Quaternions as Rotations

Let $p$ be a 3D point represented as a quaternion using its homogeneous coordinates, $p=[w,\mathbf{v}]$, and let $q$ be any non-zero quaternion. Then:

Theorem: The product $qp{q}^{-1}$ takes $p=[w,\mathbf{v}]$ to $p^{\prime }=[w,{\mathbf{v}}^{\prime }]$.

Before proving this theorem, let’s make the following observation. We can express $q$ as a multiplication of a scalar quaternion $s$ and a unit quaternion $\mathbf{U}q$, so $q=s\mathbf{U}q$. Then $qp{q}^{-1}=s\mathbf{U}qp(s\mathbf{U}q{)}^{-1}=s\mathbf{U}qp\mathbf{U}{q}^{-1}{s}^{-1}$. Because the scalar multiplication is commutative, $\mathbf{U}qp\mathbf{U}{q}^{-1}s{s}^{-1}=\mathbf{U}qp\mathbf{U}{q}^{-1}$, so the product doesn’t change irrespective of whether $q$ is a unit quaternion or not. Finally, notice that $\mathbf{U}{q}^{-1}=\mathbf{U}{q}^{\ast }$, so we can write the action as $qp{q}^{\ast }$. Note that from now on, $q$ is assumed to be a unit quaternion without loss of generality.

Next, let’s prove that the scalar part of $qp{q}^{\ast }$ is the same as the scalar of $p$ (we can use the formula to find the scalar component of a quaternion):

Therefore, the scalar part of $p$ remains constant in the operation, i.e., if $p=[w,\mathbf{v}]$, then $p^{\prime }=qp{q}^{\ast }=[w,{\mathbf{v}}^{\prime }]$. And because multiplication preserves norms, then $|p|=|p^{\prime }|$ and also $|v|=|v^{\prime }|$. $◼$

Theorem: If $|q|=1$, then $q=[\cos \theta ,\hat{\mathbf{v}}\sin \theta ]$ acts to rotate around the unit axis $\hat{\mathbf{v}}$ by $2\theta$.

Let:

Be two pure quaternions (which can be represented in 3D space), and an arbitrary quaternion $q$ which has the form:

Let $\theta$ be the angle between ${\mathbf{v}}_{\mathbf{0}}$ and ${\mathbf{v}}_{\mathbf{1}}$. Then ${\mathbf{v}}_{\mathbf{0}}\cdot {\mathbf{v}}_{\mathbf{1}}=\cos \theta$. Also, let ${\mathbf{v}}_{\mathbf{0}}\times {\mathbf{v}}_{\mathbf{1}}=\sin \theta \hat{\mathbf{v}}$. Then $\text{(1)}$ becomes:

Let’s first prove that the product $v_2=q{v}_0{q}^{\ast }$ lies in the same plane as ${\mathbf{v}}_{\mathbf{0}}$ and ${\mathbf{v}}_{\mathbf{1}}$. We do so by first proving that the product $v_2{v}_1^{\ast }$ has the same components (dot and cross products) as $v_1{v}_0^{\ast }$:

Then, if $v_2{v}_1^{\ast }=v_1{v}_0^{\ast }$, that means that $v_2=q{v}_0{q}^{\ast }$ lies in the same plane as $v_0$ and $v_1$. Also, $v_2$ forms an angle of $\theta$ with $v_1$. Furthermore, ${\mathbf{v}}_{\mathbf{1}}\times {\mathbf{v}}_{\mathbf{2}}=\hat{\mathbf{v}}\sin \theta$. Finally, if the angle between $v_0$ and $v_1$ is $\theta$, then the angle between $v_0$ and $v_2$ is $2\theta$, which confirms what’s seen on the image above.

Furthermore, the same can be said of $q$ acting on $v_1$. Let $v_3=q{v}_1{q}^{\ast }$. Then:

Now, any vector $p$ can be represented in terms of the basis $v_0$, $v_1$, and $\hat{\mathbf{v}}$, e.g., $p=s_1{\mathbf{v}}_{\mathbf{0}}+s_1{\mathbf{v}}_{\mathbf{1}}+s_2\hat{\mathbf{v}}$. We’ve seen what $q$ does to $v_0$ and $v_1$, so let’s see what it does to $\hat{\mathbf{v}}$.

Before computing $q\hat{\mathbf{v}}{q}^{\ast }$, see that:

So $q\hat{\mathbf{v}}$ is a commutative operation because the cross product is the only term that makes the quaternion operation non-commutable, and in $q\hat{\mathbf{v}}$, that term is zero. Therefore, $q\hat{\mathbf{v}}{q}^{\ast }=\hat{\mathbf{v}}q{q}^{\ast }=\hat{\mathbf{v}}$, which means that $q$ does not modify $\hat{\mathbf{v}}$.

Thus, the action of $q$ on any vector $p$ is a rotation around $\hat{\mathbf{v}}$ by $2\theta$. $◼$

Quaternion Rotation Facts

Let $q_1$ be a quaternion that rotates the pure quaternion $p_1$ to $p_2$, and also let $q_2$ be a quaternion that rotates the vector $p_2$ to $p_3$. Then $p_3$ will have the form:

Therefore, the combination of rotation $q_1$ followed by $q_2$ is given by $q=q_2{q}_1$.

When the rotations $q_1,q_2,\dots ,q_n$ are applied to the pure quaternion $p$, the result is equal to $qp{q}^{\ast }$, where $q=q_n{q}_{n-1}\dots q_2{q}_1$.