Answer key · Unit 9

Vectors and transformations.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 3

Lesson 1 of 3 · About 30 minutes

Matrices that move the plane

A linear map moves the whole plane at once, keeping lines straight and parallel: where the unit vectors go, how the rotation matrix is built, where any point goes, and why a shear does not change area.

Open the lesson

The 7 steps and their answers

Step 1 of 7
Open this step
The unit vectors
Practice · Predict, then look
Question
Stretch is $1$ ⁠, Shear $0$ ⁠ and Turn $90^\circ$ ⁠. Where will the pink arrow, $\hat\imath = (1, 0)$ ⁠, land?

Correct answer
- Correct $(0, 1)$ ⁠
Setup, choices and feedback
The Scene at this step
Stretch
1fixed
Shear
0fixed
Turn
90 °fixed
Apply the transformation
Beforefixed
Choices
- Correct $(0, 1)$ ⁠
- Wrong $(0, -1)$ ⁠
  Misconception: Turns the wrong way: takes a positive angle as clockwise, so the sine terms of a rotation get the wrong sign.
  Feedback from the catalogue
  In mathematics a positive angle turns anticlockwise, as on the unit circle. A turn of $90^\circ$ ⁠ takes $\hat\imath = (1, 0)$ ⁠ to $(0, 1)$ ⁠, up, and $\hat\jmath = (0, 1)$ ⁠ to $(-1, 0)$ ⁠, left. That is why the rotation matrix $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ ⁠ has its minus sign at the top right.
- Wrong $(-1, 0)$ ⁠
  Feedback
  $(-1, 0)$ ⁠ is a half turn, $180^\circ$ ⁠. A quarter turn takes $\hat\imath$ ⁠ only as far as the vertical axis.
- Wrong $(1, 0)$ ⁠: it does not move
  Feedback
  A turn moves every vector except the zero vector. Only the centre stays put.
Once the prediction is locked in, the Scene changes
Apply the transformation
Afterfixed
The Scene then shows: 1 · Transform: M’s bottom-left entry, the y of M·î/nodes/board/outputs/matrixC
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$(0, 1)$ ⁠: a quarter turn anticlockwise takes $\hat\imath$ ⁠ up, and $\hat\jmath$ ⁠ left, to $(-1, 0)$ ⁠. The matrix’s columns are exactly these two places: $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ ⁠.

Worked solution
1. A positive angle turns anticlockwise.
2. $\hat\imath$ ⁠ points at $0^\circ$ ⁠; after a $90^\circ$ ⁠ turn it points at $90^\circ$ ⁠, straight up.
3. $(\cos 90^\circ, \sin 90^\circ) = (0, 1)$ ⁠.
If you chose $(0, -1)$ ⁠, you turned clockwise.
Step 2 of 7
Open this step
The rotation matrix
Practice · Calculate
Question
Turn is $30^\circ$ ⁠, with no stretch and no shear. What is the top-right entry of $M$ ⁠, the horizontal component of where $\hat\jmath$ ⁠ lands? Answer, then drag Apply the transformation all the way.

Correct answer
-0.5 ± 0.01
Setup, choices and feedback
The Scene at this step
Stretch
1fixed
Shear
0fixed
Turn
30 °fixed
Apply the transformation
Before
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Transform: M’s top-right entry, the x of M·ĵ/nodes/board/outputs/matrixB
Wrong answers it recognises
- 0.5 ± 0.01
  Misconception: Turns the wrong way: takes a positive angle as clockwise, so the sine terms of a rotation get the wrong sign.
  Feedback from the catalogue
  In mathematics a positive angle turns anticlockwise, as on the unit circle. A turn of $90^\circ$ ⁠ takes $\hat\imath = (1, 0)$ ⁠ to $(0, 1)$ ⁠, up, and $\hat\jmath = (0, 1)$ ⁠ to $(-1, 0)$ ⁠, left. That is why the rotation matrix $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ ⁠ has its minus sign at the top right.
- 0.866 ± 0.01
  Misconception: Swaps cos and sin in a rotation: takes the x of a turned vector from sin θ.
  Feedback from the catalogue
  A vector of length $1$ ⁠ at angle $\theta$ ⁠ is $(\cos\theta, \sin\theta)$ ⁠: the cosine is the horizontal component, the side adjacent to the angle, and the sine the vertical one. When $\hat\jmath$ ⁠ turns by $\theta$ ⁠ it lands on $(-\sin\theta, \cos\theta)$ ⁠: a quarter turn further on.
Feedback for a right answer
$-\sin 30^\circ = -0.5$ ⁠. $\hat\jmath$ ⁠ points up, at $90^\circ$ ⁠; after the turn it is at $120^\circ$ ⁠, leaning left: $(\cos 120^\circ, \sin 120^\circ) = (-0.5, 0.866)$ ⁠.

Hint
After the turn $\hat\jmath$ ⁠ leans left. What sign does its horizontal component have?

Worked solution
1. $\hat\jmath$ ⁠ points at $90^\circ$ ⁠. A $30^\circ$ ⁠ turn brings it to $120^\circ$ ⁠.
2. A vector of length $1$ ⁠ at $120^\circ$ ⁠ is $(\cos 120^\circ, \sin 120^\circ)$ ⁠.
3. $\cos 120^\circ = -\sin 30^\circ = -0.5$ ⁠.
If you got $0.5$ ⁠, you turned clockwise. If you got $0.866$ ⁠, you took the vertical component.
Step 3 of 7
Open this step
Where any point goes
Practice · Calculate
Question
Stretch is $2$ ⁠, Shear $0.5$ ⁠ and Turn $0^\circ$ ⁠, so $M = \begin{pmatrix} 2 & 0.5 \\ 0 & 1 \end{pmatrix}$ ⁠. Where does the unit square’s corner $(1, 1)$ ⁠ land? Give its $x$ ⁠. Answer, then drag Apply the transformation.

Correct answer
2.5 ± 0.01
Setup, choices and feedback
The Scene at this step
Stretch
2fixed
Shear
0.5fixed
Turn
0 °fixed
Apply the transformation
Before
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Transform: the x of M·(1, 1), the unit square’s far corner, a + b/nodes/board/outputs/cornerX
Wrong answers it recognises
- 2 ± 0.01
  Misconception: Takes M·(1, 1) from one column only, forgetting that (1, 1) = î + ĵ.
  Feedback from the catalogue
  $(1, 1) = \hat\imath + \hat\jmath$ ⁠, and a linear map keeps sums: $M(1, 1) = M\hat\imath + M\hat\jmath$ ⁠, the sum of the two columns. Every point $(x, y)$ ⁠ goes to $x$ ⁠ times the first column plus $y$ ⁠ times the second.
Feedback for a right answer
$M(1, 1) = (2 \cdot 1 + 0.5 \cdot 1, \; 0 \cdot 1 + 1 \cdot 1) = (2.5, 1)$ ⁠: the first column plus the second.

Worked solution
1. $(1, 1) = \hat\imath + \hat\jmath$ ⁠.
2. $M\hat\imath$ ⁠ is the first column, $(2, 0)$ ⁠, and $M\hat\jmath$ ⁠ the second, $(0.5, 1)$ ⁠.
3. $M(1, 1) = (2, 0) + (0.5, 1) = (2.5, 1)$ ⁠.
If you got $2$ ⁠, you took only $M\hat\imath$ ⁠.
Step 4 of 7
Open this step
Shear
Practice · Predict, then look
Question
Shear is $1$ ⁠, with no stretch and no turn: every row slides right by its own height. The unit square becomes a parallelogram. What will its area be?

Correct answer
- Correct $1$ ⁠, the same as the square
Setup, choices and feedback
The Scene at this step
Stretch
1fixed
Shear
1fixed
Turn
0 °fixed
Apply the transformation
Beforefixed
Choices
- Correct $1$ ⁠, the same as the square
- Wrong $\sqrt2 \approx 1.41$ ⁠, because the side grew longer
  Misconception: Thinks a shear changes area because it makes the shape longer.
  Feedback from the catalogue
  A shear slides every row sideways, more the further it is from the axis, like a deck of cards pushed into a slant. The parallelogram keeps the square’s base and height, and so its area: $1 \cdot 1 = 1$ ⁠. The sides do grow longer, but the shape also grows thinner.
- Wrong $2$ ⁠
  Feedback
  The parallelogram spans $2$ ⁠ units across, but only from tip to tip. Its base is $1$ ⁠ and its height $1$ ⁠.
- Wrong $0.5$ ⁠
  Feedback
  A shear takes nothing away from the square: it only slides its rows.
Once the prediction is locked in, the Scene changes
Apply the transformation
Afterfixed
The Scene then shows: 1 · Transform: det M = ad − bc, the factor every area is scaled by (negative when flipped)/nodes/board/outputs/determinant
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$1$ ⁠. The base stays $1$ ⁠ and the height stays $1$ ⁠, so the area does not change. The determinant on the card is $1$ ⁠.

Worked solution
1. $M = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ ⁠: $\hat\imath$ ⁠ stays at $(1, 0)$ ⁠, and $\hat\jmath$ ⁠ goes to $(1, 1)$ ⁠.
2. The parallelogram sits on the same base, $1$ ⁠, with its top at the same height, $1$ ⁠.
3. Area = base × height = $1$ ⁠. Also: $\det M = 1 \cdot 1 - 1 \cdot 0 = 1$ ⁠.
Step 5 of 7
Open this step
Build a matrix
Practice · Reach the target
Question
Build a matrix that sends $\hat\imath$ ⁠ to $(0, 2)$ ⁠: straight up, length $2$ ⁠. Shear is locked; change Stretch and Turn, then press Check.

Goal
- 0 ± 0.01
  At the start of the step: 1
- 2 ± 0.01
  At the start of the step: 0
Setup, choices and feedback
The Scene at this step
Stretch
1
Shear
0fixed
Turn
0 °
Apply the transformation
Afterfixed
Read from the Scene at
Transform: M’s top-left entry, the x of M·î/nodes/board/outputs/matrixA
Read from the Scene at
Transform: M’s bottom-left entry, the y of M·î/nodes/board/outputs/matrixC
One setting that reaches it
Stretch
-2
Turn
-90 °
Feedback for a right answer
Stretch $2$ ⁠ lengthens $\hat\imath$ ⁠ to $(2, 0)$ ⁠, and Turn $90^\circ$ ⁠ stands it up at $(0, 2)$ ⁠. The shear changes only $\hat\jmath$ ⁠, so it would not have changed the answer.

Feedback for a wrong answer
Not yet. The first column of $M$ ⁠ is $M\hat\imath$ ⁠: it must be $(0, 2)$ ⁠.

Worked solution
1. $M\hat\imath$ ⁠ is set by the stretch and the turn alone: $\hat\imath$ ⁠ is first stretched to $(s, 0)$ ⁠, then turned.
2. Its length must be $2$ ⁠, so $s = 2$ ⁠.
3. It must point up, so the turn is $90^\circ$ ⁠.
$s = -2$ ⁠ with a turn of $-90^\circ$ ⁠ works too: $(-2, 0)$ ⁠ turned clockwise lands on $(0, 2)$ ⁠.
Step 6 of 7
Open this step
$\hat\jmath$ after a turn
Practice · Calculate
Question
Stretch $1$ ⁠, Shear $0$ ⁠ and Turn $60^\circ$ ⁠. What is the bottom-right entry of $M$ ⁠, the vertical component of where $\hat\jmath = (0, 1)$ ⁠ lands?

Correct answer
0.5 ± 0.01
Setup, choices and feedback
The Scene at this step
Stretch
1fixed
Shear
0fixed
Turn
60 °fixed
Apply the transformation
Before
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Transform: M’s bottom-right entry, the y of M·ĵ/nodes/board/outputs/matrixD
Wrong answers it recognises
- 0.866 ± 0.01
  Misconception: Swaps cos and sin in a rotation: takes the x of a turned vector from sin θ.
  Feedback from the catalogue
  A vector of length $1$ ⁠ at angle $\theta$ ⁠ is $(\cos\theta, \sin\theta)$ ⁠: the cosine is the horizontal component, the side adjacent to the angle, and the sine the vertical one. When $\hat\jmath$ ⁠ turns by $\theta$ ⁠ it lands on $(-\sin\theta, \cos\theta)$ ⁠: a quarter turn further on.
Feedback for a right answer
$0.5$ ⁠: $\cos 60^\circ$ ⁠. $\hat\jmath$ ⁠ leans left, and its height falls to a half.

Hint
A turn by $\theta$ ⁠ sends $\hat\jmath$ ⁠ to $(-\sin\theta, \cos\theta)$ ⁠.

Worked solution
1. The second column of $M$ ⁠ is where $\hat\jmath$ ⁠ lands: $(-\sin 60^\circ, \cos 60^\circ)$ ⁠.
2. Its vertical component: $\cos 60^\circ = 0.5$ ⁠.
3. The whole matrix: $M = \begin{pmatrix} 0.5 & -0.866 \\ 0.866 & 0.5 \end{pmatrix}$ ⁠.
If you answered $0.866$ ⁠, you swapped the sine and the cosine. After a small turn $\hat\jmath$ ⁠ still points almost straight up, so its vertical component is near $1$ ⁠ for a small angle: that is the cosine.
Step 7 of 7
Open this step
The diagonal
Practice · Calculate
Question
Stretch $1$ ⁠, Shear $0$ ⁠ and Turn $45^\circ$ ⁠. The corner $(1, 1)$ ⁠ of the unit square turns with the plane. How high, $y$ ⁠, does it land?

Correct answer
1.414 ± 0.01
Setup, choices and feedback
The Scene at this step
Stretch
1fixed
Shear
0fixed
Turn
45 °fixed
Apply the transformation
Before
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Transform: the y of M·(1, 1), c + d/nodes/board/outputs/cornerY
Wrong answers it recognises
- 1 ± 0.01
  Misconception: Takes M·(1, 1) from one column only, forgetting that (1, 1) = î + ĵ.
  Feedback from the catalogue
  $(1, 1) = \hat\imath + \hat\jmath$ ⁠, and a linear map keeps sums: $M(1, 1) = M\hat\imath + M\hat\jmath$ ⁠, the sum of the two columns. Every point $(x, y)$ ⁠ goes to $x$ ⁠ times the first column plus $y$ ⁠ times the second.
Feedback for a right answer
$\sqrt{2} \approx 1.414$ ⁠: the diagonal stands upright on the vertical axis.

Hint
$M \cdot (1, 1)$ ⁠ is the sum of $M$ ⁠’s two columns: where $\hat\imath$ ⁠ lands plus where $\hat\jmath$ ⁠ lands.

Worked solution
1. $\hat\imath$ ⁠ lands at $(\cos 45^\circ, \sin 45^\circ) \approx (0.707, 0.707)$ ⁠.
2. $\hat\jmath$ ⁠ lands at $(-\sin 45^\circ, \cos 45^\circ) \approx (-0.707, 0.707)$ ⁠.
3. The sum: $(0, 1.414)$ ⁠. The corner lands exactly on the vertical axis, at height $\sqrt{2}$ ⁠.
That makes sense: the square’s diagonal is $\sqrt{2}$ ⁠ long and tilted at $45^\circ$ ⁠, and a $45^\circ$ ⁠ turn stands it upright. If you answered $1$ ⁠, you added only one column.

Lesson 2 of 3 · About 20 minutes

The determinant

Every linear map scales every area by the same factor, the determinant ad − bc: why turns and shears keep area, what a negative determinant says, and what happens when it is zero.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
The area factor
Practice · Calculate
Question
$M$ ⁠ is built from a stretch of $1.5$ ⁠, a shear of $0.5$ ⁠ and a turn of $30^\circ$ ⁠. By what factor will every area in the plane grow? Answer, then drag Apply the transformation and read $\det M$ ⁠ on the card.

Correct answer
1.5 ± 0.01
Setup, choices and feedback
The Scene at this step
Stretch
1.5fixed
Shear
0.5fixed
Turn
30 °fixed
Apply the transformation
Before
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Transform: det M = ad − bc, the factor every area is scaled by (negative when flipped)/nodes/board/outputs/determinant
Wrong answers it recognises
- 1.677 ± 0.01
  Misconception: Takes the area scale as the product of the lengths of the two image vectors, ignoring the angle between them.
  Feedback from the catalogue
  A parallelogram’s area is base times height, not times the other side. The lengths of $M\hat\imath$ ⁠ and $M\hat\jmath$ ⁠ give the area only when they are perpendicular. Otherwise the area is smaller, and it is $|\det M| = |ad - bc|$ ⁠.
Feedback for a right answer
$\det M = 1.5$ ⁠. The turn and the shear change no area, so only the stretch is left.

Worked solution
1. $M$ ⁠ is three steps: stretch, shear, turn. The determinant of a product is the product of the determinants.
2. Stretching $\hat\imath$ ⁠ by $1.5$ ⁠: $\det = 1.5$ ⁠. The shear: $\det = 1$ ⁠. The turn: $\det = \cos^2 30^\circ + \sin^2 30^\circ = 1$ ⁠.
3. $\det M = 1.5 \cdot 1 \cdot 1 = 1.5$ ⁠.
If you got about $1.68$ ⁠, you multiplied the two arrows’ lengths: that is right only when they are perpendicular.
Step 2 of 5
Open this step
Every shape
Practice · Calculate
Question
Before the transformation the F’s area is about $1.055$ ⁠. What will its area be after the same $M$ ⁠?

Correct answer
1.583 ± 1%
Setup, choices and feedback
The Scene at this step
Stretch
1.5fixed
Shear
0.5fixed
Turn
30 °fixed
Apply the transformation
After
Accepted within
± 1%
Computed from the Scene at this step’s setting
Transform: the F’s area after M, |det M| times its area before/nodes/board/outputs/shapeArea
Feedback for a right answer
$1.5 \cdot 1.055 \approx 1.58$ ⁠. The determinant scales the area of every shape, not just the unit square: the F too can be filled with tiny squares, and each of them grows $1.5$ ⁠ times.

Worked solution
1. A linear map scales every area by the same factor, $|\det M|$ ⁠.
2. $\det M = 1.5$ ⁠.
3. The new area: $1.5 \cdot 1.055 \approx 1.58$ ⁠.
Step 3 of 5
Open this step
A negative determinant
Practice · Predict, then look
Question
Stretch is $-1$ ⁠, with no shear and no turn: $\hat\imath$ ⁠ goes to $(-1, 0)$ ⁠ and $\hat\jmath$ ⁠ stays put. What will happen to the F?

Correct answer
- Correct It will flip into its mirror image, showing its back
Setup, choices and feedback
The Scene at this step
Stretch
-1fixed
Shear
0fixed
Turn
0 °fixed
Apply the transformation
Beforefixed
Choices
- Correct It will flip into its mirror image, showing its back
- Wrong It will turn $180^\circ$ ⁠, onto its head
  Misconception: Confuses a reflection with a half turn.
  Feedback from the catalogue
  A half turn stands the F on its head, but it is still the same F, and another turn brings it back. A reflection makes it its own mirror image, a backwards F, and no turn will undo that. So a turn has determinant $1$ ⁠, and a reflection $-1$ ⁠.
- Wrong It will be squashed onto a line
  Feedback
  Squashing onto a line happens at a stretch of $0$ ⁠. At $-1$ ⁠ the lengths are kept; only the direction turns round.
- Wrong Nothing: no length changes
  Feedback
  The lengths are indeed kept, but left and right swap: $(x, y)$ ⁠ goes to $(-x, y)$ ⁠.
Once the prediction is locked in, the Scene changes
Apply the transformation
Afterfixed
The Scene then shows: -1 · Transform: det M = ad − bc, the factor every area is scaled by (negative when flipped)/nodes/board/outputs/determinant
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The F flips into its mirror image, like a page turned over, and $\det M = -1$ ⁠. The minus sign says the map has reversed the plane’s orientation: what went anticlockwise now goes clockwise.

Worked solution
1. $M = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$ ⁠ sends $(x, y)$ ⁠ to $(-x, y)$ ⁠: a reflection in the vertical axis.
2. $\det M = -1 \cdot 1 - 0 = -1$ ⁠: the area is kept and the orientation reversed.
3. A half turn sends $(x, y)$ ⁠ to $(-x, -y)$ ⁠, with $\det = 1$ ⁠: a different map.
Step 4 of 5
Open this step
A zero determinant
Practice · Reach the target
Question
Squash the whole plane onto one line: change the controls so that $\det M = 0$ ⁠, then press Check.

Goal
- 0 ± 0.001
  At the start of the step: 1.5
Setup, choices and feedback
The Scene at this step
Stretch
1.5
Shear
0.5
Turn
30 °
Apply the transformation
Afterfixed
Read from the Scene at
Transform: det M = ad − bc, the factor every area is scaled by (negative when flipped)/nodes/board/outputs/determinant
One setting that reaches it
Stretch
0
Shear
-1.5
Turn
-180 °
Feedback for a right answer
Stretch $0$ ⁠ crushes $\hat\imath$ ⁠ to zero, and the whole plane falls onto the line of $M\hat\jmath$ ⁠. Every shape now has area $0$ ⁠, and the map cannot be undone: many points land on the same one.

Feedback for a wrong answer
Not yet. The shear and the turn change no area. Which control does?

Worked solution
1. Here $\det M$ ⁠ equals the stretch, because the shear and the turn keep area.
2. $\det M = 0$ ⁠ when the stretch is $0$ ⁠.
3. Set Stretch to $0$ ⁠.
Step 5 of 5
Open this step
What changes area
Practice · Choose
Question
Which of the controls change the F’s area? Select all that do. Try each on its own.

Correct answer
- Correct Stretch
Setup, choices and feedback
The Scene at this step
Stretch
1
Shear
0
Turn
0 °
Apply the transformation
After
Choices
Shown in a random order.
- Correct Stretch
- Wrong Shear
  Misconception: Thinks a shear changes area because it makes the shape longer.
  Feedback from the catalogue
  A shear slides every row sideways, more the further it is from the axis, like a deck of cards pushed into a slant. The parallelogram keeps the square’s base and height, and so its area: $1 \cdot 1 = 1$ ⁠. The sides do grow longer, but the shape also grows thinner.
- Wrong Turn
  Feedback
  A turn moves the F as a rigid body: no length and no angle changes, so neither does the area.
Feedback for a right answer
Only Stretch: $\det M$ ⁠ equals it. Shears and turns keep area, so the determinant on the card does not move when you change them.

Worked solution
1. A turn has $\det = \cos^2\theta + \sin^2\theta = 1$ ⁠.
2. A shear $\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$ ⁠ has $\det = 1$ ⁠.
3. A stretch $\begin{pmatrix} s & 0 \\ 0 & 1 \end{pmatrix}$ ⁠ has $\det = s$ ⁠.
4. So only the stretch changes area.

Lesson 3 of 3 · About 30 minutes

Symmetry

A symmetry is a move that leaves a shape as it was: turns and reflections, how many symmetries a pattern has, how two reflections make a turn, and why a floor can repeat only turns of order 1, 2, 3, 4 or 6.

Open the lesson

The 8 steps and their answers

Step 1 of 8
Open this step
Counting symmetries
Practice · Calculate
Question
The pattern is a snowflake: $6$ ⁠ turns and $6$ ⁠ mirrors. How many moves leave it exactly as it was, doing nothing included?

Correct answer
12 ± 0
Setup, choices and feedback
The Scene at this step
Rotation order
6 foldfixed
Mirrors
onfixed
Find the symmetry
Patternfixed
Accepted within
± 0
Computed from the Scene at this step’s setting
How many moves leave the plate unchanged, doing nothing included: n turns, and n mirrors more with mirrors on/nodes/plate/outputs/symmetries
Wrong answers it recognises
- 6 ± 0
  Misconception: Counts only the turns of a pattern and forgets its reflections.
  Feedback from the catalogue
  A reflection in a mirror is a symmetry too: it leaves the pattern as it was. A pattern with $n$ ⁠ turns and $n$ ⁠ mirrors, $D_n$ ⁠, has $2n$ ⁠ symmetries: $n$ ⁠ turns (including the turn by zero, doing nothing) and $n$ ⁠ reflections.
Feedback for a right answer
$12$ ⁠: the turns by $0^\circ, 60^\circ, \dots, 300^\circ$ ⁠ and six reflections. The group is called $D_6$ ⁠.

Worked solution
1. The smallest turn is $\frac{360^\circ}{6} = 60^\circ$ ⁠, so there are $6$ ⁠ turns: $0^\circ$ ⁠ up to $300^\circ$ ⁠.
2. There are $6$ ⁠ mirrors, and reflecting in each is a symmetry.
3. Together: $6 + 6 = 12$ ⁠.
If you answered $6$ ⁠, you forgot the reflections.
Step 2 of 8
Open this step
Same size, different shapes
Practice · Reach the target
Question
Change Rotation order and Mirrors so that the pattern has exactly $8$ ⁠ symmetries, then press Check.

Goal
- 8 ± 0.1
  At the start of the step: 12
Setup, choices and feedback
The Scene at this step
Rotation order
6 fold
Mirrors
on
Find the symmetry
Patternfixed
Read from the Scene at
How many moves leave the plate unchanged, doing nothing included: n turns, and n mirrors more with mirrors on/nodes/plate/outputs/symmetries
One setting that reaches it
Rotation order
8 fold
Mirrors
off
Feedback for a right answer
There are two ways: a square, $D_4$ ⁠, with $4$ ⁠ turns and $4$ ⁠ mirrors, or $C_8$ ⁠, with $8$ ⁠ turns and no mirrors. Two different groups of the same size.

Feedback for a wrong answer
Not yet. Without mirrors there are $n$ ⁠ symmetries, and with mirrors $2n$ ⁠.

Worked solution
1. With mirrors, $D_n$ ⁠ has $2n$ ⁠ symmetries: $2n = 8$ ⁠ gives $n = 4$ ⁠.
2. Without mirrors, $C_n$ ⁠ has $n$ ⁠: $n = 8$ ⁠.
3. Set $4$ ⁠ with Mirrors on, or $8$ ⁠ with them off.
Step 3 of 8
Open this step
The angle between mirrors
Practice · Calculate
Question
The starfish has $5$ ⁠ turns and $5$ ⁠ mirrors, $D_5$ ⁠. What is the angle between two neighbouring mirrors?

Correct answer
36 ° ± 0.5 °
Setup, choices and feedback
The Scene at this step
Rotation order
5 foldfixed
Mirrors
onfixed
Find the symmetry
Patternfixed
Accepted within
± 0.5 °
Computed from the Scene at this step’s setting
The angle between neighbouring mirror lines, 180° / n; NaN with fewer than two mirrors/nodes/plate/outputs/mirrorAngle
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 72 ° ± 0.5 °
  Misconception: Expects the mirrors of Dₙ to be 360°/n apart, as the turns are, instead of 180°/n.
  Feedback from the catalogue
  Each mirror is a whole line through the centre, so it crosses the circle twice: $n$ ⁠ lines cut it into $2n$ ⁠ wedges. The angle between neighbouring mirrors is $\frac{180^\circ}{n}$ ⁠, half the smallest turn.
Feedback for a right answer
$\frac{180^\circ}{5} = 36^\circ$ ⁠. The five mirror lines pass through the centre, so they cut the circle into ten wedges.

Worked solution
1. Each mirror is a whole line through the centre, meeting the plate’s rim twice.
2. $5$ ⁠ lines cut $360^\circ$ ⁠ into $10$ ⁠ equal wedges.
3. $\frac{360^\circ}{10} = 36^\circ$ ⁠, that is $\frac{180^\circ}{n}$ ⁠.
If you answered $72^\circ$ ⁠, that is the smallest turn, $\frac{360^\circ}{5}$ ⁠.
Step 4 of 8
Open this step
Two reflections
Practice · Choose
Question
Reflect the motif in one mirror of $D_5$ ⁠, then in the neighbouring mirror, $36^\circ$ ⁠ away. Which single move does the same?

Correct answer
- Correct A turn of $72^\circ$ ⁠
Setup, choices and feedback
The Scene at this step
Rotation order
5 foldfixed
Mirrors
onfixed
Find the symmetry
Pattern
Choices
Shown in a random order.
- Correct A turn of $72^\circ$ ⁠
- Wrong A turn of $36^\circ$ ⁠
  Feedback
  Two reflections turn by twice the angle between the mirrors: $2 \cdot 36^\circ = 72^\circ$ ⁠.
- Wrong A reflection in a third mirror
  Feedback
  Each reflection turns the motif into its mirror image. Two of them bring it back the right way round, so the result is a turn.
- Wrong Doing nothing
  Feedback
  That is so only when both reflections are in the same mirror.
Feedback for a right answer
A turn of $72^\circ$ ⁠: two reflections in mirrors at an angle $\alpha$ ⁠ make a turn of $2\alpha$ ⁠. So the turns of $D_n$ ⁠ already follow from its mirrors.

Worked solution
1. Each reflection reverses the plane’s orientation ( $\det = -1$ ⁠), and two restore it ( $(-1)(-1) = 1$ ⁠): the result is a turn.
2. A point on the first mirror is carried by the second reflection to $2 \cdot 36^\circ$ ⁠ from where it was.
3. So it is a turn of $72^\circ$ ⁠, one of the turns of $D_5$ ⁠.
Step 5 of 8
Open this step
Tilings
Practice · Choose
Question
Which of these rotation orders can a floor of identical tiles, covering it completely, repeat? Select all that can. Answer, then move Find the symmetry to Tiling and check each one.

Correct answer
- Correct $3$ ⁠
- Correct $4$ ⁠
- Correct $6$ ⁠
Setup, choices and feedback
The Scene at this step
Rotation order
5 fold
Mirrors
on
Find the symmetry
Pattern
Choices
Shown in a random order. More than one choice is correct: every one must be selected.
- Correct $3$ ⁠
- Correct $4$ ⁠
- Wrong $5$ ⁠
  Misconception: Thinks a floor can repeat a turn of any order, five- and eight-fold included.
  Feedback from the catalogue
  A floor repeats itself by sliding, so its symmetries must fit a lattice. A turn of $\frac{360^\circ}{n}$ ⁠ fits a lattice only when $2\cos\frac{360^\circ}{n}$ ⁠ is a whole number: $n = 1, 2, 3, 4, 6$ ⁠ only. Pentagons leave gaps, and octagons overlap.
- Correct $6$ ⁠
- Wrong $8$ ⁠
  Misconception: Thinks a floor can repeat a turn of any order, five- and eight-fold included.
  Feedback from the catalogue
  A floor repeats itself by sliding, so its symmetries must fit a lattice. A turn of $\frac{360^\circ}{n}$ ⁠ fits a lattice only when $2\cos\frac{360^\circ}{n}$ ⁠ is a whole number: $n = 1, 2, 3, 4, 6$ ⁠ only. Pentagons leave gaps, and octagons overlap.
Feedback for a right answer
$3$ ⁠, $4$ ⁠ and $6$ ⁠: triangles, squares and hexagons. A floor can repeat turns of order $1$ ⁠, $2$ ⁠, $3$ ⁠, $4$ ⁠ or $6$ ⁠ only. That is why no crystal has five-fold symmetry, and snowflakes are always six-sided.

Worked solution
1. Round a corner of a floor of regular tiles, the angles must make up $360^\circ$ ⁠.
2. A triangle has $60^\circ$ ⁠ (six at a corner), a square $90^\circ$ ⁠ (four), a hexagon $120^\circ$ ⁠ (three).
3. A pentagon has $108^\circ$ ⁠: three make $324^\circ$ ⁠, leaving a gap. An octagon has $135^\circ$ ⁠: three make $405^\circ$ ⁠, an overlap.
4. In general a turn of $\frac{360^\circ}{n}$ ⁠ fits a lattice only when $2\cos\frac{360^\circ}{n}$ ⁠ is a whole number: $n = 1, 2, 3, 4, 6$ ⁠.
Step 6 of 8
Open this step
Eight times round
Practice · Calculate
Question
Rotation order is locked at $8$ ⁠: the pattern comes back to itself eight times in a full turn. What is the smallest turn that brings it back exactly as it was?

Correct answer
45 ° ± 0.1 °
Setup, choices and feedback
The Scene at this step
Rotation order
8 foldfixed
Mirrors
on
Find the symmetry
Pattern
Accepted within
± 0.1 °
Computed from the Scene at this step’s setting
The smallest turn that brings the pattern back, 360° / n/nodes/plate/outputs/turnAngle
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 22.5 ° ± 0.1 °
  Misconception: Expects the mirrors of Dₙ to be 360°/n apart, as the turns are, instead of 180°/n.
  Feedback from the catalogue
  Each mirror is a whole line through the centre, so it crosses the circle twice: $n$ ⁠ lines cut it into $2n$ ⁠ wedges. The angle between neighbouring mirrors is $\frac{180^\circ}{n}$ ⁠, half the smallest turn.
Feedback for a right answer
$45^\circ$ ⁠: an eighth of a turn.

Hint
Eight equal turns together make a full turn.

Worked solution
1. A full turn is $360^\circ$ ⁠.
2. The pattern comes back $8$ ⁠ times, so the smallest turn is $\dfrac{360}{8} = 45^\circ$ ⁠.
3. $90^\circ$ ⁠, $135^\circ$ ⁠ and so on bring it back too: every multiple of $45^\circ$ ⁠.
If you answered $22.5^\circ$ ⁠, you gave the angle between two neighbouring mirrors, $\dfrac{180}{8}$ ⁠. The turn is twice that.
Step 7 of 8
Open this step
A pinwheel
Practice · Reach the target
Question
Build a pinwheel: a pattern that comes back to itself after a quarter turn, but has no mirror line at all. Change the Rotation order and Mirrors, then press Check.

Goal
- 90 ± 0.1
  At the start of the step: 60
- 0 ± 0
  At the start of the step: 6
Setup, choices and feedback
The Scene at this step
Rotation order
6 fold
Mirrors
on
Find the symmetry
Pattern
Read from the Scene at
The smallest turn that brings the pattern back, 360° / n/nodes/plate/outputs/turnAngle
Read from the Scene at
Mirror lines through the centre: n with mirrors on, 0 without/nodes/plate/outputs/mirrorLines
One setting that reaches it
Rotation order
4 fold
Mirrors
off
Feedback for a right answer
Order $4$ ⁠ with no mirrors: four turns, and no reflection.

Hint
How many times does a pattern that comes back after a quarter turn come back in a full turn?

Worked solution
1. A quarter turn is $90^\circ = \dfrac{360^\circ}{4}$ ⁠, so the Rotation order is $4$ ⁠.
2. To have no mirror line, turn Mirrors off.
3. $4$ ⁠ symmetries are left: turns of $0^\circ$ ⁠, $90^\circ$ ⁠, $180^\circ$ ⁠ and $270^\circ$ ⁠.
A pinwheel has a direction: it spins one way. Its mirror image spins the other way, so it has no mirror symmetry.
Step 8 of 8
Open this step
The largest that tiles
Practice · Reach the target
Question
At $5$ ⁠ no floor can be tiled. Find the largest Rotation order that identical tiles covering a whole floor can repeat, and press Check.

Goal
- 1 ± 0
  At the start of the step: 0
- 60 ± 0.1
  At the start of the step: 72
Setup, choices and feedback
The Scene at this step
Rotation order
5 fold
Mirrors
on
Find the symmetry
Tiling
Read from the Scene at
1 when a floor of tiles can repeat this many-fold turn (1, 2, 3, 4 or 6), 0 when it cannot/nodes/plate/outputs/tiles
Read from the Scene at
The smallest turn that brings the pattern back, 360° / n/nodes/plate/outputs/turnAngle
One setting that reaches it
Rotation order
6 fold
Mirrors
off
Feedback for a right answer
$6$ ⁠, as in a honeycomb: no order beyond it tiles a floor.

Hint
Try orders above $5$ ⁠ too, and drag Find the symmetry to Tiling to see whether they fit.

Worked solution
1. A tiling repeats itself by shifts, and two neighbouring centres of turning must fit the lattice of shifts.
2. Only $1$ ⁠, $2$ ⁠, $3$ ⁠, $4$ ⁠ and $6$ ⁠ fit: the orders for which $2\cos\dfrac{360^\circ}{n}$ ⁠ is a whole number.
3. $7$ ⁠ and $8$ ⁠ do not, so the largest is $6$ ⁠: a honeycomb.
This is the crystallographic restriction: crystals, which are tilings in three dimensions, have no five-, seven- or eightfold symmetry.

Across the unit · 9 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Matrices that move the plane

The unit vectors

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

The rotation matrix

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Where any point goes

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Worked solution

Shear

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Build a matrix

Question

Goal

The Scene at this step

One setting that reaches it

Worked solution

$\hat\jmath$ after a turn

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The diagonal

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The determinant

The area factor

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Worked solution

Every shape

Question

Correct answer

The Scene at this step

Worked solution

A negative determinant

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

A zero determinant

Question

Goal

The Scene at this step

One setting that reaches it

Worked solution

What changes area

Question

Correct answer

The Scene at this step

Choices

Worked solution

Symmetry