Exercises — Quaternion rotation formula — rotating vector v by quaternion q

HUM KYA KARTE HAIN: $q=(w,\mathbf{v})=(\cos \frac{\theta }{2},\sin \frac{\theta }{2}\hat{n})$ se match karo. Scalar part hai $w=\cos \frac{\theta }{2}=\frac{\sqrt{3}}{2}$, toh $\frac{\theta }{2}=30^{\circ }$ aur $\theta =60^{\circ }.$ Vector part hai $(0,0,\frac{1}{2})=\sin \frac{\theta }{2}\hat{n}$. Kyunki $\sin 30^{\circ }=\frac{1}{2}$, axis hai $\hat{n}=(0,0,1)$ — yani $z$-axis. Answer: $+z$ ke around $60^{\circ }$ rotation.

Ek rotation quaternion ki length $1$ honi chahiye. Compute karo: $\Vert q\Vert =\sqrt{1^2+1^2+0^2+0^2}=\sqrt{2}\ne 1.$ Toh nahi, yeh valid nahi hai. Length se divide karke normalize karo: $q_{\text{unit}}=\frac{1}{\sqrt{2}}(1,1,0,0)=\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},0,0\right).$ Ab $w=\cos \frac{\theta }{2}=\frac{\sqrt{2}}{2}\Rightarrow \theta =90^{\circ }$ $x$-axis ke around.

Unit quaternion ke liye, inverse = conjugate = sirf vector part ka sign flip karo: $q^{-1}=q^{\ast }=\left(\frac{\sqrt{2}}{2},0,-\frac{\sqrt{2}}{2},0\right).$ Geometric read: $q$, $+y$ ke around $+90^{\circ }$ rotate karta hai; $q^{-1}$, $+y$ ke around $-90^{\circ }$ rotate karta hai — yani undo.

Rodrigues use karo $\theta =90^{\circ }$ ke saath ($\cos =0,\sin =1$), $\hat{n}=(0,0,1)$, $\mathbf{v}=(1,0,0)$.

• $\hat{n}\cdot \mathbf{v}=0$ (toh last term vanish ho jaata hai),
• $\hat{n}\times \mathbf{v}=(0,0,1)\times (1,0,0)=(0,1,0)$. ${\mathbf{v}}^{\prime }=\mathbf{v}\cos 90^{\circ }+(\hat{n}\times \mathbf{v})\sin 90^{\circ }=0+(0,1,0)=(0,1,0).$ Dikhta kya hai: $+x$ arrow quarter-turn counter-clockwise ghoom ke $+y$ par aa jaata hai (neeche figure dekho).

Yahan $\mathbf{v}$ axis $\hat{n}=(0,0,1)$ ke along point karta hai.

• $\hat{n}\times \mathbf{v}=0$, $\hat{n}\cdot \mathbf{v}=5$. ${\mathbf{v}}^{\prime }=(0,0,5)\cos 90^{\circ }+0+\hat{n}(5)(1-\cos 90^{\circ })=(0,0,0)+(0,0,5)=(0,0,5).$ Unchanged. $z$ ke around spin karne se kuch jo already $z$ ke along point kar raha hai, move nahi ho sakta.

$\theta =120^{\circ }$: $\cos 120^{\circ }=-\frac{1}{2}$, $\sin 120^{\circ }=\frac{\sqrt{3}}{2}$. $\hat{n}=\frac{1}{\sqrt{3}}(1,1,1)$, $\mathbf{v}=(1,0,0)$.

• $\hat{n}\cdot \mathbf{v}=\frac{1}{\sqrt{3}}$.
• $\hat{n}\times \mathbf{v}=\frac{1}{\sqrt{3}}(1,1,1)\times (1,0,0)=\frac{1}{\sqrt{3}}(0,1,-1)$.

Term by term: $\mathbf{v}\cos \theta =(-\frac{1}{2},0,0),$ $(\hat{n}\times \mathbf{v})\sin \theta =\frac{1}{\sqrt{3}}(0,1,-1)\cdot \frac{\sqrt{3}}{2}=(0,\frac{1}{2},-\frac{1}{2}),$ $\hat{n}(\hat{n}\cdot \mathbf{v})(1-\cos \theta )=\frac{1}{\sqrt{3}}(1,1,1)\cdot \frac{1}{\sqrt{3}}\cdot \frac{3}{2}=(\frac{1}{2},\frac{1}{2},\frac{1}{2}).$ Add karo: ${\mathbf{v}}^{\prime }=(-\frac{1}{2}+0+\frac{1}{2},0+\frac{1}{2}+\frac{1}{2},0-\frac{1}{2}+\frac{1}{2})=(0,1,0).$ Dikhta kya hai: body-diagonal ke around $120^{\circ }$ turn cyclically map karta hai $x\to y\to z\to x$; sach mein $+x\to +y$. ✓

Sandwich hai $(-q)v(-q{)}^{-1}$. Kyunki $(-q{)}^{-1}=-q^{-1}$ (har entry flip karne se inverse bhi flip hota hai), $(-q)v(-q{)}^{-1}=(-1)(-1)qv{q}^{-1}=qv{q}^{-1}.$ Dono minus signs cancel ho jaate hain. Numerically dono L2.1 se $(0,1,0)$ dete hain. Matlab: $q$ aur $-q$ same physical rotation hain — quaternions rotations ko double-cover karte hain.

$q_x=\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},0,0\right)$, $q_z=\left(\frac{\sqrt{2}}{2},0,0,\frac{\sqrt{2}}{2}\right)$.

(a) pehle $x$. $(0,0,1)$ ko $x$ ke around $90^{\circ }$ rotate karo: $\hat{n}=(1,0,0)$, $\hat{n}\times \mathbf{v}=(0,-1,0)$, $\hat{n}\cdot \mathbf{v}=0$ ⇒ $(0,0,1)\to (0,-1,0)$. Phir $z$ ke around $90^{\circ }$: $\hat{n}=(0,0,1)$, $(0,0,1)\times (0,-1,0)=(1,0,0)$ ⇒ $(0,-1,0)\to (1,0,0)$. Result (a): $(1,0,0)$.

(b) pehle $z$. $(0,0,1)$ ko $z$ ke around $90^{\circ }$ rotate karo: yeh on-axis hai ⇒ $(0,0,1)$ hi rehta hai. Phir $x$ ke around $90^{\circ }$: $(0,0,1)\to (0,-1,0)$. Result (b): $(0,-1,0)$.

$(1,0,0)\ne (0,-1,0)$ ⇒ order matter karta hai. Single-quaternion form mein yeh clearly dikhta hai: (a) hai $q_z{q}_x$, (b) hai $q_x{q}_z$, aur $q_z{q}_x\ne q_x{q}_z$.

Product ka conjugate lo (yeh order reverse karta hai): $(qv{q}^{-1}{)}^{\ast }=(q^{-1}{)}^{\ast }{v}^{\ast }{q}^{\ast }$. Unit $q$ ke liye: $q^{-1}=q^{\ast }$, toh $(q^{-1}{)}^{\ast }=q$ aur $q^{\ast }=q^{-1}$. Ek pure quaternion satisfy karta hai $v^{\ast }=-v$. Substitute karo: $(qv{q}^{-1}{)}^{\ast }=q(-v)q^{-1}=-qv{q}^{-1}.$ Jo quaternion apne conjugate ke minus ke barabar hota hai, uska scalar part $w$ aisa hota hai ki $w=-w\Rightarrow w=0$. Isliye $v^{\prime }$ pure hai — ek legal 3D vector. ∎

Axis: dono vectors ke perpendicular ⇒ unke cross product ka direction. $\hat{n}\propto (1,0,0)\times (0,1,0)=(0,0,1)$, already unit hai. Angle: $(1,0,0)$ aur $(0,1,0)$ ke beech angle $90^{\circ }$ hai (dot product $=0$). Half-angle build: $q=\left(\cos 45^{\circ },\sin 45^{\circ }(0,0,1)\right)=\left(\frac{\sqrt{2}}{2},0,0,\frac{\sqrt{2}}{2}\right).$ L2.1 se check karo: yeh $q$ sach mein $(1,0,0)\to (0,1,0)$ bhejta hai. ✓

$q_1=\left(\frac{\sqrt{2}}{2},0,0,\frac{\sqrt{2}}{2}\right)$, $q_2=\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},0,0\right)$. Hamilton product $q_2{q}_1=(w_2{w}_1-{\mathbf{v}}_2\cdot {\mathbf{v}}_1,w_2{\mathbf{v}}_1+w_1{\mathbf{v}}_2+{\mathbf{v}}_2\times {\mathbf{v}}_1)$ with $w_1=w_2=\frac{\sqrt{2}}{2}$, ${\mathbf{v}}_1=(0,0,\frac{\sqrt{2}}{2})$, ${\mathbf{v}}_2=(\frac{\sqrt{2}}{2},0,0)$:

• scalar: $\frac{1}{2}-0=\frac{1}{2}$.
• $w_2{\mathbf{v}}_1=(0,0,\frac{1}{2})$, $w_1{\mathbf{v}}_2=(\frac{1}{2},0,0)$, ${\mathbf{v}}_2\times {\mathbf{v}}_1=(\frac{\sqrt{2}}{2},0,0)\times (0,0,\frac{\sqrt{2}}{2})=(0,-\frac{1}{2},0).$
• vector part: $(\frac{1}{2},-\frac{1}{2},\frac{1}{2})$.

Toh $q=(\frac{1}{2},\frac{1}{2},-\frac{1}{2},\frac{1}{2})$. Norm check karo: $4\cdot \frac{1}{4}=1$ ✓. Axis–angle: $\cos \frac{\theta }{2}=\frac{1}{2}\Rightarrow \theta =120^{\circ }$; vector length $\sin \frac{\theta }{2}=\frac{\sqrt{3}}{2}$, toh $\hat{n}=\frac{(\frac{1}{2},-\frac{1}{2},\frac{1}{2})}{\sqrt{3}/2}=\frac{1}{\sqrt{3}}(1,-1,1).$ Read: do $90^{\circ }$ turns milkar ek $120^{\circ }$ turn bante hain $\frac{1}{\sqrt{3}}(1,-1,1)$ ke around.

Pehle $q$ lagao phir $q^{-1}$: combined quaternion (rightmost first) hai $q^{-1}q$. Lekin $q^{-1}q=(1,0,0,0)$, identity quaternion. Tab $v^{\prime }=(q^{-1}q)v(q^{-1}q{)}^{-1}=1\cdot v\cdot 1=v.$ Har vector ghar wapas aata hai. Isliye $q^{-1}$ literally "undo" rotation hai.

$\cos \frac{\theta }{2}=1\Rightarrow \theta =0$: identity (koi rotation nahi). Vector part hai $(0,0,0)$, toh $\sin \frac{\theta }{2}=0$ aur axis undefined hai — tum zero se divide nahi kar sakte. Interpretation: zero-angle rotation ka koi preferred axis nahi hota; har axis kaam karta hai kyunki kuch move nahi hota. Software ko yeh special-case karna padta hai (jaise identity axis return karo) zero se $(0,0,0)/0$ divide karne ki jagah.

$\cos 90^{\circ }=0$, $\sin 90^{\circ }=1$, toh $q=(0,\hat{n})$ — ek pure quaternion (zero scalar). Yeh woh borderline hai jahan $q$ aur $-q$ ko alag karna sabse mushkil hota hai, aur jahan scalar koi angle info nahi deta; axis poori tarah vector part mein rehta hai. $\hat{n}=(0,0,1)$ ke saath, Rodrigues: $\cos 180^{\circ }=-1,\sin 180^{\circ }=0$, $\hat{n}\cdot \mathbf{v}=0$: ${\mathbf{v}}^{\prime }=(1,0,0)(-1)+0+0=(-1,0,0).$ $z$ ke around half-turn $+x$ ko $-x$ kar deta hai. ✓ (Figure dekho.)

L2.3 se, $\hat{x}=(1,0,0)\to (0,1,0)$. Usi cyclic symmetry se, $\hat{y}=(0,1,0)\to (0,0,1)$.

• Lengths: $|(0,1,0)|=1$, $|(0,0,1)|=1$ — preserved.
• Dot product: $(0,1,0)\cdot (0,0,1)=0$ — abhi bhi perpendicular. Kyun aisa hona hi chahiye: $qv{q}^{-1}$ dot products preserve karta hai (yeh ek orthogonal transformation hai), toh angles aur lengths untouched rehte hain — yahi cheez isse genuine rotation banati hai na ki stretch. Yeh quaternion route hai Rotation matrices & orthogonality ki taraf.

Constant $\mathbf{\omega }$ ke saath $z$ ke along, $\dot{q}=\frac{1}{2}q\omega$ ka solution ek steady rotation hai jiska angle $\theta (t)=\Omega t$ ke roop mein $z$ ke around badhta hai: $q(t)=\left(\cos \frac{\Omega t}{2},0,0,\sin \frac{\Omega t}{2}\right).$ $t=0$ par yeh $(1,0,0,0)$ hai ✓. Half-angle phir se dikha — rate $q$ ke andar $\frac{1}{2}\Omega$ hai lekin body physically $\Omega$ par turn karti hai. Yeh connect hota hai > Angular velocity & quaternion kinematics $\dot q=\tfrac12 q\omega$ aur > Spacecraft attitude determination (GNC) se.

$q$ half-angle se banao, sandwich $qv{q}^{-1}$ se rotate karo (ya Rodrigues full angle ke saath), composition mein order respect karo, aur degenerate ($\theta =0$) aur $180^{\circ }$ axes ko special-case karo. Related: Rodrigues' rotation formula, SLERP — quaternion interpolation, Euler angles & gimbal lock, Quaternion algebra & Hamilton product.