Exercises — Quaternion product — Hamilton product

Dot product ::: $\mathbf{a}\cdot \mathbf{b}=a_x{b}_x+a_y{b}_y+a_z{b}_z$ — ek number, symmetric ($\mathbf{a}\cdot \mathbf{b}=\mathbf{b}\cdot \mathbf{a}$). Cross product ::: $\mathbf{a}\times \mathbf{b}=(a_y{b}_z-a_z{b}_y,a_z{b}_x-a_x{b}_z,a_x{b}_y-a_y{b}_x)$ — ek vector, anti-symmetric ($\mathbf{a}\times \mathbf{b}=-\mathbf{b}\times \mathbf{a}$). Conjugate ::: $\bar{q}=(w,-\mathbf{v})$ — vector part ka sign flip kar do. Norm ::: $|q|=\sqrt{w^2+x^2+y^2+z^2}$.

Scalar part hai $w_1{w}_2-{\mathbf{v}}_1\cdot {\mathbf{v}}_2$.

• $w_1{w}_2=2\cdot 3=6$.
• ${\mathbf{v}}_1\cdot {\mathbf{v}}_2=(1)(0)+(0)(2)+(-1)(1)=-1$.
• Scalar $=6-(-1)=\boxed{7}$.

Humne kya kiya: sirf formula ka pehla slot use kiya. Kyun: question ne sirf scalar maanga tha, toh vector slot yahan irrelevant hai.

Vector part $=w_1{\mathbf{v}}_2+w_2{\mathbf{v}}_1+{\mathbf{v}}_1\times {\mathbf{v}}_2$.

• $w_1{\mathbf{v}}_2=2(0,2,1)=(0,4,2)$.
• $w_2{\mathbf{v}}_1=3(1,0,-1)=(3,0,-3)$.
• ${\mathbf{v}}_1\times {\mathbf{v}}_2$ with ${\mathbf{v}}_1=(1,0,-1)$, ${\mathbf{v}}_2=(0,2,1)$:
  • $x:(0)(1)-(-1)(2)=2$
  • $y:(-1)(0)-(1)(1)=-1$
  • $z:(1)(2)-(0)(0)=2$
  • $\Rightarrow (2,-1,2)$.
• Sum: $(0+3+2,4+0-1,2-3+2)=\boxed{(5,3,1)}$.

Toh $pq=(7,5,3,1)$.

$\theta =180^{\circ }\Rightarrow \frac{\theta }{2}=90^{\circ }$.

• $\cos 90^{\circ }=0$, $\sin 90^{\circ }=1$.
• $\hat{n}=(0,1,0)$.
• $q=(0,(0,1,0))=\boxed{(0,0,1,0)}$, yaani unit $j$.

Half-angle kyun: ek vector ko rotate karne ke liye sandwich $qu{q}^{-1}$ use hota hai, jo $q$ ko effectively "do baar" apply karta hai, isliye har $q$ ko sirf aadha angle carry karna chahiye. (Dekho Axis-Angle & Euler Rodrigues.)

Har quaternion banao ($\frac{\theta }{2}=45^{\circ }$, $\cos 45^{\circ }=\sin 45^{\circ }=\frac{1}{\sqrt{2}}$):

• $q_x=(\frac{1}{\sqrt{2}},(\frac{1}{\sqrt{2}},0,0))$
• $q_y=(\frac{1}{\sqrt{2}},(0,\frac{1}{\sqrt{2}},0))$

Yahan "first" $=q_x$, "last" $=q_y$, toh $q_y{q}_x$ compute karo jahan $p=q_y,q=q_x$:

• Scalar: $w_1{w}_2-{\mathbf{v}}_1\cdot {\mathbf{v}}_2=\frac{1}{2}-(0,\frac{1}{\sqrt{2}},0)\cdot (\frac{1}{\sqrt{2}},0,0)=\frac{1}{2}-0=\frac{1}{2}$.
• Vector:
  • $w_1{\mathbf{v}}_2=\frac{1}{\sqrt{2}}(\frac{1}{\sqrt{2}},0,0)=(\frac{1}{2},0,0)$
  • $w_2{\mathbf{v}}_1=\frac{1}{\sqrt{2}}(0,\frac{1}{\sqrt{2}},0)=(0,\frac{1}{2},0)$
  • ${\mathbf{v}}_1\times {\mathbf{v}}_2=(0,\frac{1}{\sqrt{2}},0)\times (\frac{1}{\sqrt{2}},0,0)$:
    • $x:(\frac{1}{\sqrt{2}})(0)-(0)(0)=0$
    • $y:(0)(\frac{1}{\sqrt{2}})-(0)(0)=0$
    • $z:(0)(0)-(\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}})=-\frac{1}{2}$
    • $\Rightarrow (0,0,-\frac{1}{2})$
  • Sum: $(\frac{1}{2},\frac{1}{2},-\frac{1}{2})$.
• $q_y{q}_x=\boxed{(\frac{1}{2},\frac{1}{2},\frac{1}{2},-\frac{1}{2})}$.
• Norm check: $\sqrt{4\cdot \frac{1}{4}}=1$ ✅.

Figure mein kaisa dikhta hai: red arrow pehle apply hota hai ($x$ ke around), mint arrow doosra ($y$ ke around); $q_y{q}_x$ mein right-to-left padhna us timeline se match karta hai.

$pq$ ka scalar: $w_1{w}_2-{\mathbf{v}}_1\cdot {\mathbf{v}}_2$. $qp$ ka scalar: $w_2{w}_1-{\mathbf{v}}_2\cdot {\mathbf{v}}_1$. Kyunki ordinary multiplication commute karta hai ($w_1{w}_2=w_2{w}_1$) aur dot symmetric hai (${\mathbf{v}}_1\cdot {\mathbf{v}}_2={\mathbf{v}}_2\cdot {\mathbf{v}}_1$), dono scalars identical hain. ✅

$pq$ ka vector: $w_1{\mathbf{v}}_2+w_2{\mathbf{v}}_1+{\mathbf{v}}_1\times {\mathbf{v}}_2$. $qp$ ka vector: $w_2{\mathbf{v}}_1+w_1{\mathbf{v}}_2+{\mathbf{v}}_2\times {\mathbf{v}}_1$. Pehle do terms match karte hain (addition commutes). Difference hai ${\mathbf{v}}_1\times {\mathbf{v}}_2-{\mathbf{v}}_2\times {\mathbf{v}}_1=2({\mathbf{v}}_1\times {\mathbf{v}}_2),$ anti-symmetry ${\mathbf{v}}_2\times {\mathbf{v}}_1=-{\mathbf{v}}_1\times {\mathbf{v}}_2$ use karke. Toh $\boxed{pq-qp=(0,2{\mathbf{v}}_1\times {\mathbf{v}}_2)}$ — ek pure quaternion. Cross product hi akela culprit hai.

Parallel ka matlab hai ${\mathbf{v}}_2=\lambda {\mathbf{v}}_1$ kisi scalar $\lambda$ ke liye. Tab ${\mathbf{v}}_1\times {\mathbf{v}}_2={\mathbf{v}}_1\times (\lambda {\mathbf{v}}_1)=\lambda ({\mathbf{v}}_1\times {\mathbf{v}}_1)=0.$ L3.1 se, $pq-qp=(0,2\cdot 0)=(0,0)$, toh $pq=qp$. ✅

Physical meaning: same axis ke around do rotations commute karte hain ($z$ ke around $30^{\circ }$ phir $50^{\circ }$ = $80^{\circ }$ kisi bhi order mein). Sirf alag-axis wale rotations commute karne se mana karte hain — yahi Cross Product & Right-Hand Rule ka geometric core hai.

$\mathbf{u}$ ko pure quaternion $u=(0,(1,0,0))$ ke roop mein embed karo. Kyunki $q$ unit hai, $q^{-1}=\bar{q}=(\frac{1}{\sqrt{2}},(0,0,-\frac{1}{\sqrt{2}}))$.

Step 1 — inner product $t=qu$ ($p=q$, $q_{\text{arg}}=u$):

• Scalar: $w_1{w}_2-{\mathbf{v}}_1\cdot {\mathbf{v}}_2=\frac{1}{\sqrt{2}}\cdot 0-(0,0,\frac{1}{\sqrt{2}})\cdot (1,0,0)=0$.
• Vector: $w_1{\mathbf{v}}_2+w_2{\mathbf{v}}_1+{\mathbf{v}}_1\times {\mathbf{v}}_2$
  • $=\frac{1}{\sqrt{2}}(1,0,0)+0+(0,0,\frac{1}{\sqrt{2}})\times (1,0,0)$
  • cross: $(0,0,\frac{1}{\sqrt{2}})\times (1,0,0)=(0\cdot 0-\frac{1}{\sqrt{2}}\cdot 0,\frac{1}{\sqrt{2}}\cdot 1-0\cdot 0,0-0)=(0,\frac{1}{\sqrt{2}},0)$
  • $=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0)$.
• $t=(0,(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0))$.

Step 2 — outer product $u^{\prime }=t\bar{q}$ ($p=t$, $q_{\text{arg}}=\bar{q}$):

• $w_1=0,{\mathbf{v}}_1=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0)$; $w_2=\frac{1}{\sqrt{2}},{\mathbf{v}}_2=(0,0,-\frac{1}{\sqrt{2}})$.
• Scalar: $0\cdot \frac{1}{\sqrt{2}}-(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0)\cdot (0,0,-\frac{1}{\sqrt{2}})=0-0=0$ ✅ (rotated vector pure rehta hai).
• Vector: $w_1{\mathbf{v}}_2+w_2{\mathbf{v}}_1+{\mathbf{v}}_1\times {\mathbf{v}}_2$
  • $=0+\frac{1}{\sqrt{2}}(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0)+(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0)\times (0,0,-\frac{1}{\sqrt{2}})$
  • scaled term: $(\frac{1}{2},\frac{1}{2},0)$
  • cross: $x:\frac{1}{\sqrt{2}}\cdot (-\frac{1}{\sqrt{2}})-0=-\frac{1}{2}$; $y:0-\frac{1}{\sqrt{2}}\cdot (-\frac{1}{\sqrt{2}})=\frac{1}{2}$; $z:0-0=0\Rightarrow (-\frac{1}{2},\frac{1}{2},0)$
  • sum: $(\frac{1}{2}-\frac{1}{2},\frac{1}{2}+\frac{1}{2},0)=(0,1,0)$.
• $\boxed{u^{\prime }=(0,1,0)}$ — $x$-axis rotate hokar $y$-axis ban gaya. ✅

Kaisa dikhta hai: $z$ ke around $90^{\circ }$ counter-clockwise turn $\hat{x}\to \hat{y}$ bhejta hai, bilkul waise jaise figure mein dikhta hai.

Jis claim par lean karna hai: kisi bhi quaternions ke liye, $|pq|=|p||q|$. (Yeh isliye follow karta hai kyunki norm-squared $|q{|}^2=q\bar{q}$ multiplicative hai: $\overline{pq}=\bar{q}\bar{p}$, toh $|pq{|}^2=pq\overline{pq}=pq\bar{q}\bar{p}=p|q{|}^2\bar{p}=|q{|}^{2}p\bar{p}=|p{|}^2|q{|}^2$.)

$p=q_k$, $q=\delta q$ ke saath apply karo: $|q_{k+1}|=|q_k\delta q|=|q_k||\delta q|.$ Kyunki $\delta q$ ek unit quaternion hai, $|\delta q|=1$, isliye $|q_{k+1}|=|q_k|\cdot 1=|q_k|.\boxed{}$ $|q_0|=1$ se start karke, har $|q_k|=1$ hamesha ke liye — exact math mein. Quaternion Kinematics — $\dot q = \tfrac12 q\,\omega$ aur Multiplicative EKF (MEKF) dekho ki filters isko kaise exploit karte hain.

Maano $c=\cos 1^{\circ }\approx 0.9998477$, $s=\sin 1^{\circ }\approx 0.0174524$, $r=\frac{1}{\sqrt{2}}\approx 0.7071068$. $p=q_k=(r,(0,0,r))$, $q=\delta q=(c,(s,0,0))$.

• Scalar: $rc-(0,0,r)\cdot (s,0,0)=rc-0=rc\approx 0.7069991$.
• Vector: $r(s,0,0)+c(0,0,r)+(0,0,r)\times (s,0,0)$
  • $r(s,0,0)=(rs,0,0)\approx (0.0123407,0,0)$
  • $c(0,0,r)=(0,0,rc)\approx (0,0,0.7069991)$
  • cross $(0,0,r)\times (s,0,0)=(0\cdot 0-r\cdot 0,r\cdot s-0\cdot 0,0-0)=(0,rs,0)\approx (0,0.0123407,0)$
  • sum: $(rs,rs,rc)\approx (0.0123407,0.0123407,0.7069991)$
• $q_{k+1}\approx (0.7069991,0.0123407,0.0123407,0.7069991)$.
• Norm${}^2=(rc{)}^2+(rs{)}^2+(rs{)}^2+(rc{)}^2=2r^2{c}^2+2r^2{s}^2=2r^2(c^2+s^2)=2\cdot \frac{1}{2}\cdot 1=1$.
• $|q_{k+1}|=\boxed{1.000000}$ ✅.

Insight: exact $c^2+s^2=1$ collapse isliye hota hai ki unit-ness survive kare — lekin float mein, $c^2+s^2$ sirf machine precision tak $1$ hota hai, toh real filters periodically renormalize karte hain.