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

WHAT we do: match $q=(w,\mathbf{v})=(\cos \frac{\theta }{2},\sin \frac{\theta }{2}\hat{n})$. The scalar part is $w=\cos \frac{\theta }{2}=\frac{\sqrt{3}}{2}$, so $\frac{\theta }{2}=30^{\circ }$ and $\theta =60^{\circ }.$ The vector part is $(0,0,\frac{1}{2})=\sin \frac{\theta }{2}\hat{n}$. Since $\sin 30^{\circ }=\frac{1}{2}$, the axis is $\hat{n}=(0,0,1)$ — the $z$-axis. Answer: a $60^{\circ }$ rotation about $+z$.

A rotation quaternion must have length $1$. Compute $\Vert q\Vert =\sqrt{1^2+1^2+0^2+0^2}=\sqrt{2}\ne 1.$ So no, it is not valid. Normalize by dividing by the length: $q_{\text{unit}}=\frac{1}{\sqrt{2}}(1,1,0,0)=\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},0,0\right).$ Now $w=\cos \frac{\theta }{2}=\frac{\sqrt{2}}{2}\Rightarrow \theta =90^{\circ }$ about the $x$-axis.

For a unit quaternion, inverse = conjugate = flip the sign of the vector part only: $q^{-1}=q^{\ast }=\left(\frac{\sqrt{2}}{2},0,-\frac{\sqrt{2}}{2},0\right).$ Geometric read: $q$ rotates $+90^{\circ }$ about $+y$; $q^{-1}$ rotates $-90^{\circ }$ about $+y$ — the undo.

Use Rodrigues with $\theta =90^{\circ }$ ($\cos =0,\sin =1$), $\hat{n}=(0,0,1)$, $\mathbf{v}=(1,0,0)$.

• $\hat{n}\cdot \mathbf{v}=0$ (so the last term vanishes),
• $\hat{n}\times \mathbf{v}=(0,0,1)\times (1,0,0)=(0,1,0)$. ${\mathbf{v}}^{\prime }=\mathbf{v}\cos 90^{\circ }+(\hat{n}\times \mathbf{v})\sin 90^{\circ }=0+(0,1,0)=(0,1,0).$ What it looks like: the $+x$ arrow swings a quarter-turn counter-clockwise onto $+y$ (see figure below).

Here $\mathbf{v}$ points along the axis $\hat{n}=(0,0,1)$.

• $\hat{n}\times \mathbf{v}=0$, $\hat{n}\cdot \mathbf{v}=5$. ${\mathbf{v}}^{\prime }=(0,0,5)\cos 90^{\circ }+0+\hat{n}(5)(1-\cos 90^{\circ })=(0,0,0)+(0,0,5)=(0,0,5).$ Unchanged. A spin about $z$ cannot move something already pointing along $z$.

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

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

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: ${\mathbf{v}}^{\prime }=(-\frac{1}{2}+0+\frac{1}{2},0+\frac{1}{2}+\frac{1}{2},0-\frac{1}{2}+\frac{1}{2})=(0,1,0).$ What it looks like: a $120^{\circ }$ turn about the body-diagonal cyclically maps $x\to y\to z\to x$; indeed $+x\to +y$. ✓

The sandwich is $(-q)v(-q{)}^{-1}$. Because $(-q{)}^{-1}=-q^{-1}$ (flipping every entry flips the inverse too), $(-q)v(-q{)}^{-1}=(-1)(-1)qv{q}^{-1}=qv{q}^{-1}.$ The two minus signs cancel. Numerically both give $(0,1,0)$ from L2.1. Meaning: $q$ and $-q$ are the same physical rotation — quaternions double-cover rotations.

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

(a) $x$ first. Rotate $(0,0,1)$ by $90^{\circ }$ about $x$: $\hat{n}=(1,0,0)$, $\hat{n}\times \mathbf{v}=(0,-1,0)$, $\hat{n}\cdot \mathbf{v}=0$ ⇒ $(0,0,1)\to (0,-1,0)$. Then $90^{\circ }$ about $z$: $\hat{n}=(0,0,1)$, $(0,0,1)\times (0,-1,0)=(1,0,0)$ ⇒ $(0,-1,0)\to (1,0,0)$. Result (a): $(1,0,0)$.

(b) $z$ first. Rotate $(0,0,1)$ by $90^{\circ }$ about $z$: it's on-axis ⇒ stays $(0,0,1)$. Then $90^{\circ }$ about $x$: $(0,0,1)\to (0,-1,0)$. Result (b): $(0,-1,0)$.

$(1,0,0)\ne (0,-1,0)$ ⇒ order matters. The single-quaternion form makes it obvious: (a) is $q_z{q}_x$, (b) is $q_x{q}_z$, and $q_z{q}_x\ne q_x{q}_z$.

Take the conjugate of a product (it reverses order): $(qv{q}^{-1}{)}^{\ast }=(q^{-1}{)}^{\ast }{v}^{\ast }{q}^{\ast }$. For unit $q$: $q^{-1}=q^{\ast }$, so $(q^{-1}{)}^{\ast }=q$ and $q^{\ast }=q^{-1}$. A pure quaternion satisfies $v^{\ast }=-v$. Substitute: $(qv{q}^{-1}{)}^{\ast }=q(-v)q^{-1}=-qv{q}^{-1}.$ A quaternion equal to minus its own conjugate has scalar part $w$ with $w=-w\Rightarrow w=0$. Hence $v^{\prime }$ is pure — a legal 3D vector. ∎

Axis: perpendicular to both vectors ⇒ their cross product direction. $\hat{n}\propto (1,0,0)\times (0,1,0)=(0,0,1)$, already unit. Angle: between $(1,0,0)$ and $(0,1,0)$ the angle is $90^{\circ }$ (dot product $=0$). Half-angle build: $q=\left(\cos 45^{\circ },\sin 45^{\circ }(0,0,1)\right)=\left(\frac{\sqrt{2}}{2},0,0,\frac{\sqrt{2}}{2}\right).$ Check with L2.1: this $q$ indeed sends $(1,0,0)\to (0,1,0)$. ✓

$q_1=\left(\frac{\sqrt{2}}{2},0,0,\frac{\sqrt{2}}{2}\right)$, $q_2=\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},0,0\right)$. 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: $(\frac{1}{2},-\frac{1}{2},\frac{1}{2})$.

So $q=(\frac{1}{2},\frac{1}{2},-\frac{1}{2},\frac{1}{2})$. Check norm: $4\cdot \frac{1}{4}=1$ ✓. Axis–angle: $\cos \frac{\theta }{2}=\frac{1}{2}\Rightarrow \theta =120^{\circ }$; vector length $\sin \frac{\theta }{2}=\frac{\sqrt{3}}{2}$, so $\hat{n}=\frac{(\frac{1}{2},-\frac{1}{2},\frac{1}{2})}{\sqrt{3}/2}=\frac{1}{\sqrt{3}}(1,-1,1).$ Read: two $90^{\circ }$ turns fuse into one $120^{\circ }$ turn about $\frac{1}{\sqrt{3}}(1,-1,1)$.

Apply $q$ then $q^{-1}$: the combined quaternion (rightmost first) is $q^{-1}q$. But $q^{-1}q=(1,0,0,0)$, the identity quaternion. Then $v^{\prime }=(q^{-1}q)v(q^{-1}q{)}^{-1}=1\cdot v\cdot 1=v.$ Every vector comes home. This is why $q^{-1}$ is literally the "undo" rotation.

$\cos \frac{\theta }{2}=1\Rightarrow \theta =0$: the identity (no rotation). The vector part is $(0,0,0)$, so $\sin \frac{\theta }{2}=0$ and the axis is undefined — you cannot divide by zero. Interpretation: a zero-angle rotation has no preferred axis; every axis works because nothing moves. Software must special-case this (e.g. return the identity axis) instead of dividing $(0,0,0)/0$.

$\cos 90^{\circ }=0$, $\sin 90^{\circ }=1$, so $q=(0,\hat{n})$ — a pure quaternion (zero scalar). This is the borderline where $q$ and $-q$ are hardest to tell apart, and where the scalar carries no angle info; the axis lives entirely in the vector part. With $\hat{n}=(0,0,1)$, Rodrigues: $\cos 180^{\circ }=-1,\sin 180^{\circ }=0$, $\hat{n}\cdot \mathbf{v}=0$: ${\mathbf{v}}^{\prime }=(1,0,0)(-1)+0+0=(-1,0,0).$ A half-turn about $z$ flips $+x$ to $-x$. ✓ (See figure.)

From L2.3, $\hat{x}=(1,0,0)\to (0,1,0)$. By the same cyclic symmetry, $\hat{y}=(0,1,0)\to (0,0,1)$.

• Lengths: $|(0,1,0)|=1$, $|(0,0,1)|=1$ — preserved.
• Dot product: $(0,1,0)\cdot (0,0,1)=0$ — still perpendicular. Why this must happen: $qv{q}^{-1}$ preserves dot products (it's an orthogonal transformation), so angles and lengths are untouched — this is what makes it a genuine rotation and not a stretch. This is the quaternion route to Rotation matrices & orthogonality.

With constant $\mathbf{\omega }$ along $z$, the solution of $\dot{q}=\frac{1}{2}q\omega$ is a steady rotation whose angle grows as $\theta (t)=\Omega t$ about $z$: $q(t)=\left(\cos \frac{\Omega t}{2},0,0,\sin \frac{\Omega t}{2}\right).$ At $t=0$ this is $(1,0,0,0)$ ✓. The half-angle reappears — the rate is $\frac{1}{2}\Omega$ inside $q$ but the body physically turns at $\Omega$. This connects to > Angular velocity & quaternion kinematics $\dot q=\tfrac12 q\omega$ and > Spacecraft attitude determination (GNC).

Build $q$ with the half-angle, rotate with the sandwich $qv{q}^{-1}$ (or Rodrigues with the full angle), respect order in composition, and special-case the degenerate ($\theta =0$) and $180^{\circ }$ axes. Related: Rodrigues' rotation formula, SLERP — quaternion interpolation, Euler angles & gimbal lock, Quaternion algebra & Hamilton product.