Quaternions — definition q = (q₀, q₁, q₂, q₃), unit quaternion constraint: Exercises

Before we start, one picture to fix the whole game in your head. A unit quaternion is a point sitting exactly one unit away from the centre in a 4-dimensional space — it lives on a sphere we cannot fully draw, so we draw a slice of it.

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

Look at the blue circle: it is the 2D shadow of the unit 3-sphere $S^{3}$ . The yellow dot is a valid rotation quaternion (it sits on the circle). The red dot has drifted off — its four squares no longer add to $1$ , so it is not a rotation until we renormalize it (the red arrow snaps it back).

L1 — Recognition

These test whether you can read a quaternion and check the one law it must obey.

Recall Solution 1.1

The test is the unit constraint: $q_{0}^{2} + q_{1}^{2} + q_{2}^{2} + q_{3}^{2} = 1$ . Compute each sum of squares.

(a) $1 + 0 + 0 + 0 = 1$ → valid ✔ (this is the identity, "do nothing").
(b) $0.25 \times 4 = 1$ → valid ✔.
(c) $0.36 + 0.64 = 1$ → valid ✔.
(d) $1 + 1 = 2 \neq = 1$ → not valid ✗ (norm is $2$ , must be normalized first).

Recall Solution 1.2

Scalar part = $q_{0} = 0.6$ (the "how much twist" piece).
Vector part $q_{v} = (q_{1}, q_{2}, q_{3}) = (0, 0.8, 0)$ (the "which axis" piece).
Conjugate flips only the vector part: $q^{*} = (0.6, 0, - 0.8, 0)$ .

L2 — Application

Now you build quaternions from an angle and axis, using the half-angle rule.

Recall Solution 2.1

Formula: $q = (cos \frac{θ}{2}, \overset{n}{^} sin \frac{θ}{2})$ . The half-angle is used because rotating a vector applies $q$ twice in the sandwich $q v q^{- 1}$ . Half-angle: $\frac{6 0 ^{\circ}}{2} = 3 0^{\circ}$ , so $cos 3 0^{\circ} = \frac{3}{2} \approx 0.8660$ , $sin 3 0^{\circ} = 0.5$ . $q = (\frac{3}{2}, 0, 0.5, 0) \approx (0.8660, 0, 0.5, 0) .$ Norm check: $(\frac{3}{2})^{2} + 0. 5^{2} = 0.75 + 0.25 = 1$ ✔.

Recall Solution 2.2

Match to $q = (cos \frac{θ}{2}, \overset{n}{^} sin \frac{θ}{2})$ .

Scalar: $cos \frac{θ}{2} = 0.7071 = \frac{2}{2}$ ⇒ $\frac{θ}{2} = 4 5^{\circ}$ ⇒ $θ = 9 0^{\circ}$ .
Vector part $= (0.7071, 0, 0)$ , and its length must equal $sin \frac{θ}{2} = sin 4 5^{\circ} = 0.7071$ . It does. So the unit axis is $\overset{n}{^} = (1, 0, 0)$ , the $x$ -axis. This is a $9 0^{\circ}$ rotation about $x$ . (See Axis-Angle Representation for the reverse conversion in general.)

L3 — Analysis

Take quaternions apart: conjugates, inverses, and the non-commuting product.

Recall Solution 3.1

Since $q$ is a unit quaternion, $∥ q ∥ = 1$ , so $q^{- 1} = q^{*} = (q_{0}, - q_{v})$ — just flip the vector signs: $q^{- 1} = (0.8660, 0, - 0.5, 0) .$ Why it must be the conjugate: $q \otimes q^{*} = (q_{0}^{2} + ∣ q_{v} ∣^{2}, 0) = (∥ q ∥^{2}, 0) = (1, 0)$ exactly when the norm is $1$ . Undoing a $6 0^{\circ}$ rotation about $+ y$ is a $6 0^{\circ}$ rotation about $- y$ , which is what negating the $y$ -component says. ✔

Recall Solution 3.2

Hamilton product: $p \otimes q = (p_{0} q_{0} - p \cdot q, p_{0} q + q_{0} p + p \times q)$ . Here $p_{0} = q_{0} = 0$ , $p = (1, 0, 0)$ , $q = (0, 1, 0)$ .

Scalar: $0 - (p \cdot q) = - (0) = 0$ .
Vector: $0 + 0 + p \times q = (1, 0, 0) \times (0, 1, 0) = (0, 0, 1)$ . $p \otimes q = (0, 0, 0, 1) = k .$ Now swap. The cross product flips sign: $q \times p = (0, 0, - 1)$ . $q \otimes p = (0, 0, 0, - 1) = - k .$ They differ by a sign — a direct picture of $ij = k$ but $j i = - k$ . The $p \times q$ term is exactly why rotations do not commute either.

L4 — Synthesis

Combine several ideas at once: composing rotations, the double cover, renormalization inside a chain.

Recall Solution 4.1

Step 1 — build each factor (half-angle, $4 5^{\circ}$ , $cos = sin = \frac{2}{2} \approx 0.7071$ ): $q_{1} = (\frac{2}{2}, 0, 0, \frac{2}{2}) (z -axis), q_{2} = (\frac{2}{2}, \frac{2}{2}, 0, 0) (x -axis) .$ Step 2 — Hamilton product with $p = q_{2}$ , $q = q_{1}$ . Let $c = \frac{2}{2}$ , so $c^{2} = \frac{1}{2}$ . $p = (c, 0, 0)$ , $q = (0, 0, c)$ .

Scalar: $p_{0} q_{0} - p \cdot q = c \cdot c - 0 = c^{2} = \frac{1}{2}$ .
$p_{0} q = (0, 0, c^{2}) = (0, 0, \frac{1}{2})$ .
$q_{0} p = (c^{2}, 0, 0) = (\frac{1}{2}, 0, 0)$ .
$p \times q = (c, 0, 0) \times (0, 0, c) = (0, - c^{2}, 0) = (0, - \frac{1}{2}, 0)$ . Sum the vector parts: $(\frac{1}{2}, - \frac{1}{2}, \frac{1}{2})$ . $q_{2} \otimes q_{1} = (\frac{1}{2}, \frac{1}{2}, - \frac{1}{2}, \frac{1}{2}) .$ Step 3 — sanity check norm: $4 \times (\frac{1}{2})^{2} = 4 \times \frac{1}{4} = 1$ ✔. The composed object is still a valid unit quaternion, so it is a legal single rotation — one of the great advantages over chaining Euler angles (Euler Angles and Gimbal Lock).

Recall Solution 4.2

Both have $\sum q_{i}^{2} = 4 \times 0.25 = 1$ ✔, so both are unit quaternions. For $q$ : $cos \frac{θ}{2} = 0.5 \Rightarrow \frac{θ}{2} = 6 0^{\circ} \Rightarrow θ = 12 0^{\circ}$ about axis $\overset{n}{^} \propto (0.5, - 0.5, 0.5)$ , i.e. $\overset{n}{^} = \frac{1}{3} (1, - 1, 1)$ . For $- q$ : $cos \frac{θ}{2} = - 0.5 \Rightarrow \frac{θ}{2} = 12 0^{\circ} \Rightarrow θ = 24 0^{\circ}$ about axis $\propto (- 0.5, 0.5, - 0.5) = - \frac{1}{3} (1, - 1, 1)$ . A $24 0^{\circ}$ turn about $- \overset{n}{^}$ is a $12 0^{\circ}$ turn about $+ \overset{n}{^}$ (going the long way round the opposite direction lands you at the same place). In the sandwich, the two minus signs cancel: $(- q) v (- q)^{- 1} = q v q^{- 1}$ . Same physical rotation. This is the $S^{3} \to S O (3)$ double cover.

L5 — Mastery

Do exactly what a flight computer does each timestep: integrate, drift, and repair.

Recall Solution 5.1

(a) Sum of squares: $0.994 5^{2} + 0.052 3^{2} + 0.069 8^{2} + 0.034 9^{2}$ $= 0.989030 + 0.002735 + 0.004872 + 0.001218 = 0.997855$ . $∥ q ∥ = 0.997855 \approx 0.998927$ . (b) Divide each component by the norm: $q \leftarrow \frac{q}{∥ q ∥} \approx (0.99557, 0.05236, 0.06987, 0.03494) .$ New sum of squares $= 1$ by construction ✔. (c) Discrete integration accumulates rounding error, so $q$ slowly leaves the unit sphere; dividing by the norm projects it straight back onto $S^{3}$ , keeping it a valid rotation for the next Rotation Matrices SO(3) conversion the guidance loop needs.

Recall Solution 5.2

(a) Half-angle $\frac{18 0 ^{\circ}}{2} = 9 0^{\circ}$ : $cos 9 0^{\circ} = 0$ , $sin 9 0^{\circ} = 1$ . $q = (0, \overset{n}{^} \cdot 1) = (0, \frac{1}{2}, \frac{1}{2}, 0) \approx (0, 0.7071, 0.7071, 0) .$ Norm: $0 + \frac{1}{2} + \frac{1}{2} + 0 = 1$ ✔ (a "pure" quaternion, scalar part $0$ , since $18 0^{\circ}$ turns always have $cos 9 0^{\circ} = 0$ ). (b) Unit ⇒ $q^{- 1} = q^{*} = (0, - \frac{1}{2}, - \frac{1}{2}, 0)$ . (c) $q \otimes q^{- 1} = (q_{0}^{2} + ∣ q_{v} ∣^{2}, 0) = (0 + 1, 0) = (1, 0, 0, 0)$ = identity ✔. Note a curiosity of $18 0^{\circ}$ : here $q = (0, q_{v})$ and $- q = (0, - q_{v}) = q^{*} = q^{- 1}$ , so this rotation is its own inverse — turning $18 0^{\circ}$ twice returns you home.

Active Recall

Recall One-line: how do you test if

$(a, b, c, d)$ is a rotation quaternion? Check $a^{2} + b^{2} + c^{2} + d^{2} = 1$ .

Recall One-line: given

$θ$ and $\overset{n}{^}$ , write $q$ . $q = (cos \frac{θ}{2}, \overset{n}{^} sin \frac{θ}{2})$ — half the angle.

Recall One-line: inverse of a unit quaternion and why.

$q^{- 1} = q^{*}$ (flip vector part), because $q \otimes q^{*} = ∥ q ∥^{2} = 1$ .