Exercises — Quaternion product — Hamilton product

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

The scalar part is $w_1{w}_2-{\mathbf{v}}_1\cdot {\mathbf{v}}_2$.

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

What we did: used only the first slot of the formula. Why: the question asked for the scalar alone, so the vector slot is irrelevant here.

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: $(0+3+2,4+0-1,2-3+2)=\boxed{(5,3,1)}$.

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

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

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

Why the half-angle: rotating a vector uses the sandwich $qu{q}^{-1}$, which applies $q$ effectively "twice", so each $q$ must carry only half the angle. (See Axis-Angle & Euler Rodrigues.)

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

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

Here "first" $=q_x$, "last" $=q_y$, so compute $q_y{q}_x$ with $p=q_y,q=q_x$:

• 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: $\sqrt{4\cdot \frac{1}{4}}=1$ ✅.

What it looks like (figure): the red arrow is applied first (about $x$), the mint arrow second (about $y$); reading right-to-left in $q_y{q}_x$ matches that timeline.

Scalar of $pq$: $w_1{w}_2-{\mathbf{v}}_1\cdot {\mathbf{v}}_2$. Scalar of $qp$: $w_2{w}_1-{\mathbf{v}}_2\cdot {\mathbf{v}}_1$. Since ordinary multiplication commutes ($w_1{w}_2=w_2{w}_1$) and dot is symmetric (${\mathbf{v}}_1\cdot {\mathbf{v}}_2={\mathbf{v}}_2\cdot {\mathbf{v}}_1$), the two scalars are identical. ✅

Vector of $pq$: $w_1{\mathbf{v}}_2+w_2{\mathbf{v}}_1+{\mathbf{v}}_1\times {\mathbf{v}}_2$. Vector of $qp$: $w_2{\mathbf{v}}_1+w_1{\mathbf{v}}_2+{\mathbf{v}}_2\times {\mathbf{v}}_1$. The first two terms match (addition commutes). The difference is ${\mathbf{v}}_1\times {\mathbf{v}}_2-{\mathbf{v}}_2\times {\mathbf{v}}_1=2({\mathbf{v}}_1\times {\mathbf{v}}_2),$ using anti-symmetry ${\mathbf{v}}_2\times {\mathbf{v}}_1=-{\mathbf{v}}_1\times {\mathbf{v}}_2$. So $\boxed{pq-qp=(0,2{\mathbf{v}}_1\times {\mathbf{v}}_2)}$ — a pure quaternion. The cross product is the sole culprit.

Parallel means ${\mathbf{v}}_2=\lambda {\mathbf{v}}_1$ for some scalar $\lambda$. Then ${\mathbf{v}}_1\times {\mathbf{v}}_2={\mathbf{v}}_1\times (\lambda {\mathbf{v}}_1)=\lambda ({\mathbf{v}}_1\times {\mathbf{v}}_1)=0.$ By L3.1, $pq-qp=(0,2\cdot 0)=(0,0)$, so $pq=qp$. ✅

Physical meaning: two rotations about the same axis commute (spin $30^{\circ }$ then $50^{\circ }$ about $z$ = $80^{\circ }$ either way). Only differently-axised rotations refuse to commute — the geometric heart of Cross Product & Right-Hand Rule.

Embed $\mathbf{u}$ as a pure quaternion $u=(0,(1,0,0))$. Since $q$ is unit, $q^{-1}=\bar{q}=(\frac{1}{\sqrt{2}},(0,0,-\frac{1}{\sqrt{2}}))$.

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: $0\cdot \frac{1}{\sqrt{2}}-(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0)\cdot (0,0,-\frac{1}{\sqrt{2}})=0-0=0$ ✅ (a rotated vector stays pure).
• 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: $(\frac{1}{2}-\frac{1}{2},\frac{1}{2}+\frac{1}{2},0)=(0,1,0)$.
• $\boxed{u^{\prime }=(0,1,0)}$ — the $x$-axis rotated to the $y$-axis. ✅

What it looks like: a $90^{\circ }$ counter-clockwise turn about $z$ sends $\hat{x}\to \hat{y}$, exactly as the figure shows.

Claim to lean on: for any quaternions, $|pq|=|p||q|$. (This follows because the norm-squared $|q{|}^2=q\bar{q}$ is multiplicative: $\overline{pq}=\bar{q}\bar{p}$, so $|pq{|}^2=pq\overline{pq}=pq\bar{q}\bar{p}=p|q{|}^2\bar{p}=|q{|}^{2}p\bar{p}=|p{|}^2|q{|}^2$.)

Apply with $p=q_k$, $q=\delta q$: $|q_{k+1}|=|q_k\delta q|=|q_k||\delta q|.$ Since $\delta q$ is a unit quaternion, $|\delta q|=1$, hence $|q_{k+1}|=|q_k|\cdot 1=|q_k|.\boxed{}$ Starting from $|q_0|=1$, every $|q_k|=1$ forever — in exact math. See Quaternion Kinematics — $\dot q = \tfrac12 q\,\omega$ and Multiplicative EKF (MEKF) for how filters exploit this.

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

• 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: the exact $c^2+s^2=1$ collapse is why unit-ness survives — but in float, $c^2+s^2$ is $1$ only to machine precision, so real filters renormalize periodically.