quaternions

The Quaternions

The collection of quaternions are: $$ \mathbb{ H } := \left\{ a + b \mathbf{ i } + c \mathbf{ j }+ d \mathbf{ k } \right\} $$ where

a, b, c, d \in R

and

i^{2} = j^{2} = k^{2} = i j k = - 1

Real Part of the Quaternion

Given a quaternion

q = a + b i + c j + d k

, we say that

a

is the scalar part of the quaternion (or real part), and is denoted by

r (q)

Vector Part of the Quaternion

Given a quaternion

q = a + b i + c j + d k

, we say that

b i + c j + d k

is the vector part of the quaternion, and is denoted by

v (q)

Note that we think of the vector part of a quaternion as an element of $R^{3}$ , and so it inherits all the operations from that space, such as equality, as noted in the next corollary.

Quaternion from a Real Number

We define the map

t o_q u a t : ℝ \to ℍ

by $$ \operatorname{to\_quat}(a) := a + 0\mathbf{i} + 0\mathbf{j} + 0\mathbf{k} $$

Pure Quaternion from a 3D Vector

We define the map

t o_q u a t : ℝ^{3} \to ℍ

by $$ \operatorname{to\_quat}(b, c, d) := 0 + b\mathbf{i} + c\mathbf{j} + d\mathbf{k} $$

Quaternion from a 4D Vector

We define the map

t o_q u a t : ℝ^{4} \to ℍ

by $$ \operatorname{to\_quat}(a, b, c, d) := a + b\mathbf{i} + c\mathbf{j} + d\mathbf{k} $$

to_quat from

ℝ^{3}

Produces a Pure Quaternion

For any

𝐮 \in ℝ^{3}

t o_q u a t (𝐮)

is a pure quaternion, i.e.

r (t o_q u a t (𝐮)) = 0

Let

𝐮 = (b, c, d)

. Then

t o_q u a t (𝐮) = 0 + b 𝐢 + c 𝐣 + d 𝐤

, so

r (t o_q u a t (𝐮)) = 0

Vector Part of to_quat from

ℝ^{3}

is the Identity

For any

𝐮 \in ℝ^{3}

v (t o_q u a t (𝐮)) = 𝐮

Let

𝐮 = (b, c, d)

. Then

v (t o_q u a t (𝐮)) = v (0 + b 𝐢 + c 𝐣 + d 𝐤) = (b, c, d) = 𝐮

to_quat from

ℝ

Has Zero Vector Part

For any

a \in ℝ

v (t o_q u a t (a)) = 0

and

r (t o_q u a t (a)) = a

t o_q u a t (a) = a + 0 𝐢 + 0 𝐣 + 0 𝐤

, so

v (t o_q u a t (a)) = (0, 0, 0) = 0

and

r (t o_q u a t (a)) = a

Quaternion Decomposes as Sum of to_quats

For any

q \in ℍ

, $$ q = \operatorname{to\_quat}(r(q)) + \operatorname{to\_quat}(v(q)) $$

Let

q = a + b i + c j + d k

. Then

t o_q u a t (r (q)) + t o_q u a t (v (q)) = (a + 0 𝐢 + 0 𝐣 + 0 𝐤) + (0 + b 𝐢 + c 𝐣 + d 𝐤) = a + b i + c j + d k = q

Equality by Vector and Real Equality

For any

p, q \in H

we have $$ p = q \iff \left( r\left( p \right) = r\left( q \right) \land v\left( p \right) = v\left( q \right) \right) $$

Suppose that

p = a + b i + c j + d k

and

q = u + x i + y j + z k

then

p = q

if and only if

a = u

b = x

c = y

d = z

which is true if and only if

r (p) = r (q)

and

v (p) = v (q)

, as needed.

Quaternion Addition

$$ (a_1+b_1\mathbf i + c_1\mathbf j + d_1\mathbf k) + (a_2 + b_2\mathbf i + c_2\mathbf j + d_2\mathbf k) = (a_1 + a_2) + (b_1 + b_2)\mathbf i + (c_1 + c_2)\mathbf j + (d_1 + d_2) \mathbf k $$

Scalar Quaternion Multiplication

$$ \lambda(a + b\,\mathbf i + c\,\mathbf j + d\,\mathbf k) = \lambda a + (\lambda b)\,\mathbf i + (\lambda c)\,\mathbf j + (\lambda d)\,\mathbf k. $$

Identity Quaternion

$$ e := 1 + 0\mathbf{i} + 0\mathbf{j} + 0\mathbf{k} $$

Conjugate Quaternion

Given the quaternion

q = a + b i + c j + d k

, its conjugate is given by $$ \overline{ q } := a - b \mathbf{ i }- c \mathbf{ j } - d \mathbf{ k } $$

Quaternion Norm

Given a quaternion

q = a + b i + c j + d k

, its norm is defined as $$ \| q \| := \sqrt{a^2 + b^2 + c^2 + d^2} $$

Unit Quaternion

A quaternion

q \in ℍ

is said to be a unit quaternion if

‖ q ‖ = 1

Pure Quaternion

A quaternion

p \in H

is said to be pure iff

r (p) = 0

Pure Unit Quaternion

A quaternion

q \in ℍ

is said to be a pure unit quaternion if it is both pure and unit, i.e.

r (q) = 0

and

‖ q ‖ = 1

Product of Two Quaternions

Suppose that

p, q \in H

, then we define $$ p q := r \left( p \right) r \left( q \right) - v\left( p \right) \cdot v\left( q \right) + r \left( p \right) v\left( q \right) + r\left( q \right) v\left( p \right) + v \left( p \right) \times v \left( q \right) $$ More precisely: $$ pq = \operatorname{to\_quat}\!\Big(r(p)r(q) - v(p) \cdot v(q)\Big) + \operatorname{to\_quat}\!\Big(r(p)\,v(q) + r(q)\,v(p) + v(p) \times v(q)\Big) $$

Real Part of Quaternion Product

For any

p, q \in ℍ

, $$ r(pq) = r(p)\,r(q) - v(p) \cdot v(q) $$

By the definition of the quaternion product, the scalar part is

r (p) r (q) - v (p) \cdot v (q)

and the vector part is

r (p) v (q) + r (q) v (p) + v (p) \times v (q)

. Therefore

r (p q) = r (p) r (q) - v (p) \cdot v (q)

Vector Part of Quaternion Product

For any

p, q \in ℍ

, $$ v(pq) = r(p)\,v(q) + r(q)\,v(p) + v(p) \times v(q) $$

Product of Two Pure Quaternions

For any two pure quaternions

p, q \in ℍ

(i.e.

r (p) = r (q) = 0

), we have $$ pq = \operatorname{to\_quat}\!\big({-v(p) \cdot v(q)}\big) + \operatorname{to\_quat}\!\big(v(p) \times v(q)\big) $$

Since

p

and

q

are pure,

r (p) = 0

and

r (q) = 0

. By the definition of the quaternion product: $$ \begin{align} pq &= \operatorname{to\_quat}\!\Big(r(p)r(q) - v(p) \cdot v(q)\Big) + \operatorname{to\_quat}\!\Big(r(p)\,v(q) + r(q)\,v(p) + v(p) \times v(q)\Big) \\ &= \operatorname{to\_quat}\!\Big(0 \cdot 0 - v(p) \cdot v(q)\Big) + \operatorname{to\_quat}\!\Big(0 \cdot v(q) + 0 \cdot v(p) + v(p) \times v(q)\Big) \\ &= \operatorname{to\_quat}\!\big({-v(p) \cdot v(q)}\big) + \operatorname{to\_quat}\!\big(v(p) \times v(q)\big) \end{align} $$

Real Part of Product of Two Pure Quaternions

For any two pure quaternions

p, q \in ℍ

, $$ r(pq) = -v(p) \cdot v(q) $$

By the product of two pure quaternions, the real part is

- v (p) \cdot v (q)

Vector Part of Product of Two Pure Quaternions

For any two pure quaternions

p, q \in ℍ

, $$ v(pq) = v(p) \times v(q) $$

By the product of two pure quaternions, the vector part is

v (p) \times v (q)

Square of a Pure Unit Quaternion is Minus One

For any pure unit quaternion

\hat{n} \in ℍ

, we have $$ \hat{n}^2 = -e $$ where

e

is the identity quaternion.

Since

\hat{n}

is pure and unit,

r (\hat{n}) = 0

and

‖ v (\hat{n}) ‖ = 1

. By the product of two pure quaternions: $$ \hat{n}^2 = \operatorname{to\_quat}(-v(\hat{n}) \cdot v(\hat{n})) + \operatorname{to\_quat}(v(\hat{n}) \times v(\hat{n})) $$ Since

v (\hat{n}) \cdot v (\hat{n}) = ‖ v (\hat{n}) ‖^{2} = 1

and

v (\hat{n}) \times v (\hat{n}) = 0

, we get: $$ \hat{n}^2 = \operatorname{to\_quat}(-1) + \operatorname{to\_quat}(\mathbf{0}) = -1 + 0\mathbf{i} + 0\mathbf{j} + 0\mathbf{k} = -e $$

Perpendicular Pure Quaternions Anticommute

For any two pure quaternions

p, q \in ℍ

with

v (p) \cdot v (q) = 0

, we have $$ pq = -qp $$

By the product of two pure quaternions: $$ pq = \operatorname{to\_quat}(-v(p) \cdot v(q)) + \operatorname{to\_quat}(v(p) \times v(q)) $$ Since

v (p) \cdot v (q) = 0

: $$ pq = \operatorname{to\_quat}(0) + \operatorname{to\_quat}(v(p) \times v(q)) = \operatorname{to\_quat}(v(p) \times v(q)) $$ Similarly: $$ qp = \operatorname{to\_quat}(0) + \operatorname{to\_quat}(v(q) \times v(p)) = \operatorname{to\_quat}(v(q) \times v(p)) $$ Since the cross product is anti-commutative,

v (q) \times v (p) = - (v (p) \times v (q))

, so

q p = - p q

, i.e.

p q = - q p

Quaternion Multiplication is Associative

For any

p, q, s \in ℍ

, we have $$ (pq)s = p(qs) $$

Write $p = a_{1} + b_{1} 𝐢 + c_{1} 𝐣 + d_{1} 𝐤$ , $q = a_{2} + b_{2} 𝐢 + c_{2} 𝐣 + d_{2} 𝐤$ , and $s = a_{3} + b_{3} 𝐢 + c_{3} 𝐣 + d_{3} 𝐤$ .

Since quaternion multiplication is defined to be distributive over addition and is determined by the multiplication rules on the basis elements ${1, 𝐢, 𝐣, 𝐤}$ , it suffices to verify associativity on all triples of basis elements. There are $4^{3} = 64$ such triples.

We verify the key non-trivial cases. Recall the basis multiplication rules derived from $𝐢^{2} = 𝐣^{2} = 𝐤^{2} = i j k = - 1$ : $$ \mathbf{ij} = \mathbf{k}, \quad \mathbf{jk} = \mathbf{i}, \quad \mathbf{ki} = \mathbf{j} $$ $$ \mathbf{ji} = -\mathbf{k}, \quad \mathbf{kj} = -\mathbf{i}, \quad \mathbf{ik} = -\mathbf{j} $$

Case $(𝐢, 𝐣, 𝐤)$ : $(i j) 𝐤 = 𝐤 \cdot 𝐤 = - 1$ and $𝐢 (j k) = 𝐢 \cdot 𝐢 = - 1$ . ✓

Case $(𝐢, 𝐤, 𝐣)$ : $(i k) 𝐣 = (- 𝐣) 𝐣 = - (𝐣^{2}) = 1$ and $𝐢 (k j) = 𝐢 (- 𝐢) = - 𝐢^{2} = 1$ . ✓

Case $(𝐢, 𝐢, 𝐣)$ : $(i i) 𝐣 = (- 1) 𝐣 = - 𝐣$ and $𝐢 (i j) = 𝐢 𝐤 = - 𝐣$ . ✓

Case $(𝐣, 𝐤, 𝐢)$ : $(j k) 𝐢 = 𝐢 \cdot 𝐢 = - 1$ and $𝐣 (k i) = 𝐣 \cdot 𝐣 = - 1$ . ✓

All other non-trivial cases follow by similar computation. Any triple involving the identity $1$ is immediate since $1$ commutes with everything and acts as the identity. Since associativity holds on all basis triples and quaternion multiplication distributes over addition, associativity holds for all quaternions.

Product with a Real Quaternion

For any

q \in ℍ

and

a \in ℝ

, $$ \operatorname{to\_quat}(a) \cdot q = q \cdot \operatorname{to\_quat}(a) = a \, q $$ where

a q

denotes scalar multiplication.

Let

p = t o_q u a t (a)

, so

r (p) = a

and

v (p) = 0

. By the quaternion product definition: $$ \begin{align} pq &= a \cdot r(q) - \mathbf{0} \cdot v(q) + a \, v(q) + r(q) \cdot \mathbf{0} + \mathbf{0} \times v(q) \\ &= a \, r(q) + a \, v(q) = a(r(q) + v(q)) = a \, q \end{align} $$ Similarly

q p = r (q) \cdot a - v (q) \cdot 0 + r (q) \cdot 0 + a v (q) + v (q) \times 0 = a r (q) + a v (q) = a q

Conjugation is Additive

Suppose that

p, q \in H

then $$ \overline{ p + q } = \overline{ p } + \overline{ q } $$

Suppose that

p = a + b i + c j + d k

and

q = u + x i + y j + z k

then $$ \begin{align} \overline{ p + q } &= \overline{ \left( a + u \right) + \left( b + x \right) i + \left( c + y \right) j + \left( d + z \right) k } \\ &= \left( a + u \right) - \left( b + x \right) i - \left( c + y \right) j - \left( d + z \right) k \\ &= a - bi - cj - dk + u - xi - yj - zk \\ &= \overline{ p } + \overline{ q } \end{align} $$

Vector Part of the Conjugate is Minus 1 Times the Original

For any

p \in H

we have $$ v \left( \overline{ p } \right) = -v \left( p \right) $$

Suppose that

p = a + b i + c j + d k

. Then $$ v \left( \overline{ p } \right) = v(a - bi - cj - dk) = (-b, -c, -d) = -(b, c, d) = -v \left( p \right) $$ as needed.

Real Part of the Conjugate Doesn't Change

For any

p \in H

we have $$ r \left( \overline{ p } \right) = r \left( p \right) $$

Suppose that

p = a + b i + c j + d k

. Then $$ r \left( \overline{ p } \right) = r(a - bi - cj - dk) = a = r \left( p \right) $$ as needed.

Conjugate Fixes Real Quaternions

For any

p \in H

, such that

v (p) = 0

, then $$ \overline{ p } = p $$

Since

v (p) = 0

, we have

p = r (p) + 0 𝐢 + 0 𝐣 + 0 𝐤

. Then

\overset{―}{p} = r (p) - 0 𝐢 - 0 𝐣 - 0 𝐤 = p

Conjugate Only Applies to Vector Part

For any

p \in H

we have $$ \overline{ p } = \operatorname{to\_quat}(r(p)) + \operatorname{to\_quat}(-v(p)) $$

Conjugate of a Pure Quaternion is its Negation

For any pure quaternion

p \in ℍ

, $$ \overline{p} = -p $$

Since

p

is pure,

r (p) = 0

. Then by the conjugate formula: $$ \overline{p} = \operatorname{to\_quat}(0) + \operatorname{to\_quat}(-v(p)) = \operatorname{to\_quat}(-v(p)) = -p $$

Product Commutes in the Real Part

For any

p, q \in H

we have $$ r \left( pq \right) = r\left( qp \right) $$

Quaternion Times its Conjugate

For any

q \in ℍ

, $$ q \overline{q} = \overline{q} q = \operatorname{to\_quat}(\| q \|^2) $$

Let

q = a + b i + c j + d k

. Then

r (q) = a

v (q) = (b, c, d)

r (\overset{―}{q}) = a

v (\overset{―}{q}) = (- b, - c, - d)

. $$ \begin{align} r(q\overline{q}) &= r(q)\,r(\overline{q}) - v(q) \cdot v(\overline{q}) \\ &= a \cdot a - (b, c, d) \cdot (-b, -c, -d) \\ &= a^2 - (-b^2 - c^2 - d^2) \\ &= a^2 + b^2 + c^2 + d^2 = \| q \|^2 \end{align} $$ $$ \begin{align} v(q\overline{q}) &= r(q)\,v(\overline{q}) + r(\overline{q})\,v(q) + v(q) \times v(\overline{q}) \\ &= a(-b, -c, -d) + a(b, c, d) + (b, c, d) \times (-b, -c, -d) \\ &= \mathbf{0} + \mathbf{0} = \mathbf{0} \end{align} $$ where

(b, c, d) \times (- b, - c, - d) = - (b, c, d) \times (b, c, d) = 0

since the cross product of any vector with itself is zero. Therefore

q \overset{―}{q} = t o_q u a t (‖ q ‖^{2})

. The proof that

\overset{―}{q} q = t o_q u a t (‖ q ‖^{2})

is analogous.

Unit Quaternion Inverse is its Conjugate

For any unit quaternion

q \in ℍ

, $$ q^{-1} = \overline{q} $$

Since

‖ q ‖ = 1

, by the previous proposition,

q \overset{―}{q} = t o_q u a t (1) = e

and

\overset{―}{q} q = e

. Therefore

\overset{―}{q}

is the multiplicative inverse of

q

Dot Product of Two Quaternions

Suppose that

p, q \in ℍ

, then we define $$ p \cdot q := r(p)\,r(q) + v(p) \cdot v(q) $$

Conjugation is a Homomorphism in the Real Part

For any

p, q \in H

we have $$ r \left( \overline{ pq } \right) = r\left( \overline{ p } \; \overline{ q } \right) $$

First of all, we know conjugation doesn't change the real part, so we have: $$ r \left( \overline{ pq } \right) = r\left( pq \right) = r \left( p \right) r \left( q \right) - v\left( p \right) \cdot v\left( q \right) $$ and then we know that $$ r \left( \overline{ p } \; \overline{ q } \right) = r \left( \overline{ p } \right) r \left( \overline{ q } \right) - v\left( \overline{ p } \right) \cdot v\left( \overline{ q } \right) $$ Since minus comes out of the vector part, and constants can get pulled out of the dot product we have that $$ r \left( \overline{ p } \right) r \left( \overline{ q } \right) - v\left( \overline{ p } \right) \cdot v\left( \overline{ q } \right) = r\left( p \right) r \left( q \right) - \left( -1 \right) \left( -1 \right) v\left( p \right) \cdot v\left( q \right) = r\left( p \right) r \left( q \right) - v\left( p \right) \cdot v\left( q \right) $$ therefore we have that

r (\overset{―}{p q}) = r (\overset{―}{p} \overset{―}{q})

Conjugation Swaps Order in the Vector Part

For any

p, q \in H

we have $$ v \left( \overline{ pq } \right) = v\left( \overline{ q } \; \overline{ p } \right) $$

We recall that, and so $$ v \left( \overline{ pq } \right) = -v\left( pq \right) $$ and then we compute: $$ \begin{align} v\left( \overline{ q } \; \overline{ p } \right) &= r \left( \overline{ q } \right) v\left( \overline{ p } \right) + r\left( \overline{ p } \right) v\left( \overline{ q } \right) + v \left( \overline{ q } \right) \times v \left( \overline{ p } \right) \\ &= r(q)(-v(p)) + r(p)(-v(q)) + (-v(q)) \times (-v(p)) \\ &= -r(q)\,v(p) - r(p)\,v(q) + v(q) \times v(p) \\ &= -r(p)\,v(q) - r(q)\,v(p) - v(p) \times v(q) \\ &= -\big(r(p)\,v(q) + r(q)\,v(p) + v(p) \times v(q)\big) \\ &= -v(pq) \end{align} $$ where we used that

v (q) \times v (p) = - (v (p) \times v (q))

. Therefore

v (\overset{―}{p q}) = - v (p q) = v (\overset{―}{q} \overset{―}{p})

Conjugation Distributes by Swapping

For any

p, q \in H

we have $$ \overline{ pq } = \overline{ q } \; \overline{ p } $$

We show they are equal by checking both parts. For the real part: $$ r\left( \overline{ pq } \right) = r(pq) = r(qp) = r\left( \overline{qp} \right) = r\left( \overline{q}\;\overline{p} \right) $$ where the first and third equalities use that conjugation preserves the real part, the second uses that the product commutes in the real part, and the last uses the conjugation homomorphism lemma. For the vector part, we have by the conjugation swaps order lemma that

v (\overset{―}{p q}) = v (\overset{―}{q} \overset{―}{p})

. Therefore

\overset{―}{p q} = \overset{―}{q} \overset{―}{p}

Conjugation Preserves Norm

For any

q \in ℍ

, $$ \| \overline{q} \| = \| q \| $$

Let

q = a + b i + c j + d k

. Then

\overset{―}{q} = a - b i - c j - d k

, so $$ \| \overline{q} \| = \sqrt{a^2 + (-b)^2 + (-c)^2 + (-d)^2} = \sqrt{a^2 + b^2 + c^2 + d^2} = \| q \| $$

Quaternion Product is Norm-Multiplicative

For any

p, q \in ℍ

, $$ \| pq \| = \| p \| \cdot \| q \| $$

We compute: $$ \begin{align} \| pq \|^2 &= r(pq \cdot \overline{pq}) \\ &= r(pq \cdot \overline{q}\;\overline{p}) \end{align} $$ by conjugation distributing by swapping. By associativity: $$ pq \cdot \overline{q}\;\overline{p} = p(q\overline{q})\overline{p} = p \cdot \operatorname{to\_quat}(\| q \|^2) \cdot \overline{p} = \| q \|^2 (p\overline{p}) = \| q \|^2 \cdot \operatorname{to\_quat}(\| p \|^2) $$ where we used that

q \overset{―}{q} = t o_q u a t (‖ q ‖^{2})

and that real quaternions commute with everything. Therefore

‖ p q ‖^{2} = ‖ p ‖^{2} ‖ q ‖^{2}

, and taking square roots gives

‖ p q ‖ = ‖ p ‖ \cdot ‖ q ‖

Conjugation of a Pure Quaternion by a Unit Quaternion is Pure

For any pure quaternion

p \in ℍ

and unit quaternion

q \in ℍ

, the quaternion

q p \overset{―}{q}

is pure, i.e. $$ r(qp\overline{q}) = 0 $$

We use the fact that a quaternion $x$ is pure if and only if $\overset{―}{x} = - x$ . We verify this for $x = q p \overset{―}{q}$ :

$$ \begin{align} \overline{qp\overline{q}} &= \overline{\overline{q}} \; \overline{p} \; \overline{q} \\ &= q(-p)\overline{q} \\ &= -(qp\overline{q}) \end{align} $$

where we used conjugation distributes by swapping (applied twice), the fact that $\overset{―}{\overset{―}{q}} = q$ , and the fact that $\overset{―}{p} = - p$ since $p$ is pure. Since $\overset{―}{q p \overset{―}{q}} = - (q p \overset{―}{q})$ , the quaternion $q p \overset{―}{q}$ is pure.

Conjugation by a Unit Quaternion Preserves the Norm of the Vector Part

For any pure quaternion

p \in ℍ

and unit quaternion

q \in ℍ

, $$ \| qp\overline{q} \| = \| p \| $$

By norm-multiplicativity: $$ \| qp\overline{q} \| = \| q \| \cdot \| p \| \cdot \| \overline{q} \| = 1 \cdot \| p \| \cdot 1 = \| p \| $$ where we used that

‖ q ‖ = 1

and

‖ \overset{―}{q} ‖ = ‖ q ‖ = 1

Parallel and Perpendicular Decomposition

Given a vector

𝐰 \in ℝ^{3}

and a unit vector

\hat{n} \in ℝ^{3}

, we define the parallel and perpendicular components of

𝐰

with respect to

\hat{n}

as: $$ \mathbf{w}_\parallel := (\hat{n} \cdot \mathbf{w})\,\hat{n}, \qquad \mathbf{w}_\perp := \mathbf{w} - \mathbf{w}_\parallel $$ so that

𝐰 = 𝐰_{∥} + 𝐰_{⟂}

, where

\hat{n} \cdot 𝐰_{⟂} = 0

Pure Unit Quaternion Times a Parallel Pure Quaternion

Let

\hat{n}

be a pure unit quaternion and let

p_{∥}

be a pure quaternion such that

v (p_{∥}) = λ v (\hat{n})

for some

λ \in ℝ

. Then $$ \hat{n}\, p_\parallel = \operatorname{to\_quat}(-\lambda) $$ In particular,

\hat{n} p_{∥}

is a real quaternion.

Since both are pure, by the product of two pure quaternions: $$ \begin{align} \hat{n}\, p_\parallel &= \operatorname{to\_quat}(-v(\hat{n}) \cdot v(p_\parallel)) + \operatorname{to\_quat}(v(\hat{n}) \times v(p_\parallel)) \\ &= \operatorname{to\_quat}(-v(\hat{n}) \cdot \lambda\,v(\hat{n})) + \operatorname{to\_quat}(v(\hat{n}) \times \lambda\,v(\hat{n})) \\ &= \operatorname{to\_quat}(-\lambda\,\| v(\hat{n}) \|^2) + \operatorname{to\_quat}(\lambda\,(v(\hat{n}) \times v(\hat{n}))) \\ &= \operatorname{to\_quat}(-\lambda) + \operatorname{to\_quat}(\mathbf{0}) \\ &= \operatorname{to\_quat}(-\lambda) \end{align} $$ where we used

‖ v (\hat{n}) ‖ = 1

and

v (\hat{n}) \times v (\hat{n}) = 0

Pure Unit Quaternion Times a Perpendicular Pure Quaternion

Let

\hat{n}

be a pure unit quaternion and let

p_{⟂}

be a pure quaternion such that

v (\hat{n}) \cdot v (p_{⟂}) = 0

. Then $$ \hat{n}\, p_\perp = \operatorname{to\_quat}(v(\hat{n}) \times v(p_\perp)) $$ In particular,

\hat{n} p_{⟂}

is a pure quaternion.

Pure Unit Quaternion Sandwich is a Reflection

Let

\hat{n} \in ℍ

be a pure unit quaternion, and let

p \in ℍ

be a pure quaternion. Then $$ \hat{n}\, p\, \hat{n} = \operatorname{to\_quat}\!\big(-v(p)_\parallel + v(p)_\perp\big) $$ where

v (p)_{∥} = (v (\hat{n}) \cdot v (p)) v (\hat{n})

and

v (p)_{⟂} = v (p) - v (p)_{∥}

are the parallel and perpendicular components of

v (p)

with respect to

v (\hat{n})

. This is the reflection of

v (p)

through the plane perpendicular to

v (\hat{n})

Decompose $p = p_{∥} + p_{⟂}$ where $p_{∥} = t o_q u a t (v (p)_{∥})$ and $p_{⟂} = t o_q u a t (v (p)_{⟂})$ . Both are pure quaternions, and $v (\hat{n}) \cdot v (p_{⟂}) = 0$ , while $v (p_{∥}) = λ v (\hat{n})$ where $λ = v (\hat{n}) \cdot v (p)$ .

Parallel part: We compute $\hat{n} p_{∥} \hat{n}$ . By the parallel product lemma, $\hat{n} p_{∥} = t o_q u a t (- λ)$ . Since this is a real quaternion, by the real product property: $$ \hat{n}\, p_\parallel\, \hat{n} = \operatorname{to\_quat}(-\lambda) \cdot \hat{n} = -\lambda\, \hat{n} = \operatorname{to\_quat}(-\lambda\, v(\hat{n})) = \operatorname{to\_quat}(-v(p)_\parallel) $$ The parallel component is negated.

Perpendicular part: We compute $\hat{n} p_{⟂} \hat{n}$ . Since $v (\hat{n}) \cdot v (p_{⟂}) = 0$ , the quaternions $\hat{n}$ and $p_{⟂}$ anticommute: $\hat{n} p_{⟂} = - p_{⟂} \hat{n}$ . Therefore: $$ \hat{n}\, p_\perp\, \hat{n} = (-p_\perp\, \hat{n})\, \hat{n} = -p_\perp\, (\hat{n}\, \hat{n}) = -p_\perp\, \hat{n}^2 $$ By the square of a pure unit quaternion, ${\hat{n}}^{2} = - e$ , so: $$ -p_\perp(-e) = p_\perp \cdot e = p_\perp = \operatorname{to\_quat}(v(p)_\perp) $$ The perpendicular component is preserved.

Combining: By associativity and the distributive law: $$ \hat{n}\, p\, \hat{n} = \hat{n}(p_\parallel + p_\perp)\hat{n} = \hat{n}\, p_\parallel\, \hat{n} + \hat{n}\, p_\perp\, \hat{n} = \operatorname{to\_quat}(-v(p)_\parallel) + \operatorname{to\_quat}(v(p)_\perp) = \operatorname{to\_quat}(-v(p)_\parallel + v(p)_\perp) $$ This is exactly the reflection of $v (p)$ through the plane perpendicular to $v (\hat{n})$ .

Unit Quaternion Factors into Two Pure Unit Quaternions

Let

q = \cos (\frac{θ}{2}) + \hat{n} \sin (\frac{θ}{2})

be a unit quaternion, where

\hat{n}

is a pure unit quaternion. Then there exist pure unit quaternions

\hat{a}, \hat{b} \in ℍ

with

v (\hat{a}) \cdot v (\hat{n}) = 0

and

v (\hat{b}) \cdot v (\hat{n}) = 0

, such that: $$ q = \hat{b}\,\hat{a} $$ Moreover, the angle between

v (\hat{a})

and

v (\hat{b})

\frac{θ}{2}

, and the two reflection planes (perpendicular to

v (\hat{a})

and

v (\hat{b})

respectively) intersect along the axis

v (\hat{n})

🏗️ ΘρϵηΠατπ🚧 (under construction)

🏗️ $Θ ρ ϵ η Π α τ π$ 🚧 (under construction)