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

False. Sirf unit quaternions ($\sum q_i^2=1$) rotations represent karte hain; ek general 4-tuple jaise $(2,0,0,0)$ ka norm $2$ hai aur use pehle normalize karna padega.

True. Kyunki $q_0=\cos \frac{\theta }{2}$ aur $\frac{\theta }{2}$ ka range $[0,\pi ]$ hai jab $\theta \in [0,2\pi ]$, toh $\cos \frac{\theta }{2}$ poora interval $[-1,1]$ cover karta hai.

False. Vector part mein $\vec{p}\times \vec{q}$ hota hai, jo sign flip kar leta hai jab tum $p$ aur $q$ swap karo, isliye ye alag hote hain jab tak axes parallel na hon — yahi reflect karta hai ki 3D rotations commute nahi karte.

False. Ye same rotation describe karte hain: sandwich $qv{q}^{-1}$ mein dono minus signs cancel ho jaate hain. Yahi double cover $S^3\to SO(3)$ hai.

False. $(-1,0,0,0)$ ka bhi zero vector part aur unit norm hai; ye ek $360^{\circ }$ rotation ke barabar hai, jo physically identity jaisi hi orientation hai.

True. $q^{-1}=q^{\ast }/\Vert q{\Vert }^2$, aur $\Vert q{\Vert }^2=1$ hone par ye $q^{-1}=q^{\ast }=(q_0,-{\vec{q}}_v)$ ho jaata hai.

Saamaanyatah False. Scalar part $\cos \frac{\theta }{2}=0$ hone par $\theta =180^{\circ }$ force hota hai, isliye sirf unit pure quaternions ek specific $180^{\circ }$ rotation represent karte hain; ek non-unit pure quaternion kuch bhi represent nahi karta.

True. Quaternion norm multiplicative hai, isliye do unit quaternions ka product ek aur unit quaternion deta hai — rotations compose karne par tum $S^3$ par hi rehte ho.

Error: full angle use kiya. Quaternions half-angle lete hain, isliye $q=(\cos 45^{\circ },0,0,\sin 45^{\circ })=(\frac{\sqrt{2}}{2},0,0,\frac{\sqrt{2}}{2})$.

Error. $-q$ same rotation hai, inverse nahi. Inverse sirf vector part ko conjugate karta hai: $q^{-1}=(q_0,-q_1,-q_2,-q_3)$.

Error. Stored vector part $\hat{n}\sin \frac{\theta }{2}$ hai, yaani axis $\sin \frac{\theta }{2}$ se scaled hai. $\hat{n}$ recover karne ke liye $(q_1,q_2,q_3)$ ko uski khud ki length se divide karo.

Error. Discrete integration mein rounding accumulate hoti hai aur dheere dheere $S^3$ se drift ho jaata hai; tumhe har timestep par renormalize karna hoga. (Dekho Quaternion Kinematics — dq/dt.)

Error jab tak unit na ho. $q\otimes q^{\ast }=(\Vert q{\Vert }^2,\vec{0})$. Sirf tab jab $\Vert q\Vert =1$ hoga, ye $(1,0,0,0)$ ke barabar hoga; warna ye ek scalar hai jo $\ne 1$ hai.

Error (order). Pehle $p$ phir $q$ apply karna $q\otimes p$ hai — baad wali rotation left par hoti hai, bilkul matrix multiplication order ki tarah.

4th number wo redundancy hai jo ek unit constraint se hati jaati hai ($4-1=3$); yahi smooth 4D packing exactly woh singularity (gimbal lock) avoid karti hai jo 3-number Euler angles mein hoti hai.

Ek vector ko double-sided product $v^{\prime }=q\otimes v\otimes q^{-1}$ se rotate kiya jaata hai, isliye $q$ effectively do baar apply hota hai; har factor ko poore angle ka aadha carry karna padta hai taaki sandwich poora $\theta$ deliver kare.

Ye 3-sphere $S^3$ ki smooth, closed surface par rehte hain jisme "align" hone ke liye koi coordinate axes nahi hain — Euler Angles and Gimbal Lock mein singularity ek buri chart ki wajah se aati hai, rotations se nahi.

Quaternion inverse bas teen signs negate karta hai ($q^{\ast }$), jabki matrix route mein transpose aur orthogonality checks chahiye — limited spacecraft compute par yeh ek real faayda hai.

Real 3D rotations commute nahi karte (spin-then-tilt $\ne$ tilt-then-spin); $\vec{p}\times \vec{q}$ term exactly is order-dependence ko encode karta hai.

Har unit quaternion 4D mein origin se $1$ distance par hota hai; ek drifted $q$ ko uski length se divide karna use radially $S^3$ par project karta hai bina us direction (aur isliye rotation) ko change kiye jis taraf wo point karta hai.

2D mein ek unit complex number $\cos \theta +i\sin \theta$ plane ko rotate karta hai; quaternions do aur imaginary units $(j,k)$ add karte hain taaki ek single object 3D space ko usi clean tarike se rotate kar sake.

Identity $(1,0,0,0)$: $\theta =0$ se $\cos 0=1$ aur $\hat{n}\sin 0=\vec{0}$ milta hai; isse multiply karne par koi bhi quaternion unchanged rehta hai, jaise $1$ se multiply karna.

$\cos 90^{\circ }=0$, isliye $q=(0,\hat{n})$ ek pure quaternion hai; axis poori tarah visible hai lekin do half-rotations $q$ aur $-q$ same flip ke liye maximally separated 4-tuples ban jaate hain.

$\theta =360^{\circ }$ se $\cos 180^{\circ }=-1$ milta hai, isliye $q=(-1,0,0,0)$. Physically identity jaisi hi orientation hai, phir bhi quaternion identity ka $-1\times$ hai — double cover ki yahi pehchaan hai.

Nahi. Norm derivation $n_x^2+n_y^2+n_z^2=1$ par rely karta tha; non-unit axis unit constraint tod deta hai, isliye $q$ build karne se pehle $\hat{n}$ normalize karna zaroori hai.

Haan. Vector part $\hat{n}\sin 0=\vec{0}$ hota hai chahe koi bhi axis choose karo, isliye undefined axis harmless hai aur $q=(1,\vec{0})$ unambiguous hai.

Haan — exactly antipodal pair $q$ aur $-q$. Koi aur coincidences nahi hote, isliye map $S^3\to SO(3)$ precisely two-to-one hai.