Question bank — Quaternion rotation formula — rotating vector v by quaternion q

False. Akela $qv$ generally nonzero scalar part produce karta hai, isliye result pure quaternion (ek 3D vector) bhi nahi hota. Ek rotation matrix $R$ ek $3\times 3$ grid hai jo already 3-vectors ko 3-vectors pe map karta hai, isliye ek multiply $R\mathbf{v}$ kaafi hai; quaternion ko output valid vector rakhne ke liye right pe conjugate chahiye.

False. Dono signs sandwich mein cancel ho jaate hain: $(-q)v(-q{)}^{-1}=qv{q}^{-1}$. Quaternions rotations ko double-cover karte hain — har physical rotation ke exactly do quaternion names hote hain.

False. $q$ half-angle store karta hai: $q=(\cos \frac{\theta }{2},\sin \frac{\theta }{2}\hat{n})$. Kyunki $q$ $qv{q}^{-1}$ mein do baar act karta hai, har copy $\theta /2$ supply karta hai, aur double-angle identities unhe $\theta$ mein jod deti hain.

True. Parallel component satisfy karta hai $\hat{n}\times {\mathbf{v}}_{\parallel }=0$, isliye algebra ${\mathbf{v}}_{\parallel }$ ko untouched return karta hai — tum us vector ko spin nahi kar sakte jo pehle se spin axis ki taraf point kar raha ho.

False. Hamilton product non-commutative hai ($q_1{q}_2\ne q_2{q}_1$), bilkul rotation matrices ki tarah, jahan do matrix multiplications ka order swap karne se result badal jaata hai. "Roll phir pitch" rotate karna "pitch phir roll" se alag hai.

False in practice. Kyunki $\Vert q\Vert =1$, inverse sirf conjugate $q^{\ast }=(w,-\mathbf{v})$ hai — tum sirf vector part ka sign flip karte ho, koi division nahi chahiye.

False. Ye ek hi rotation hai do tarike se likhi gayi; sandwich algebraically Rodrigues' rotation formula mein collapse ho jaata hai, jo $\mathbf{v}$ ko $\mathbf{v}\cos \theta$, ek turning term $(\hat{n}\times \mathbf{v})\sin \theta$, aur fixed axial part ko combine karke spin karta hai. Quaternion sirf sasta storage hai (4 numbers vs axis pe ek formula).

False. Vector ki length preserved rehti hai aur rotate karne ke tarike pe irrelevant hai; sirf axis $\hat{n}$ unit vector hona chahiye aur $q$ unit hona chahiye. Vector khud kisi bhi length ka ho sakta hai.

Wrong: inverse conjugate $q^{\ast }=(w,-\mathbf{v})$ hai, $q$ khud nahi. Sirf tab $q^{-1}=q$ hold karta agar vector part zero ho (ek pure scalar quaternion).

Wrong order. Action explicitly likho: pehle $q_1$ se rotate karo, $q_{1}v{q}_1^{-1}$ milta hai; phir use $q_2$ se rotate karo, $q_2(q_{1}v{q}_1^{-1})q_2^{-1}=(q_2{q}_1)v(q_2{q}_1{)}^{-1}$ milta hai. To combined quaternion $q_2{q}_1$ hai — jo pehle act karta hai wo vector ke nearest, right pe baith ta hai — aur $(q_2{q}_1{)}^{-1}=q_1^{-1}{q}_2^{-1}$ order reverse karta hai.

Wrong: vector ek pure quaternion $(0,\mathbf{v})$ hona chahiye jiska scalar part zero ho. $|\mathbf{v}|$ ko scalar slot mein daalna sandwich ko corrupt karta hai aur output ek plain vector nahi rehta.

Wrong framing: input ka scalar part pehle se $0$ hai (ye ek pure quaternion hai). Aur closure argument (neeche "Why must the conjugate sit on the right?" item dekho) dikhata hai output ka scalar part bhi exactly $0$ hai. Koi nonzero scalar hai hi nahi worry karne ke liye.

Wrong: ye structural hai. $c\equiv \cos \frac{\theta }{2}$ aur $s\equiv \sin \frac{\theta }{2}$ likhne par, $q$ aur $q^{-1}$ ke do half-angles ko zaruri recombine karna padta hai, aur identities $c^2-s^2=\cos \theta$, $2cs=\sin \theta$ exactly wahi machinery hai jo do half-angle contributions ko ek poore $\theta$ mein convert karti hai.

Wrong: $\theta =360^{\circ }$ half-angle $180^{\circ }$ deta hai, isliye $q=(\cos 180^{\circ },\sin 180^{\circ }\hat{n})=(-1,0)=-1$, $+1$ nahi. Ye vectors pe identity ki tarah act karta hai (signs cancel ho jaate hain), lekin quaternion khud $-1$ hai — double cover ki hallmark.

Wrong: near-unit quaternion ko renormalize karna use $\Vert q\Vert =1$ pe wapas laata hai rotation mein tiny change ke saath — yahi cheapness exactly woh reason hai kyun GNC systems quaternions prefer karte hain (dekho Spacecraft attitude determination (GNC), jahan onboard filters har update pe renormalize karte hain).

Yahan poori chain hai. Hum chahte hain output pure ho (scalar part zero) taaki ye ek genuine 3D vector ho. Sandwich ka conjugate lo rule use karke "product ka conjugate order reverse karta hai aur har factor ko conjugate karta hai": $(qv{q}^{-1}{)}^{\ast }=(q^{-1}{)}^{\ast }{v}^{\ast }{q}^{\ast }$. Unit $q$ ke liye $(q^{-1}{)}^{\ast }=q$ aur $q^{\ast }=q^{-1}$ hai; aur kyunki $v$ pure hai, $v^{\ast }=-v$. Substitute karne pe $(qv{q}^{-1}{)}^{\ast }=q(-v)q^{-1}=-(qv{q}^{-1})$ milta hai. Jo object minus apne khud ke conjugate ke barabar ho uska scalar part zero hona chahiye (kyunki conjugation sirf vector part flip karta hai; apne negation ke barabar hona unflipped scalar ko $0$ force karta hai). Isliye output pure hai. $qvq$ use karne se ye cancellation break ho jaati.

Cross product $\hat{n}\times \mathbf{v}$ tab vanish ho jaata hai jab $\mathbf{v}\parallel \hat{n}$, isliye parallel piece ke paas rotate hone ke liye kuch nahi hota. Sirf perpendicular piece rotation ke plane mein rehta hai, jahan wo swing karta hai aur $\sin \theta (\hat{n}\times {\mathbf{v}}_{\perp })$ turning term gain karta hai.

Quaternions mein gimbal lock nahi hota aur 1 constraint ke saath 4 numbers use karte hain, jabki Euler angles do rotation axes line up hone par ek degree of freedom lose kar sakte hain. Quaternions smoothly interpolate bhi karte hain — do orientations ke beech ek shortest-arc blend, jo SLERP — quaternion interpolation mein cover hai.

Do pure quaternions multiply karo aur units dekho. Lo $\mathbf{a}=a_{1}i+a_{2}j+a_{3}k$ aur $\mathbf{b}=b_{1}i+b_{2}j+b_{3}k$. Same-axis terms $i^2=j^2=k^2=-1$ use karte hain: jaise $a_{1}i\cdot b_{1}i=a_1{b}_1{i}^2=-a_1{b}_1$, ek pure scalar; teeno ko sum karne par $-(a_1{b}_1+a_2{b}_2+a_3{b}_3)=-(\mathbf{a}\cdot \mathbf{b})$ milta hai — wo dot product hai scalar slot mein. Cross-axis terms $ij=k,jk=i,ki=j$ use karte hain: jaise $a_{1}i\cdot b_{2}j=a_1{b}_{2}k$, $k$ ke along ek pure vector; aaise saare terms collect karne par exactly $\mathbf{a}\times \mathbf{b}$ vector slot mein milta hai. To split sirf $i,j,k$ table hai "scalar-making" aur "vector-making" products mein sort hua — dekho Quaternion algebra & Hamilton product.

Ek unit quaternion ka norm 1 hota hai, aur quaternion multiplication norms pe multiplicative hoti hai: $\Vert qv{q}^{-1}\Vert =\Vert q\Vert \Vert v\Vert \Vert q^{-1}\Vert =\Vert v\Vert$. Length conserved rehti hai, jaisa koi bhi true rotation demand karta hai.

Kyunki $q_1$ nested sandwich $q_2(q_{1}v{q}_1^{-1})q_2^{-1}$ mein $v$ ke paas baith ta hai, isliye wo vector ko pehle touch karta hai. Ye general operator-action rule hai: $qv{q}^{-1}$ mein jo quaternion vector ke nearest hai wo pehle act karta hai, aur sandwiches stack karna inward-to-outward nest karta hai. Left-to-right reading order aur application order isliye opposite hain, bilkul function composition $f(g(x))$ ki tarah jahan $g$ pehle run karta hai.

$\theta =0$ pe quaternion $(1,0)$ hai — vector part vanish ho jaata hai, isliye $\hat{n}$ undefined hai. Koi rotation hi nahi hai, isliye koi distinguished axis nahi hai, jaisa ek zero rotation ki koi meaningful direction nahi hoti.

$q=(\cos 0,\sin 0\hat{n})=(1,0)$, identity. Sandwich $v$ ko unchanged return karta hai — koi axis bhi nahi chahiye kyunki kuch move hi nahi hota.

Haan — $q=(0,\hat{n})$ ek pure quaternion hai, isliye axis clearly readable hai. Perpendicular component sign flip karta hai ($\cos 180^{\circ }=-1$, $\sin 180^{\circ }=0$), $v_{\perp }$ axis se reflect ho jaata hai.

Zero vector, hamesha: $q(0,0)q^{-1}=(0,0)$. Origin pe ek point kisi bhi rotation se move nahi ho sakta jo us se guzarne wale axis ke around ho.

Nahi — half-angle form $q=(\cos \frac{\theta }{2},\sin \frac{\theta }{2}\hat{n})$ assume karta hai $\Vert \hat{n}\Vert =1$. Non-unit axis $\Vert q\Vert \ne 1$ banata hai, $q^{-1}=q^{\ast }$ break karta hai aur rotation distort hoti hai. Pehle axis normalize karo.

Sandwich ab vector ko thoda scale karta hai (length preserved nahi rehti) kyunki $\Vert q\Vert \ne 1$. Fix sasta hai: renormalization $q/\Vert q\Vert$, jo accumulating rotation matrices ke comparison mein quaternions ka ek key GNC advantage hai.

Vector exactly apne aap wapas aata hai (ek poora turn 3D vectors pe identity hai). Quaternion nahi aata: wo $-1$ ban jaata hai, $+1$ pe wapas aane ke liye $720^{\circ }$ chahiye — double cover ka geometric fingerprint.

Same rotation. Dono axis sign aur angle sign flip karne se $\sin \frac{\theta }{2}\hat{n}$ unchanged rehta hai aur $\cos \frac{\theta }{2}$ unchanged rehta hai, isliye quaternion — aur physical spin — identical hai.