Quaternion product — Hamilton product: Worked examples

3.5.7 · D3 · Physics › Guidance, Navigation & Control (GNC) › Quaternion product — Hamilton product

Intuition Yeh page kya hai

Parent note ne rule banaya tha. Yahan hum usse drill karte hain. Har quaternion product jo tum GNC (Guidance, Navigation & Control — woh on-board maths jo spacecraft ya drone ko sahi direction mein rakhti hai) mein kabhi bhi dekhoge, woh kuch giney-chune case classes mein se ek hoga — parallel axes, perpendicular axes, identity, pure vectors, non-commuting pairs, sandwich rotation, aur woh limiting "tiny angle" case jisme gyro integrators rehte hain. Hum har class ka ek fully-solved example karte hain taaki jab tumhara filter raat ke 3 baje koi case de, tumne uski shape pehle se dekhi ho.

Notation reminder (kuch naya nahi assume kiya): ek quaternion hai $q = (w, v)$ jahan $w$ ek plain number hai (scalar part) aur $v = (x, y, z)$ 3D mein ek arrow hai (vector part). Ek hi quaternion likhne ke do tarike neeche aate hain — hamesha same component order mein:

paired form $q = (w, (x, y, z))$ — pehle scalar, phir arrow ke teen components brackets mein;
flat form $q = (w, x, y, z)$ — wahi chaar numbers bina inner brackets ke.

Ye same object hain; $(w, x, y, z)$ ka matlab bas $(w, (x, y, z))$ hai. Scalar hamesha pehli slot mein hota hai, aur vector components hamesha $x, y, z$ order mein hote hain. Kisi letter par hat, jaise $\overset{n}{^}$ , ka matlab hai "is arrow ki length exactly $1$ hai" — ek pure direction bina kisi size ke; coordinate directions $\overset{x}{^} = (1, 0, 0)$ , $\overset{y}{^} = (0, 1, 0)$ , $\overset{z}{^} = (0, 0, 1)$ axes ke saath standard unit arrows hain. Product rule, jise hum neeche baar baar USE karte hain, hai $pq = (scalar w_{1} w_{2} - v_{1} \cdot v_{2}, vector w_{1} v_{2} + w_{2} v_{1} + v_{1} \times v_{2})$

Scenario matrix

Har Hamilton-product situation is table ki ek row hai. Neeche ke examples har cell ko cover karte hain. ( $\overset{q}{ˉ}$ = conjugate $(w, - v)$ jo neeche wale box mein define hai.)

#	Case class	Kya special hai	Example
A	Identity element	ek factor $1 = (1, 0)$ hai	Ex 1
B	Pure imaginary × pure imaginary, perpendicular	dono scalars $0$ , axes $⊥$ → dot $= 0$	Ex 2
C	Pure imaginary, parallel/equal axes	dono scalars $0$ , axes parallel → cross $= 0$	Ex 3
D	Order swap (non-commutativity)	$pq$ aur $q p$ compute karo, compare karo	Ex 4
E	Do real rotations, general axes	full formula, teeno terms nonzero	Ex 5
F	Sandwich (rotate a vector) $q u \overset{q}{ˉ}$	conjugate $\overset{q}{ˉ}$ neeche wale box se; do products, length preserved	Ex 6
G	Sign / negative-scalar (half-angle $> 18 0^{\circ}$ )	$cos \frac{θ}{2} < 0$ ; double-cover $q \equiv - q$	Ex 7
H	Limiting case: infinitesimal rotation	$θ \to 0$ , gyro integration se link	Ex 8
I	Exam twist: unknown factor solve karo	diya $q p = r$ , inverse se $p$ nikalo	Ex 9

Do building blocks jo hum baar baar use karte hain:

Case A — identity element

Worked example Ex 1 —

$1$ se multiply karo (dono sides par) $q \cdot 1$ aur $1 \cdot q$ compute karo jahan $q = (w, v)$ arbitrary hai aur $1 = (1, (0, 0, 0))$ .

Forecast: guess karo — kya "number one" se multiply karne par kuch change hoga, aur kya side matter karti hai?

$q \cdot 1$ scalar part $= w \cdot 1 - v \cdot 0 = w$ . Yeh step kyun? Kisi bhi arrow ko zero arrow ke saath dot karne par $0$ milta hai, toh koi scalar andar nahi aata.
$q \cdot 1$ vector part $= w 0 + 1 v + v \times 0 = v$ . Yeh step kyun? Zero arrow ke saath cross zero hota hai; ek hi survivor hai $1 \cdot v$ . Toh $q \cdot 1 = (w, v) = q$ .
$1 \cdot q$ scalar part $= 1 \cdot w - 0 \cdot v = w$ . Yeh step kyun? Ab $1$ pehla factor hai; uska vector part $0$ hai, toh dot term phir bhi $0$ hai. Dot ka order matter nahi karta (dot symmetric hai).
$1 \cdot q$ vector part $= 1 v + w 0 + 0 \times v = v$ . Yeh step kyun? Cross term hai $0 \times v = 0$ — yahi ek term tha jo symmetry tod sakti thi, aur woh vanish ho gayi. Toh $1 \cdot q = (w, v) = q$ bhi.

Verify: $q \cdot 1 = 1 \cdot q = q$ . Identity $1$ woh ek factor hai jo sab kuch ke saath commute karta hai — isliye nahi ki non-commutativity band ho gayi, balki isliye ki iska cross term hamesha zero hota hai. Baaki har pair mein order-dependence rehti hai jo hum Ex 4 mein exploit karte hain.

Case B — perpendicular pure imaginaries

Worked example Ex 2 — formula se

$i j$ $p = i = (0, (1, 0, 0))$ , $q = j = (0, (0, 1, 0))$ .

Forecast: parent kehta hai $ij = k$ . Formula ka kaunsa term ise produce karta hai — dot ya cross?

Scalar $= 0 \cdot 0 - (1, 0, 0) \cdot (0, 1, 0) = 0 - 0 = 0$ . Yeh step kyun? Dono axes $9 0^{\circ}$ par hain; perpendicular arrows ka dot product zero hota hai, toh koi scalar nahi aata.
Vector $= 0 \cdot (0, 1, 0) + 0 \cdot (1, 0, 0) + (1, 0, 0) \times (0, 1, 0)$ . Yeh step kyun? Dono scalars $0$ hain, toh do weighting terms vanish ho jaate hain; bas cross bachta hai.
$(1, 0, 0) \times (0, 1, 0) = (0, 0, 1)$ right-hand rule se ( $\overset{x}{^} \times \overset{y}{^} = \overset{z}{^}$ ).
Result $= (0, (0, 0, 1)) = k$ . ✅

Figure — kya dekh rahe ho: $x$ ke saath black arrow $i$ hai, $y$ ke saath black arrow $j$ hai. Jahan ye milte hain wahan ek small right-angle mark hai jis par " $9 0^{\circ}$ (dot $= 0$ )" likha hai. Red arrow result $k = i \times j$ hai, $x$ – $y$ plane se bahar right-hand rule se point kar raha hai. Red = woh answer jo humne banaya.

Verify: multiplication table $ij = k$ se match karta hai. Length: $∣ k ∣ = 1$ , aur $∣ i ∣∣ j ∣ = 1$ , toh length preserved hai — ek healthy sign.

Case C — parallel / equal axes

Worked example Ex 3 —

$i \cdot i$ (same axis) $p = q = i = (0, (1, 0, 0))$ .

Forecast: do identical pure vectors — ek vector ko khud se cross karne par kya milta hai?

Scalar $= 0 - (1, 0, 0) \cdot (1, 0, 0) = - 1$ . Yeh step kyun? Ek arrow ko khud se dot karne par uska length-squared milta hai $= 1$ ; formula isse subtract karta hai, giving $- 1$ . Yahaan se $i^{2} = - 1$ geometry se paida hota hai.
Vector $= 0 + 0 + (1, 0, 0) \times (1, 0, 0) = (0, 0, 0)$ . Yeh step kyun? Koi bhi vector khud se cross karne par zero deta hai (jo parallelogram ye banate hain uska area zero hai). Toh koi vector part nahi.
Result $= (- 1, 0) = - 1$ . ✅ ( $i^{2} = - 1$ .)

Verify: yeh Hamilton ka rule $i^{2} = - 1$ purely "dot of self $= 1$ , cross of self $= 0$ " se reproduce karta hai. Wahi $j^{2}$ aur $k^{2}$ ke liye bhi kaam karta hai. Parallel axes → cross term marta hai, dot term dominate karta hai.

Case D — order matters (non-commutativity)

Worked example Ex 4 —

$ij$ aur $j i$ compare karo Dono products compute karo aur subtract karo.

Forecast: kya $ij$ equal hoga $j i$ ke? Agar nahi, toh farq kahaan se aayega?

$ij = (0, (0, 0, 1)) = k$ (Ex 2 se).
$j i$ : scalar $= 0 - (0, 1, 0) \cdot (1, 0, 0) = 0$ (abhi bhi perpendicular). Yeh step kyun? Dot symmetric hai, toh order swap karne par badlta nahi — scalar part kabhi disagreement cause nahi kar sakta.
$j i$ ka vector $= (0, 1, 0) \times (1, 0, 0) = (0, 0, - 1)$ . Yeh step kyun? Cross product anti-symmetric hai: $b \times a = - (a \times b)$ . Swap karne par sign flip ho jaati hai.
$j i = (0, (0, 0, - 1)) = - k$ .
$ij - j i = k - (- k) = 2 k \neq = 0$ . Non-commutative. ✅

Figure — kya dekh rahe ho: black arrows $a$ ( $x$ ke saath) aur $b$ ( $y$ ke saath) do factors hain. Red arrow upar ki taraf $ij = + k$ hai; red arrow neeche ki taraf $j i = - k$ hai. Same do black inputs, opposite red outputs — jab tum order swap karte ho toh red arrow literally reverse ho jaata hai. Red = (order-dependent) answer.

Verify: $ij = - j i$ , exactly parent se match karta hai. Cross term akela culprit hai — dot term dono taraf identical tha. Yeh algebraic pedantry nahi hai: yeh kehta hai "yaw-then-roll ≠ roll-then-yaw," 3D orientation ke baare mein ek physical fact.

Case E — do real rotations, general axes

Worked example Ex 5 —

$z$ ke baare mein $9 0^{\circ}$ , phir $x$ ke baare mein $9 0^{\circ}$ Angle $θ$ ke liye unit axis $\overset{n}{^}$ ke baare mein unit quaternion (hat ka matlab $∣ \overset{n}{^} ∣ = 1$ ): $q = (cos \frac{θ}{2}, sin \frac{θ}{2} \overset{n}{^})$ . $q_{z} = (\frac{1}{2}, 0, 0, \frac{1}{2})$ , $q_{x} = (\frac{1}{2}, \frac{1}{2}, 0, 0)$ (flat form: scalar, phir $x, y, z$ ). $q_{x} q_{z}$ compute karo ("pehle $z$ , phir $x$ ").

Forecast: yahaan teeno formula terms nonzero hain — genuinely "mixed" quaternion expect karo.

Scalar $= \frac{1}{2} \cdot \frac{1}{2} - (\frac{1}{2}, 0, 0) \cdot (0, 0, \frac{1}{2}) = \frac{1}{2} - 0 = \frac{1}{2}$ . Yeh step kyun? Dono axes $\overset{x}{^}$ aur $\overset{z}{^}$ perpendicular hain → unka dot $0$ hai, toh bas $w_{1} w_{2}$ half-angle product bachta hai.
Vector $= w_{1} v_{2} \frac{1}{2} (0, 0, \frac{1}{2}) + w_{2} v_{1} \frac{1}{2} (\frac{1}{2}, 0, 0) + v_{1} \times v_{2} (\frac{1}{2}, 0, 0) \times (0, 0, \frac{1}{2})$ . Yeh step kyun? Dono scalars ab nonzero hain, toh do weighting terms cross ke saath survive karte hain.
$= (0, 0, \frac{1}{2}) + (\frac{1}{2}, 0, 0) + (0, - \frac{1}{2}, 0)$ — note $\overset{x}{^} \times \overset{z}{^} = - \overset{y}{^}$ .
Sum $= (\frac{1}{2}, - \frac{1}{2}, \frac{1}{2})$ . Toh $q_{x} q_{z} = (\frac{1}{2}, \frac{1}{2}, - \frac{1}{2}, \frac{1}{2})$ flat form mein.

Verify: norm $= 4 \cdot \frac{1}{4} = 1$ ✅ — unit quaternions ka product ek unit quaternion hota hai, toh yeh valid rotation hai. Yeh woh workhorse case hai jo ek filter har millisecond run karta hai.

Case F — sandwich: ek real vector rotate karo

Worked example Ex 6 —

$u = (1, 0, 0)$ ko $z$ ke baare mein $9 0^{\circ}$ rotate karo $u = (0, (1, 0, 0))$ embed karo. $q_{z} = (\frac{1}{2}, 0, 0, \frac{1}{2})$ aur uska conjugate $\overset{q}{ˉ}_{z} = (\frac{1}{2}, 0, 0, - \frac{1}{2})$ use karo (Conjugate box se: same scalar, negated vector). $u^{'} = q_{z} u \overset{q}{ˉ}_{z}$ compute karo.

Forecast: $x$ -axis ko $z$ ke baare mein $9 0^{\circ}$ rotate karne par $y$ -axis par land karna chahiye. Dekho length $1$ rahe.

Pehla product $a = q_{z} u$ . Scalar $= \frac{1}{2} \cdot 0 - (0, 0, \frac{1}{2}) \cdot (1, 0, 0) = 0$ . Yeh step kyun? Hum do half-steps mein rotate karte hain; $q u$ pehla squeeze hai. Axis $\overset{z}{^} ⊥ \overset{x}{^}$ toh dot $= 0$ .
$a$ ka vector $= \frac{1}{2} (1, 0, 0) + 0 + (0, 0, \frac{1}{2}) \times (1, 0, 0) = (\frac{1}{2}, 0, 0) + (0, \frac{1}{2}, 0) = (\frac{1}{2}, \frac{1}{2}, 0)$ . Toh $a = (0, (\frac{1}{2}, \frac{1}{2}, 0))$ .
Doosra product $u^{'} = a \overset{q}{ˉ}_{z}$ . Scalar $= 0 \cdot \frac{1}{2} - (\frac{1}{2}, \frac{1}{2}, 0) \cdot (0, 0, - \frac{1}{2}) = 0$ . Yeh step kyun? Result ka scalar $0$ hona chahiye ek pure vector ke liye — ek built-in sanity check.
$u^{'}$ ka vector $= 0 + \frac{1}{2} (\frac{1}{2}, \frac{1}{2}, 0) + (\frac{1}{2}, \frac{1}{2}, 0) \times (0, 0, - \frac{1}{2})$ . Yeh step kyun? Yeh doosra half-turn hai: $w_{2} v_{1}$ $a$ ke arrow ko $\overset{q}{ˉ}_{z}$ ke scalar se scale karta hai, aur cross term woh sideways twist add karta hai jo $\overset{q}{ˉ}_{z}$ contribute karta hai. Saath mein do half-turns poore $9 0^{\circ}$ mein compose ho jaate hain. Cross $= (\frac{1}{2} \cdot (- \frac{1}{2}) - 0, 0 - \frac{1}{2} \cdot (- \frac{1}{2}), 0) = (- \frac{1}{2}, \frac{1}{2}, 0)$ . Sum $= (\frac{1}{2}, \frac{1}{2}, 0) + (- \frac{1}{2}, \frac{1}{2}, 0) = (0, 1, 0)$ .
$u^{'} = (0, (0, 1, 0)) \Rightarrow u^{'} = (0, 1, 0) = \overset{y}{^}$ . ✅

Figure — kya dekh rahe ho: $x$ ke saath black arrow input $u = (1, 0, 0)$ hai; dashed black arc $z$ ke baare mein $+ 9 0^{\circ}$ sweep hai; red arrow $y$ ke saath output $u^{'} = (0, 1, 0)$ hai. Same length, quarter turn — red = jahan vector land karta hai.

Verify: $∣ u^{'} ∣ = 1 = ∣ u ∣$ (length preserved), aur $\overset{x}{^}$ , $\overset{y}{^}$ par gaya exactly jaise ek $+ 9 0^{\circ}$ turn $z$ ke baare mein hona chahiye. $q$ ke andar half-angle wahi hai jo do half-turns ko ek full rotation mein compose karta hai.

Case G — negative scalar aur double cover

Worked example Ex 7 —

$z$ ke baare mein $27 0^{\circ}$ rotation $θ = 27 0^{\circ} \Rightarrow \frac{θ}{2} = 13 5^{\circ}$ , $cos 13 5^{\circ} = - \frac{1}{2}$ , $sin 13 5^{\circ} = + \frac{1}{2}$ . Toh $q = (- \frac{1}{2}, 0, 0, \frac{1}{2})$ (flat form).

Forecast: $z$ ke baare mein $27 0^{\circ}$ turn wahi physical orientation hai jaise $- 9 0^{\circ}$ $z$ ke baare mein. Kya dono quaternions equal dikhenge — ya opposite?

$- 9 0^{\circ}$ $z$ ke baare mein quaternion: $\frac{θ}{2} = - 4 5^{\circ}$ , giving $q^{'} = (\frac{1}{2}, 0, 0, - \frac{1}{2})$ . Yeh step kyun? Hum ek hi physical turn ke do naamon ko compare kar rahe hain.
Compare karo: $q = (- \frac{1}{2}, 0, 0, \frac{1}{2})$ aur $- q^{'} = (- \frac{1}{2}, 0, 0, \frac{1}{2})$ . Ye equal hain, yaani $q = - q^{'}$ . Yeh step kyun? Har rotation ke do quaternions hote hain, $q$ aur $- q$ — yeh double cover hai. Ek negative scalar bas matlab hai ki half-angle $9 0^{\circ}$ se aage gaya.
Prove karo ki ye identically act karte hain $\overset{x}{^} = (1, 0, 0)$ ko $q$ se sandwich karke. Pehle $a = q x$ jahan $x = (0, (1, 0, 0))$ : scalar $= 0 - (0, 0, \frac{1}{2}) \cdot (1, 0, 0) = 0$ ; vector $= - \frac{1}{2} (1, 0, 0) + 0 + (0, 0, \frac{1}{2}) \times (1, 0, 0) = (- \frac{1}{2}, \frac{1}{2}, 0)$ , toh $a = (0, (- \frac{1}{2}, \frac{1}{2}, 0))$ . Yeh step kyun? "Same rotation" ka sabse honest test yeh hai: kya ye ek basis vector ko same tarike se move karte hain?
Phir $x^{'} = a \overset{q}{ˉ}$ with $\overset{q}{ˉ} = (- \frac{1}{2}, 0, 0, - \frac{1}{2})$ : scalar $= 0 - (- \frac{1}{2}, \frac{1}{2}, 0) \cdot (0, 0, - \frac{1}{2}) = 0$ ; vector $= - \frac{1}{2} (- \frac{1}{2}, \frac{1}{2}, 0) + (- \frac{1}{2}, \frac{1}{2}, 0) \times (0, 0, - \frac{1}{2}) = (\frac{1}{2}, - \frac{1}{2}, 0) + (- \frac{1}{2}, - \frac{1}{2}, 0) = (0, - 1, 0)$ . Toh $\overset{x}{^} \mapsto (0, - 1, 0) = - \overset{y}{^}$ . $z$ ke baare mein $27 0^{\circ}$ (equivalently $- 9 0^{\circ}$ ) turn $\overset{x}{^}$ ko $- \overset{y}{^}$ par bhejta hai. ✅

Verify: $∣ q ∣ = \frac{1}{2} + \frac{1}{2} = 1$ ✅, aur dono $q$ aur $q^{'} = - q$ $\overset{x}{^} \to - \overset{y}{^}$ bhejte hain (tum identical vector paate ho kyunki $(- q) u \overline{(- q)} = q u q^{- 1}$ — do minus signs cancel ho jaate hain). Negative $w$ kabhi "galat" nahi hota; yeh do valid labels mein se ek hai. Filters $q \to - q$ flip karte hain jab $w < 0$ ho, shortest-path representation rakhne ke liye.

Case H — limiting case (tiny rotation)

Worked example Ex 8 — infinitesimal update,

$θ \to 0$ Ek gyro angular velocity $ω$ report karta hai ek tiny time $Δ t$ par, giving ek small rotation of angle $θ = ∣ ω ∣Δ t$ axis $\overset{n}{^}$ ke baare mein. $θ = 0.01$ rad lao $\overset{n}{^} = (0, 0, 1)$ ke baare mein aur ise $q_{0} = 1 = (1, 0)$ par multiply karo.

Forecast: ek tiny angle ke liye, $cos \frac{θ}{2}$ aur $sin \frac{θ}{2}$ ka kya hota hai? Approximate quaternion guess karo.

Taylor-expand karo: $cos \frac{θ}{2} = 1 - \frac{1}{2} (\frac{θ}{2})^{2} + O (θ^{4})$ aur $sin \frac{θ}{2} = \frac{θ}{2} - \frac{1}{6} (\frac{θ}{2})^{3} + O (θ^{5})$ . Hum bas leading terms rakhte hain: $cos \frac{θ}{2} \approx 1$ (error $\sim θ^{2} /8$ ) aur $sin \frac{θ}{2} \approx \frac{θ}{2}$ (error $\sim θ^{3} /48$ ). Yeh step kyun? Jo terms neglect kiye hain woh $O (θ^{2})$ aur chote hain; $θ ≲ 0.1$ rad ( $\approx 6^{\circ}$ , ek typical gyro step) ke liye relative error $\frac{0. 1 ^{2}}{8} \approx 0.13%$ se kam hai, sensor noise se safely neeche — toh linearization exactly us regime mein trustworthy hai jahan use hoti hai.
Small quaternion banao. $\overset{n}{^} = (0, 0, 1)$ aur $\frac{θ}{2} = 0.005$ ke saath: $δ q = (cos \frac{θ}{2}, sin \frac{θ}{2} \overset{n}{^}) \approx (1, (0, 0, 0.005)), δ q = (1, 0, 0, 0.005) (flat form).$ Numerically $cos 0.005 \approx 0.99998750$ aur $sin 0.005 \approx 0.00499998$ , dono rakhe gaye values par round ho jaate hain. Yeh step kyun? Yeh linear approximation hai jo gyro integrators actually store karte hain: scalar $\approx 1$ , vector $\approx \frac{1}{2} ω Δ t$ .
Current attitude $q_{0} = 1$ par update apply karo: $q_{1} = δ q q_{0} = δ q \cdot 1 = δ q = (1, 0, 0, 0.005)$ . Yeh step kyun? Case A (Ex 1) se, identity $1$ se multiply karne par $δ q$ unchanged rehta hai — toh naya attitude bas $δ q$ hi hai. "No rotation" se shuru karke, ek tiny step hai woh tiny rotation.
Renormalize karo: $∣ δ q ∣ = 1^{2} + 0.00 5^{2} = 1.000025 \approx 1.0000125$ , toh $q_{1} \leftarrow q_{1} /∣ q_{1} ∣ \approx (0.99998750, 0, 0, 0.00499994)$ . Yeh step kyun? $O (θ^{2})$ term drop karne se $δ q$ thoda off-unit rah gaya (norm $> 1$ by $\sim 1.25 \times 1 0^{- 5}$ ); norm se divide karne par ek valid rotation restore hoti hai (parent ki "renormalize" warning in action).

Verify: stored vector part $\frac{1}{2} θ = 0.005$ , $\frac{1}{2} ∣ ω ∣Δ t$ se exactly match karta hai — yeh kinematics $\dot q=\tfrac12 q\,\omega$ recover karta hai. Jab $θ \to 0$ : $cos \frac{θ}{2} \to 1$ , $sin \frac{θ}{2} \to 0$ , toh $δ q \to (1, 0) = 1$ — no rotation, no change. Smooth limiting behaviour ✅.

Case I — exam twist: unknown factor solve karo

Worked example Ex 9 — diya

$q p = r$ , $p$ nikalo Tumhe total rotation $r = k = (0, (0, 0, 1))$ aur last-applied factor $q = i = (0, (1, 0, 0))$ pata hai, jabki first-applied factor $p$ unknown hai. $p$ nikalo.

Forecast: $q$ ke left multiplication ko "undo" karne ke liye, hum kya multiply karte hain, aur kis side par?

$q p = r$ se, dono sides ko left-multiply karo $q^{- 1}$ se: $q^{- 1} q p = q^{- 1} r \Rightarrow p = q^{- 1} r$ . Yeh step kyun? Multiplication non-commutative hai, toh hume $q$ ko usi side cancel karna hai jis par woh baitha hai — left. Order discipline poora point hai.
$q = i$ unit hai, toh $q^{- 1} = \overset{q}{ˉ} = (0, (- 1, 0, 0)) = - i$ (conjugate: same scalar $0$ , negated vector). Yeh step kyun? Unit quaternions ke liye inverse $=$ conjugate; vector part flip karo.
$p = (- i) (k)$ . Scalar $= 0 - (- 1, 0, 0) \cdot (0, 0, 1) = 0$ . Vector $= 0 + 0 + (- 1, 0, 0) \times (0, 0, 1) = (0, 1, 0) = \overset{y}{^}$ . Toh $p = (0, (0, 1, 0)) = j$ . Yeh step kyun? Hamilton formula ek baar phir use kiya; $(- \overset{x}{^}) \times \overset{z}{^} = + \overset{y}{^}$ .
Answer: $p = j$ .

Verify: check karo $q p = i j = k = r$ ✅ (Ex 2 se). Unknown factor exactly recover ho gaya — yeh exactly woh hai jaise ek MEKF ek measured vs. predicted attitude se error quaternion back out karta hai. Dekho Multiplicative EKF (MEKF).

Recall

Recall Har degenerate case mein kaunsa term marta hai?

Perpendicular axes ::: dot term $= 0$ (Ex 2). Parallel/equal axes ::: cross term $= 0$ (Ex 3). Ek factor $1$ hai ::: $v$ ke siwa sab marta hai; identity, dono sides commute karta hai (Ex 1). Tiny angle $θ \to 0$ ::: quaternion $\to (1, 0)$ , koi change nahi (Ex 8).

Recall Khud test karo

$ij - j i = ?$ ::: $2 k$ — doubled cross term (Ex 4). $q p = r$ mein left factor undo kaise karo? ::: $p = q^{- 1} r$ , left par multiply karo (Ex 9). Ek rotation ke liye do quaternions? ::: $q$ aur $- q$ (double cover, Ex 7). $δ q$ ka small-angle vector part? ::: $\frac{1}{2} ω Δ t$ (Ex 8).

Connections

Quaternion product — Hamilton product — woh rule jo yeh examples drill karte hain.
Cross Product & Right-Hand Rule — Ex 2, 4, 5 mein har sign flip ka source.
Axis-Angle & Euler Rodrigues — jahan se Ex 5–8 mein half-angle $\frac{θ}{2}$ aata hai.
Quaternion Kinematics — $\dot q = \tfrac12 q\,\omega$ — woh limit jo Ex 8 recover karta hai.
Multiplicative EKF (MEKF) — Ex 9 ka error-quaternion trick.
Rotation Matrices — SO(3) — Ex 6 ko rotation matrix ke against cross-check karo.

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