Quaternions — definition q = (q₀, q₁, q₂, q₃), unit quaternion constraint: Worked examples

Before we start, one reminder of the tools we lean on (all built in the parent — we never use a symbol we did not earn there):

Recall The three tools we reuse

Encoding: $q = (cos \frac{θ}{2}, \overset{n}{^} sin \frac{θ}{2})$ — angle $θ$ about unit axis $\overset{n}{^}$ .
Product: $p \otimes q = (p_{0} q_{0} - p \cdot q, p_{0} q + q_{0} p + p \times q)$ .
Sandwich rotation of a vector $v$ : treat $v$ as a pure quaternion $v = (0, v)$ , then $v^{'} = q \otimes v \otimes q^{- 1}$ , and for a unit $q$ , $q^{- 1} = q^{*} = (q_{0}, - q_{v})$ .

The scenario matrix

Every situation this topic can hand you falls into one of these cells. Each worked example below is tagged with the cell it kills.

Cell	Scenario class	What could trip you up
A	Zero / degenerate: $θ = 0$	axis is undefined, must still give identity
B	Positive angle, clean axis (Q-I style)	half-angle, norm check
C	Negative angle / opposite spin	which sign flips, $q$ vs $q^{- 1}$
D	Limiting value: $θ = 18 0^{\circ}$	$cos 9 0^{\circ} = 0$ , scalar part vanishes
E	Non-commutativity: $p \otimes q$ vs $q \otimes p$	the $p \times q$ term
F	Double cover: $q$ vs $- q$	same rotation, different tuple
G	Actually rotate a vector (sandwich)	does the arrow really move?
H	Drift & renormalize (numerical)	project back to $S^{3}$
I	Real-world word problem (spacecraft)	translate words → axis + angle
J	Exam twist: recover $θ, \overset{n}{^}$ from a given $q$	inverting cos/sin, arccos, unit axis

Cell A — the degenerate "no rotation" case

Cell B — clean positive rotation (+ figure)

Figure — Quaternions — definition q = (q₀, q₁, q₂, q₃), unit quaternion constraint

Look at the figure: the teal axis is $z$ , and the quaternion parks the angle $4 5^{\circ}$ (half of the real $9 0^{\circ}$ , in plum) into the scalar slot. The full $9 0^{\circ}$ physical turn (burnt orange arc) is what the sandwich will actually produce on a vector — we do that in Cell G.

Cell C — negative angle / the inverse

Worked example C ·

$- 9 0^{\circ}$ about $z$ , and how it relates to $+ 9 0^{\circ}$ Same axis $\overset{n}{^} = (0, 0, 1)$ but $θ = - 9 0^{\circ}$ .

Forecast: which slot changes sign compared to Cell B?

Half-angle: $- 4 5^{\circ}$ .
Scalar: $cos (- 4 5^{\circ}) = cos 4 5^{\circ} = \frac{2}{2}$ (cosine is even — unchanged). Why this step? Reversing the spin does not flip the scalar; that is why $q_{0}$ tells you the amount of turn, not its direction.
Vector: $\overset{n}{^} sin (- 4 5^{\circ}) = (0, 0, - \frac{2}{2})$ (sine is odd — sign flips).
$q_{-} = (\frac{2}{2}, 0, 0, - \frac{2}{2})$ .

Verify — is this the inverse of Cell B? The unit inverse is the conjugate: $q_{B}^{*} = (\frac{2}{2}, 0, 0, - \frac{2}{2}) = q_{-}$ ✔. So rotating by $- θ$ $=$ conjugating — exactly what "undo the rotation" should mean. Also $q_{B} \otimes q_{-}$ should be identity: scalar part $= \frac{1}{2} - (0, 0, \frac{2}{2}) \cdot (0, 0, - \frac{2}{2}) = \frac{1}{2} + \frac{1}{2} = 1$ , vector part $= 0$ ✔.

Cell D — the limiting case $θ = 18 0^{\circ}$

Cell E — non-commutativity, the whole point

Cell F — double cover, $q$ and $- q$

Cell G — actually rotate a vector (sandwich) + figure

Worked example G · turn the point

$(1, 0, 0)$ by $9 0^{\circ}$ about $z$ Use $q = (\frac{2}{2}, 0, 0, \frac{2}{2})$ (Cell B). Vector $v = (1, 0, 0)$ , written as pure quaternion $v = (0, 1, 0, 0)$ .

Forecast: a $9 0^{\circ}$ turn about $z$ should send the $x$ -arrow onto the $y$ -arrow. Predict $v^{'}$ before computing.

First product $t = q \otimes v$ . Scalar $= q_{0} \cdot 0 - (0, 0, \frac{2}{2}) \cdot (1, 0, 0) = 0$ . Vector $= q_{0} (1, 0, 0) + 0 \cdot q_{v} + (0, 0, \frac{2}{2}) \times (1, 0, 0)$ .
- $q_{0} (1, 0, 0) = (\frac{2}{2}, 0, 0)$ .
- cross $= (0, 0, \frac{2}{2}) \times (1, 0, 0) = (0, \frac{2}{2}, 0)$ .
- So $t = (0, \frac{2}{2}, \frac{2}{2}, 0)$ . Why this step? We build the sandwich left-to-right; $t$ is a temporary, not the answer.
Second product $v^{'} = t \otimes q^{- 1}$ , with $q^{- 1} = q^{*} = (\frac{2}{2}, 0, 0, - \frac{2}{2})$ .
- Scalar $= t_{0} q_{0}^{- 1} - t \cdot q^{- 1} = 0 - (\frac{2}{2}, \frac{2}{2}, 0) \cdot (0, 0, - \frac{2}{2}) = 0$ .
- Vector $= t_{0} q^{- 1} + q_{0}^{- 1} t + t \times q^{- 1}$ $= 0 + \frac{2}{2} (\frac{2}{2}, \frac{2}{2}, 0) + (\frac{2}{2}, \frac{2}{2}, 0) \times (0, 0, - \frac{2}{2})$ .
  - middle term $= (\frac{1}{2}, \frac{1}{2}, 0)$ .
  - cross $= (\frac{2}{2}, \frac{2}{2}, 0) \times (0, 0, - \frac{2}{2}) = (- \frac{1}{2}, \frac{1}{2}, 0)$ .
  - sum $= (0, 1, 0)$ .
So $v^{'} = (0, 0, 1, 0)$ , i.e. the vector $(0, 1, 0)$ .

Verify: the $x$ -arrow $(1, 0, 0)$ became the $y$ -arrow $(0, 1, 0)$ — exactly a $+ 9 0^{\circ}$ turn about $z$ ✔. Length preserved: $0^{2} + 1^{2} + 0^{2} = 1$ ✔ (rotations never stretch). Scalar part stayed $0$ , so $v^{'}$ is still a pure quaternion ✔.

The figure shows the input arrow (burnt orange, along $x$ ) swinging up to the output arrow (teal, along $y$ ), with the $9 0^{\circ}$ arc in plum. The sandwich did what geometry demanded.

Cell H — drift and renormalize

Cell I — real-world word problem

Worked example I · spacecraft slew command

Mission control says: "Slew the telescope $12 0^{\circ}$ about the body axis $\overset{n}{^} = \frac{1}{3} (1, 1, 1)$ ." Build the command quaternion.

Forecast: the axis is already unit (check!) — so which numbers land in the vector slots?

Confirm unit axis: $(\frac{1}{3})^{2} \cdot 3 = \frac{1}{3} \cdot 3 = 1$ ✔. Good, encoding requires a unit axis.
Half-angle: $12 0^{\circ} /2 = 6 0^{\circ}$ .
Scalar: $cos 6 0^{\circ} = 0.5$ .
Vector: $\overset{n}{^} sin 6 0^{\circ} = \frac{1}{3} (1, 1, 1) \cdot \frac{3}{2}$ .
- Note $\frac{1}{3} \cdot \frac{3}{2} = \frac{1}{2}$ , so vector $= (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}) = (0.5, 0.5, 0.5)$ . Why this step? The $3$ 's cancel beautifully — a $12 0^{\circ}$ turn about the diagonal gives the clean quaternion $(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})$ , a famous one (it cyclically permutes the axes $x \to y \to z$ ).
Command $q = (0.5, 0.5, 0.5, 0.5)$ .

Verify: norm $^{2} = 0.25 \cdot 4 = 1$ ✔. Applying it twice ( $q \otimes q$ , a $24 0^{\circ}$ turn) should still be unit and its scalar $= 0.25 - (0.5, 0.5, 0.5) \cdot (0.5, 0.5, 0.5) = 0.25 - 0.75 = - 0.5$ — consistent with $cos 12 0^{\circ} = - 0.5$ ✔.

Cell J — exam twist: recover $θ$ and $\overset{n}{^}$ from $q$

Worked example J · read the rotation back out

Given $q = (0.5, 0.5, 0.5, 0.5)$ (Cell I's answer, but pretend you don't know its origin). Recover the angle $θ$ and axis $\overset{n}{^}$ .

Forecast: the scalar is $0.5 = cos (something)$ — what angle, and remember to double it.

Angle from scalar: $q_{0} = cos \frac{θ}{2} = 0.5 \Rightarrow \frac{θ}{2} = arccos (0.5) = 6 0^{\circ} \Rightarrow θ = 12 0^{\circ}$ . Why this step? $arccos$ undoes cosine — it answers "which half-angle has this cosine?" We then multiply by $2$ to recover the true turn.
Axis from vector: $\overset{n}{^} = \frac{q _{v}}{sin ( θ /2 )} = \frac{( 0.5 , 0.5 , 0.5 )}{sin 6 0 ^{\circ}} = \frac{( 0.5 , 0.5 , 0.5 )}{0.86603} = (0.57735, 0.57735, 0.57735)$ . Why this step? The vector slot is $\overset{n}{^} sin \frac{θ}{2}$ , so dividing by $sin \frac{θ}{2}$ strips off the length and leaves the pure unit axis $= \frac{1}{3} (1, 1, 1)$ .
Degenerate warning: if $q_{0} = \pm 1$ then $sin \frac{θ}{2} = 0$ and step 2 divides by zero — that's the identity (Cell A), where the axis is genuinely undefined. Detect it before dividing.

Verify: recovered $θ = 12 0^{\circ}$ ✔ and $\overset{n}{^} = (0.57735, \dots)$ with norm $^{2} = 3 \cdot 0.5773 5^{2} = 1.0000$ ✔ — a unit axis, exactly matching Cell I's input.

Active Recall

Recall Cell D: what is special about the scalar part of a

$18 0^{\circ}$ rotation? $cos 9 0^{\circ} = 0$ , so $q_{0} = 0$ — it's a pure quaternion.

Recall Cell J: how do you get

$θ$ back from a quaternion? $θ = 2 arccos (q_{0})$ , then $\overset{n}{^} = q_{v} / sin \frac{θ}{2}$ (watch the $q_{0} = \pm 1$ division-by-zero).

Recall Cell E: which term makes

$p \otimes q \neq = q \otimes p$ ? The cross product $p \times q$ , which flips sign on swap.

Related build-outs: Axis-Angle Representation, Quaternion Kinematics — dq/dt, Rotation Matrices SO(3), Euler Angles and Gimbal Lock, Spacecraft Attitude Determination, Complex Numbers as 2D Rotations.

Flashcards

Recover angle from a unit quaternion.

θ = 2 arccos (q_{0})

Recover the axis from a unit quaternion.

\overset{n}{^} = q_{v} / sin (θ /2)

(undefined when

q_{0} = \pm 1

Quaternion for

12 0^{\circ}

about

\frac{1}{3} (1, 1, 1)

(0.5, 0.5, 0.5, 0.5)

Why does

q_{0} = 0

mean a half-turn?

cos (θ /2) = 0 \Rightarrow θ /2 = 9 0^{\circ} \Rightarrow θ = 18 0^{\circ}