your TI-Nspire CX II

Check (b) and (c): -([12,-5,0]+[-4,9,6]) and norm([-8,-4,-6]).
📍 Where to find it on the TI-Nspire
norm( is in menu💻 click menu on keypad▸7: Matrix & Vector▸7: Norms▸1: Norm

✅ Answer (a) (8, 4, 6) N. (b) F₃ = (−8, −4, −6) N. (c) 10.8 N. (d) (

\dfrac{12}{13}

, −

\dfrac{5}{13}

, 0). (e) m = 3, resultant (0, 22, 18).

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL HL · Vectors 1 · Vectors: size, direction & arithmetic

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL

Vectors: size, direction & arithmetic

Worked examples

Challenge questions