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

3.5.6 · D3 · Physics › Guidance, Navigation & Control (GNC) › Quaternions — definition q = (q₀, q₁, q₂, q₃), unit quaterni

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Encoding: $q = (cos \frac{θ}{2}, \overset{n}{^} sin \frac{θ}{2})$ — unit axis $\overset{n}{^}$ ke baare mein angle $θ$ .
Product: $p \otimes q = (p_{0} q_{0} - p \cdot q, p_{0} q + q_{0} p + p \times q)$ .
Vector $v$ ka Sandwich rotation: $v$ ko ek pure quaternion $v = (0, v)$ ki tarah treat karo, phir $v^{'} = q \otimes v \otimes q^{- 1}$ , aur unit $q$ ke liye, $q^{- 1} = q^{*} = (q_{0}, - q_{v})$ .

Scenario matrix

Is topic mein jo bhi situation aa sakti hai woh in cells mein se kisi ek mein aati hai. Neeche har worked example us cell ke saath tagged hai jise woh khatam karta hai.

Cell	Scenario class	Kahan trip ho sakte ho
A	Zero / degenerate: $θ = 0$	axis undefined hai, phir bhi identity deni hai
B	Positive angle, clean axis (Q-I style)	half-angle, norm check
C	Negative angle / opposite spin	kaun sa sign flip hota hai, $q$ vs $q^{- 1}$
D	Limiting value: $θ = 18 0^{\circ}$	$cos 9 0^{\circ} = 0$ , scalar part vanish ho jaata hai
E	Non-commutativity: $p \otimes q$ vs $q \otimes p$	$p \times q$ term
F	Double cover: $q$ vs $- q$	same rotation, alag tuple
G	Actually ek vector rotate karna (sandwich)	kya arrow sach mein move karta hai?
H	Drift & renormalize (numerical)	$S^{3}$ par project back karo
I	Real-world word problem (spacecraft)	words ko axis + angle mein translate karo
J	Exam twist: diye gaye $q$ se $θ, \overset{n}{^}$ recover karna	cos/sin ko invert karna, arccos, unit axis

Cell A — degenerate "no rotation" case

Cell B — clean positive rotation (+ figure)

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

Figure dekho: teal axis $z$ hai, aur quaternion angle $4 5^{\circ}$ (actual $9 0^{\circ}$ ka half, plum mein) ko scalar slot mein park karta hai. Poora $9 0^{\circ}$ physical turn (burnt orange arc) woh hai jo sandwich actually ek vector par produce karega — woh hum Cell G mein karte hain.

Cell C — negative angle / the inverse

Worked example C ·

$z$ ke baare mein $- 9 0^{\circ}$ , aur yeh $+ 9 0^{\circ}$ se kaise related hai Same axis $\overset{n}{^} = (0, 0, 1)$ lekin $θ = - 9 0^{\circ}$ .

Forecast: Cell B ke comparison mein kaun sa slot sign change karega?

Half-angle: $- 4 5^{\circ}$ .
Scalar: $cos (- 4 5^{\circ}) = cos 4 5^{\circ} = \frac{2}{2}$ (cosine even hai — unchanged). Yeh step kyun? Spin reverse karne se scalar flip nahi hoti; isliye $q_{0}$ tumhe turn ki amount batata hai, uski direction nahi.
Vector: $\overset{n}{^} sin (- 4 5^{\circ}) = (0, 0, - \frac{2}{2})$ (sine odd hai — sign flip hoti hai).
$q_{-} = (\frac{2}{2}, 0, 0, - \frac{2}{2})$ .

Verify — kya yeh Cell B ka inverse hai? Unit inverse conjugate hota hai: $q_{B}^{*} = (\frac{2}{2}, 0, 0, - \frac{2}{2}) = q_{-}$ ✔. Toh $- θ$ se rotate karna $=$ conjugate karna — exactly wahi jo "rotation undo karna" ka matlab hona chahiye. Aur $q_{B} \otimes q_{-}$ identity hona chahiye: 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 — limiting case $θ = 18 0^{\circ}$

Cell E — non-commutativity, asli baat

Worked example E ·

$p \otimes q \neq = q \otimes p$ Maano $p = (0, 1, 0, 0)$ (Cell-D style, $x$ ke baare mein $18 0^{\circ}$ ) aur $q = (0, 0, 1, 0)$ ( $y$ ke baare mein $18 0^{\circ}$ ).

Forecast: kya scalar parts match honge lekin vectors differ karenge? Ya kuch aur?

$p \otimes q$ compute karo. Scalars: $p_{0} q_{0} = 0$ ; $p \cdot q = (1, 0, 0) \cdot (0, 1, 0) = 0$ . Scalar part $= 0$ .
Vector part $= p_{0} q + q_{0} p + p \times q = 0 + 0 + (1, 0, 0) \times (0, 1, 0) = (0, 0, 1)$ . Yeh step kyun? Jab dono scalars zero hon, toh sirf cross product bachta hai — yeh woh term hai jo "order matters" ki information carry karta hai.
Toh $p \otimes q = (0, 0, 0, 1)$ .
Ab swap karo: $q \otimes p$ vector part $= (0, 1, 0) \times (1, 0, 0) = (0, 0, - 1)$ , giving $q \otimes p = (0, 0, 0, - 1)$ .

Verify: $p \otimes q = (0, 0, 0, 1) \neq = (0, 0, 0, - 1) = q \otimes p$ ✔. Woh $q_{3}$ mein ek sign se differ karte hain — exactly kyunki $p \times q = - (q \times p)$ . Rotations commute nahi karte, aur cross product isliye hai.

Cell F — double cover, $q$ aur $- q$

Cell G — actually ek vector rotate karna (sandwich) + figure

Worked example G · point

$(1, 0, 0)$ ko $z$ ke baare mein $9 0^{\circ}$ turn karo $q = (\frac{2}{2}, 0, 0, \frac{2}{2})$ use karo (Cell B). Vector $v = (1, 0, 0)$ , pure quaternion $v = (0, 1, 0, 0)$ ki tarah likha.

Forecast: $z$ ke baare mein $9 0^{\circ}$ turn $x$ -arrow ko $y$ -arrow par bhejni chahiye. Compute karne se pehle $v^{'}$ predict karo.

Pehla 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 $= (0, 0, \frac{2}{2}) \times (1, 0, 0) = (0, \frac{2}{2}, 0)$ .
- Toh $t = (0, \frac{2}{2}, \frac{2}{2}, 0)$ . Yeh step kyun? Hum sandwich left-to-right build karte hain; $t$ ek temporary hai, answer nahi.
Doosra product $v^{'} = t \otimes q^{- 1}$ , jisme $q^{- 1} = q^{*} = (\frac{2}{2}, 0, 0, - \frac{2}{2})$ .
- 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 $= (0, 1, 0)$ .
Toh $v^{'} = (0, 0, 1, 0)$ , yaani vector $(0, 1, 0)$ .

Verify: $x$ -arrow $(1, 0, 0)$ $y$ -arrow $(0, 1, 0)$ ban gaya — exactly $z$ ke baare mein $+ 9 0^{\circ}$ turn ✔. Length preserved: $0^{2} + 1^{2} + 0^{2} = 1$ ✔ (rotations kabhi stretch nahi karte). Scalar part $0$ raha, toh $v^{'}$ abhi bhi ek pure quaternion hai ✔.

Figure mein input arrow (burnt orange, $x$ ke along) output arrow (teal, $y$ ke along) tak swing karta hua dikhta hai, plum mein $9 0^{\circ}$ arc ke saath. Sandwich ne wahi kiya jo geometry ne demand kiya tha.

Cell H — drift aur renormalize

Cell I — real-world word problem

Worked example I · spacecraft slew command

Mission control kehta hai: "Telescope ko body axis $\overset{n}{^} = \frac{1}{3} (1, 1, 1)$ ke baare mein $12 0^{\circ}$ slew karo." Command quaternion banao.

Forecast: axis already unit hai (check karo!) — toh vector slots mein kaun se numbers aayenge?

Unit axis confirm karo: $(\frac{1}{3})^{2} \cdot 3 = \frac{1}{3} \cdot 3 = 1$ ✔. Acha, encoding ke liye ek unit axis chahiye.
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 karo $\frac{1}{3} \cdot \frac{3}{2} = \frac{1}{2}$ , toh vector $= (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}) = (0.5, 0.5, 0.5)$ . Yeh step kyun? $3$ beautifully cancel ho jaate hain — diagonal ke baare mein ek $12 0^{\circ}$ turn clean quaternion $(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})$ deta hai, ek famous wala (yeh axes ko cyclically permute karta hai $x \to y \to z$ ).
Command $q = (0.5, 0.5, 0.5, 0.5)$ .

Verify: norm $^{2} = 0.25 \cdot 4 = 1$ ✔. Ise do baar apply karna ( $q \otimes q$ , ek $24 0^{\circ}$ turn) phir bhi unit hona chahiye aur uska scalar $= 0.25 - (0.5, 0.5, 0.5) \cdot (0.5, 0.5, 0.5) = 0.25 - 0.75 = - 0.5$ — $cos 12 0^{\circ} = - 0.5$ ke saath consistent ✔.

Cell J — exam twist: $q$ se $θ$ aur $\overset{n}{^}$ recover karna

Worked example J · rotation wapas padhna

Diya gaya $q = (0.5, 0.5, 0.5, 0.5)$ (Cell I ka answer, lekin pretend karo ki tumhe uska origin nahi pata). Angle $θ$ aur axis $\overset{n}{^}$ recover karo.

Forecast: scalar $0.5 = cos (kuch)$ hai — kaun sa angle, aur yaad raho ise double karna hai.

Scalar se angle: $q_{0} = cos \frac{θ}{2} = 0.5 \Rightarrow \frac{θ}{2} = arccos (0.5) = 6 0^{\circ} \Rightarrow θ = 12 0^{\circ}$ . Yeh step kyun? $arccos$ cosine ko undo karta hai — yeh jawab deta hai "kis half-angle ka yeh cosine hai?" Phir hum true turn recover karne ke liye $2$ se multiply karte hain.
Vector se axis: $\overset{n}{^} = \frac{q _{v}}{sin ( θ /2 )} = \frac{( 0.5 , 0.5 , 0.5 )}{sin 6 0 ^{\circ}} = \frac{( 0.5 , 0.5 , 0.5 )}{0.86603} = (0.57735, 0.57735, 0.57735)$ . Yeh step kyun? Vector slot $\overset{n}{^} sin \frac{θ}{2}$ hai, toh $sin \frac{θ}{2}$ se divide karne par length nikal jaati hai aur pure unit axis $= \frac{1}{3} (1, 1, 1)$ bachti hai.
Degenerate warning: agar $q_{0} = \pm 1$ toh $sin \frac{θ}{2} = 0$ aur step 2 zero se divide karta hai — woh identity hai (Cell A), jahan axis genuinely undefined hai. Divide karne se pehle detect karo.

Verify: recover kiya $θ = 12 0^{\circ}$ ✔ aur $\overset{n}{^} = (0.57735, \dots)$ jiske saath norm $^{2} = 3 \cdot 0.5773 5^{2} = 1.0000$ ✔ — ek unit axis, exactly Cell I ke input se match karta hua.

Active Recall

Recall Cell D: ek

$18 0^{\circ}$ rotation ke scalar part mein kya khaas hota hai? $cos 9 0^{\circ} = 0$ , toh $q_{0} = 0$ — yeh ek pure quaternion hai.

Recall Cell J: quaternion se

$θ$ wapas kaise lete hain? $θ = 2 arccos (q_{0})$ , phir $\overset{n}{^} = q_{v} / sin \frac{θ}{2}$ ( $q_{0} = \pm 1$ division-by-zero dekho).

Recall Cell E: kaun sa term

$p \otimes q \neq = q \otimes p$ banata hai? Cross product $p \times q$ , jo swap par sign flip karta hai.

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}