Question bank — Quaternion rotation formula — rotating vector v by quaternion q

False. A lone $qv$ generally produces a nonzero scalar part, so the result isn't even a pure quaternion (a 3D vector). A rotation matrix $R$ is a $3\times 3$ grid that already maps 3-vectors to 3-vectors, so one multiply $R\mathbf{v}$ suffices; a quaternion instead needs the conjugate on the right to keep the output a valid vector.

False. Both signs cancel in the sandwich: $(-q)v(-q{)}^{-1}=qv{q}^{-1}$. Quaternions double-cover rotations — every physical rotation has exactly two quaternion names.

False. $q$ stores the half-angle: $q=(\cos \frac{\theta }{2},\sin \frac{\theta }{2}\hat{n})$. Because $q$ acts twice in $qv{q}^{-1}$, each copy supplies $\theta /2$, and double-angle identities glue them into $\theta$.

True. The parallel component satisfies $\hat{n}\times {\mathbf{v}}_{\parallel }=0$, so the algebra returns ${\mathbf{v}}_{\parallel }$ untouched — you cannot spin a vector that already points down the spin axis.

False. The Hamilton product is non-commutative ($q_1{q}_2\ne q_2{q}_1$), exactly like rotation matrices, where swapping the order of two matrix multiplies changes the result. Rotating "roll then pitch" differs from "pitch then roll."

False in practice. Since $\Vert q\Vert =1$, the inverse is just the conjugate $q^{\ast }=(w,-\mathbf{v})$ — you only flip the sign of the vector part, no division needed.

False. They are the same rotation written two ways; the sandwich algebraically collapses to Rodrigues' rotation formula, which spins $\mathbf{v}$ by combining $\mathbf{v}\cos \theta$, a turning term $(\hat{n}\times \mathbf{v})\sin \theta$, and the fixed axial part. The quaternion is just cheaper storage (4 numbers vs a formula on the axis).

False. The vector's length is preserved and irrelevant to how you rotate; only the axis $\hat{n}$ must be a unit vector and $q$ must be unit. The vector itself can be any length.

Wrong: the inverse is the conjugate $q^{\ast }=(w,-\mathbf{v})$, not $q$ itself. Only if the vector part is zero (a pure scalar quaternion) would $q^{-1}=q$ hold.

Wrong order. Write the action explicitly: rotate by $q_1$ first, giving $q_{1}v{q}_1^{-1}$; then rotate that by $q_2$, giving $q_2(q_{1}v{q}_1^{-1})q_2^{-1}=(q_2{q}_1)v(q_2{q}_1{)}^{-1}$. So the combined quaternion is $q_2{q}_1$ — the one acting first sits nearest the vector, on the right — and $(q_2{q}_1{)}^{-1}=q_1^{-1}{q}_2^{-1}$ reverses the order.

Wrong: the vector must be a pure quaternion $(0,\mathbf{v})$ with scalar part zero. Putting $|\mathbf{v}|$ in the scalar slot corrupts the sandwich and the output is no longer a plain vector.

Wrong framing: the scalar part of the input is already $0$ (it's a pure quaternion). And the closure argument (see the "Why must the conjugate sit on the right?" item below) shows the output scalar part is also exactly $0$. There is no nonzero scalar to worry about.

Wrong: it's structural. Writing $c\equiv \cos \frac{\theta }{2}$ and $s\equiv \sin \frac{\theta }{2}$, the two half-angles from $q$ and $q^{-1}$ must recombine, and the identities $c^2-s^2=\cos \theta$, $2cs=\sin \theta$ are precisely the machinery that turns two half-angle contributions into one whole $\theta$.

Wrong: $\theta =360^{\circ }$ gives half-angle $180^{\circ }$, so $q=(\cos 180^{\circ },\sin 180^{\circ }\hat{n})=(-1,0)=-1$, not $+1$. It still acts as identity on vectors (signs cancel), but the quaternion itself is $-1$ — the hallmark of the double cover.

Wrong: renormalizing a near-unit quaternion nudges it back onto $\Vert q\Vert =1$ with a tiny change in the rotation — this cheapness is exactly why GNC systems favour quaternions (see Spacecraft attitude determination (GNC), where onboard filters renormalize every update).

Here is the full chain. We want the output to be pure (scalar part zero) so it is a genuine 3D vector. Take the conjugate of the sandwich using the rule "conjugate of a product reverses order and conjugates each factor": $(qv{q}^{-1}{)}^{\ast }=(q^{-1}{)}^{\ast }{v}^{\ast }{q}^{\ast }$. For unit $q$ we have $(q^{-1}{)}^{\ast }=q$ and $q^{\ast }=q^{-1}$; and since $v$ is pure, $v^{\ast }=-v$. Substituting gives $(qv{q}^{-1}{)}^{\ast }=q(-v)q^{-1}=-(qv{q}^{-1})$. An object equal to minus its own conjugate must have zero scalar part (because conjugation flips only the vector part; equalling your own negation forces the unflipped scalar to be $0$). Hence the output is pure. Using $qvq$ instead would break this cancellation.

The cross product $\hat{n}\times \mathbf{v}$ vanishes when $\mathbf{v}\parallel \hat{n}$, so the parallel piece has nothing to rotate. Only the perpendicular piece lives in the plane of rotation, where it swings and gains the $\sin \theta (\hat{n}\times {\mathbf{v}}_{\perp })$ turning term.

Quaternions have no gimbal lock and use 4 numbers with 1 constraint, whereas Euler angles can lose a degree of freedom when two rotation axes line up. Quaternions also interpolate smoothly — a shortest-arc blend between two orientations, covered in SLERP — quaternion interpolation.

Multiply two pure quaternions and watch the units. Take $\mathbf{a}=a_{1}i+a_{2}j+a_{3}k$ and $\mathbf{b}=b_{1}i+b_{2}j+b_{3}k$. Same-axis terms use $i^2=j^2=k^2=-1$: for instance $a_{1}i\cdot b_{1}i=a_1{b}_1{i}^2=-a_1{b}_1$, a pure scalar; summing all three gives $-(a_1{b}_1+a_2{b}_2+a_3{b}_3)=-(\mathbf{a}\cdot \mathbf{b})$ — that's the dot product in the scalar slot. Cross-axis terms use $ij=k,jk=i,ki=j$: for instance $a_{1}i\cdot b_{2}j=a_1{b}_{2}k$, a pure vector along $k$; collecting all such terms gives exactly $\mathbf{a}\times \mathbf{b}$ in the vector slot. So the split is just the $i,j,k$ table sorted into "scalar-making" and "vector-making" products — see Quaternion algebra & Hamilton product.

A unit quaternion has norm 1, and quaternion multiplication is multiplicative over norms: $\Vert qv{q}^{-1}\Vert =\Vert q\Vert \Vert v\Vert \Vert q^{-1}\Vert =\Vert v\Vert$. Length is conserved, as any true rotation demands.

Because $q_1$ sits next to $v$ in the nested sandwich $q_2(q_{1}v{q}_1^{-1})q_2^{-1}$, so it touches the vector first. This is the general operator-action rule: in $qv{q}^{-1}$ the quaternion nearest the vector acts first, and stacking sandwiches nests inward-to-outward. Left-to-right reading order and application order are therefore opposite, just like function composition $f(g(x))$ where $g$ runs first.

At $\theta =0$ the quaternion is $(1,0)$ — the vector part vanishes, so $\hat{n}$ is undefined. There's no rotation to speak of, hence no distinguished axis, just as a zero rotation has no meaningful direction.

$q=(\cos 0,\sin 0\hat{n})=(1,0)$, the identity. The sandwich returns $v$ unchanged — no axis is even needed since nothing moves.

Yes — $q=(0,\hat{n})$ is a pure quaternion, so the axis is clearly readable. The perpendicular component flips sign ($\cos 180^{\circ }=-1$, $\sin 180^{\circ }=0$), reflecting $v_{\perp }$ through the axis.

The zero vector, always: $q(0,0)q^{-1}=(0,0)$. A point at the origin can't be moved by any rotation about an axis through it.

No — the half-angle form $q=(\cos \frac{\theta }{2},\sin \frac{\theta }{2}\hat{n})$ assumes $\Vert \hat{n}\Vert =1$. A non-unit axis makes $\Vert q\Vert \ne 1$, breaking $q^{-1}=q^{\ast }$ and distorting the rotation. Normalize the axis first.

The sandwich now slightly scales the vector (length no longer preserved) because $\Vert q\Vert \ne 1$. The fix is cheap renormalization $q/\Vert q\Vert$, one of quaternions' key GNC advantages over accumulating rotation matrices.

The vector returns exactly to itself (a full turn is identity on 3D vectors). The quaternion does not: it becomes $-1$, needing $720^{\circ }$ to return to $+1$ — the geometric fingerprint of the double cover.

Same rotation. Flipping both the axis sign and the angle sign leaves $\sin \frac{\theta }{2}\hat{n}$ unchanged and $\cos \frac{\theta }{2}$ unchanged, so the quaternion — and the physical spin — is identical.