Question bank — Quaternions — definition q = (q₀, q₁, q₂, q₃), unit quaternion constraint

False. Only unit quaternions ($\sum q_i^2=1$) represent rotations; a general 4-tuple like $(2,0,0,0)$ has norm $2$ and must be normalized first.

True. Since $q_0=\cos \frac{\theta }{2}$ and $\frac{\theta }{2}$ ranges over $[0,\pi ]$ for $\theta \in [0,2\pi ]$, $\cos \frac{\theta }{2}$ sweeps the full interval $[-1,1]$.

False. The vector part contains $\vec{p}\times \vec{q}$, which flips sign when you swap $p$ and $q$, so they differ unless the axes are parallel — mirroring that 3D rotations don't commute.

False. They describe the same rotation: in the sandwich $qv{q}^{-1}$ both minus signs cancel. This is the double cover $S^3\to SO(3)$.

False. $(-1,0,0,0)$ also has zero vector part and unit norm; it equals a $360^{\circ }$ rotation, which is the same physical orientation as identity.

True. $q^{-1}=q^{\ast }/\Vert q{\Vert }^2$, and $\Vert q{\Vert }^2=1$ collapses this to $q^{-1}=q^{\ast }=(q_0,-{\vec{q}}_v)$.

False in general. Scalar part $\cos \frac{\theta }{2}=0$ forces $\theta =180^{\circ }$, so only unit pure quaternions represent a specific $180^{\circ }$ rotation; a non-unit pure quaternion represents nothing.

True. The quaternion norm is multiplicative, so multiplying two unit quaternions gives another unit quaternion — composing rotations keeps you on $S^3$.

Error: full angle used. Quaternions take the half-angle, so $q=(\cos 45^{\circ },0,0,\sin 45^{\circ })=(\frac{\sqrt{2}}{2},0,0,\frac{\sqrt{2}}{2})$.

Error. $-q$ is the same rotation, not the inverse. The inverse conjugates only the vector part: $q^{-1}=(q_0,-q_1,-q_2,-q_3)$.

Error. The stored vector part is $\hat{n}\sin \frac{\theta }{2}$, i.e. the axis scaled by $\sin \frac{\theta }{2}$. You recover $\hat{n}$ by dividing $(q_1,q_2,q_3)$ by its own length.

Error. Discrete integration accumulates rounding and slowly drifts off $S^3$; you must renormalize each timestep. (See Quaternion Kinematics — dq/dt.)

Error unless unit. $q\otimes q^{\ast }=(\Vert q{\Vert }^2,\vec{0})$. Only when $\Vert q\Vert =1$ does this equal $(1,0,0,0)$; otherwise it's a scalar $\ne 1$.

Error (order). Applying $p$ first then $q$ is $q\otimes p$ — the later rotation sits on the left, just like matrix multiplication order.

The 4th number is redundancy removed by one unit constraint ($4-1=3$); this smooth 4D packing is exactly what avoids the singularity (gimbal lock) that plagues 3-number Euler angles.

A vector is rotated by the double-sided product $v^{\prime }=q\otimes v\otimes q^{-1}$, so $q$ is effectively applied twice; each factor must carry half the total angle so the sandwich delivers the full $\theta$.

They live on the smooth, closed surface of the 3-sphere $S^3$ with no coordinate axes to "align" — the singularity in Euler Angles and Gimbal Lock comes from a bad chart, not from rotations themselves.

A quaternion inverse just negates three signs ($q^{\ast }$), whereas the matrix route needs a transpose and orthogonality checks — a real win on limited spacecraft compute.

Real 3D rotations don't commute (spin-then-tilt $\ne$ tilt-then-spin); the $\vec{p}\times \vec{q}$ term encodes exactly this order-dependence.

Every unit quaternion sits at distance $1$ from the origin in 4D; dividing a drifted $q$ by its length projects it radially back onto $S^3$ without changing the direction (hence the rotation) it points to.

In 2D a unit complex number $\cos \theta +i\sin \theta$ rotates the plane; quaternions add two more imaginary units $(j,k)$ so a single object can rotate 3D space the same clean way.

The identity $(1,0,0,0)$: $\theta =0$ gives $\cos 0=1$ and $\hat{n}\sin 0=\vec{0}$; multiplying by it leaves any quaternion unchanged, like multiplying by $1$.

$\cos 90^{\circ }=0$, so $q=(0,\hat{n})$ is a pure quaternion; the axis is fully visible but the two half-rotations $q$ and $-q$ become maximally separated 4-tuples for the same flip.

$\theta =360^{\circ }$ gives $\cos 180^{\circ }=-1$, so $q=(-1,0,0,0)$. Physically the same orientation as identity, yet the quaternion is $-1\times$ the identity — the signature of the double cover.

No. The norm derivation relied on $n_x^2+n_y^2+n_z^2=1$; a non-unit axis breaks the unit constraint, so you must normalize $\hat{n}$ before building $q$.

Yes. The vector part is $\hat{n}\sin 0=\vec{0}$ regardless of which axis you picked, so the undefined axis is harmless and $q=(1,\vec{0})$ is unambiguous.

Yes — exactly the antipodal pair $q$ and $-q$. No other coincidences occur, so the map $S^3\to SO(3)$ is precisely two-to-one.