Understanding Quaternions

Understanding Quaternions

In this article I will attempt to explain the concept of Quaternions in an easy to understand way. I will explain how you might visualize a Quaternion as well as explain the different operations that can be applied to quaternions. I will also compare applications of matrices, euler angles, and quaternions and try to explain when you would want to use quaterions instead of Euler angles or matrices and when you would not.

DISCLAIMER: You cannot fully understand quaternions in just 45 minutes.

Introduction

In computer graphics, we use transformation matrices to express a position in space (translation) as well as its orientation in space (rotation). Optionally, a single transformation matrix can also be used to express the scale or “shear” of an object. We can think of this transformation matrix as a “basis space” where if you multiply a vector or a point (or even another matrix) by a transformation matrix you “transform” that vector, point or matrix into the space represented by that matrix.

In this article, I will not discuss the details of transformation matrices. For a detailed description of transformation matrices, you can refer to my previous article titled Matrices.

In this article, I want to discuss an alternative method of describing the orientation of an object (rotation) in space using quaternions.

The concept of quaterinions was realized by the Irish mathematician Sir William Rowan Hamilton on Monday October 16th 1843 in Dublin, Ireland. Hamilton was on his way to the Royal Irish Academy with his wife and as he was passing over the Royal Canal on the Brougham Bridge he made a dramatic realization that he immediately carved into the stone of the bridge.

$i^2=j^2=k^2=ijk=-1$

William Rowan Hamilton Plaque on Broome Bridge on the Royal Canal commemorating his discovery of the fundamental formula for quaternion multiplication.

Complex Numbers

Before we can fully understand quaterions, we must first understand where they came from. The root of quaternions is based on the concept of the complex number system.

In addition to the well-known number sets (Natural, Integer, Real, and Rational), the Complex Number system introduces a new set of numbers called imaginary numbers. Imaginary numbers were invented to solve certain equations that had no solutions such as:

$x^2+1=0$

To solve this expression, we must state that $x^2=-1$ which we know is not possible because the square of any number (positive or negative) is always positive.

Mathematicians generally can’t accept that an expression does not have a solution so a new term was invented called the imaginary number that can be used to solve such equations.

The imaginary number has the form:

$i^2=-1$

Don’t try to actually understand this term as there is no logical reason why it exists. We just have to accept that $i$ is just something that squares to $-1$.

The set of imaginary numbers can be represented by $\mathbb{I}$.

The set of complex numbers (represented by the symbol $\mathbb{C}$) is the sum of a real number and an imaginary number and has the form:

$z=a+bi~~a,b\in\mathbb{R},~~i^2=-1$

It could also be stated that all Real numbers are complex numbers with $b=0$ and all imaginary numbers are complex numbers with $a=0$.

Complex numbers can be added and subtracted by adding or subtracting the real, and imaginary parts.

$(a_1+b_1i)+(a_2+b_2i)=(a_1+a_2)+(b_1+b_2)i$

Subtraction:

$(a_1+b_1i)-(a_2+b_2i)=(a_1-a_2)+(b_1-b_2)i$

Multiply a Complex Number by a Scalar

A complex number is multiplied by a scalar by multiplying each term of the complex number by the scalar:

$\lambda(a+bi)=\lambda{a}+\lambda{b}i$

Product of Complex Numbers

Complex numbers can also be multiplied by applying normal algebraic rules.

$\begin{array}{rcl}z_1&=&(a_1+b_1i)\\z_2&=&(a_2+b_2i)\\z_1z_2&=&(a_1+b_1i)(a_2+b_2i)\\&=&a_1a_2+a_1b_2i+b_1a_2i+b_1b_2i^2\\&=&(a_1a_2-b_1b_2)+(a_1b_2+b_1a_2)i\end{array}$

Square of Complex Numbers

A complex number can also be squared by multiplying by itself:

$\begin{array}{rcl}z&=&(a+bi)\\z^2&=&(a+bi)(a+bi)\\&=&(a^2-b^2)+2abi\end{array}$

Complex Conjugate

The conjugate of a complex number is a complex number with the imaginary part negated and is denoted as either $\bar{z}$ or $z^*$.

$\begin{array}{rcl}z&=&(a+bi)\\z^*&=&(a-bi)\end{array}$

The product of a complex number and its conjugate gives a special result.

$\begin{array}{rcl}z&=&(a+bi)\\z^*&=&(a-bi)\\zz^*&=&(a+bi)(a-bi)\\&=&a^2-abi+abi+b^2\\&=&a^2+b^2\end{array}$

Absolute Value of a Complex Number

We can use the conjugate of a complex number to compute the absolute value (or norm, or magnitude) of a complex number. The absolute value of a complex number is the square-root of the complex number multiplied by its conjugate and is denoted $|z|$:

$\begin{array}{rcl}z&=&(a+bi)\\|z|&=&\sqrt{zz^*}\\&=&\sqrt{(a+bi)(a-bi)}\\&=&\sqrt{a^2+b^2}\end{array}$

Quotient of Two Complex Numbers

To compute the quotient of two complex numbers, we multiply the numerator and denominator by the complex conjugate of the denominator.

$\begin{array}{rcl}z_1&=&(a_1+b_1i)\\z_2&=&(a_2+b_2i)\\[0.8em]\cfrac{z_1}{z_2}&=&\cfrac{a_1+b_1i}{a_2+b_2i}\\[1.3em]&=&\cfrac{(a_1+b_1i)(a_2-b_2i)}{(a_2+b_2i)(a_2-b_2i)}\\[1.3em]&=&\cfrac{a_1a_2-a_1b_2i+b_1a_2i-b_1b_2i^2}{a_2^2+b_2^2}\\[1.3em]&=&\cfrac{a_1a_2+b_1b_2}{a_2^2+b_2^2}+\cfrac{b_1a_2-a_1b_2}{a_2^2+b_2^2}i\end{array}$

Powers of i

If we state that $i^2=-1$ then it should be possible to raise i to other powers as well.

$\begin{array}{rrrrrrr}i^0&=&&&&&1\\i^1&=&&&&&i\\i^2&=&&&&&-1\\i^3&=&ii^2&=&&&-i\\i^4&=&i^{2}i^{2}&=&&&1\\i^5&=&ii^4&=&&&i\\i^6&=&ii^5&=&i^2&=&-1\end{array}$

If we keep writing this sequence, we will see a pattern emerge $(1,i,-1,-i,1,\dots)$.

A similar pattern emerges from the increasing negative powers.

$\begin{array}{rcr}i^0&=&1\\i^{-1}&=&-i\\i^{-2}&=&-1\\i^{-3}&=&i\\i^{-4}&=&1\\i^{-5}&=&-i\\i^{-6}&=&-1\end{array}$

You may have seen a similar pattern in mathematics before but in the form $(x,y,-x,-y,x,\dots)$ which is generated by rotating a point 90 degrees counter-clockwise on a 2D Cartesian plane and the sequence $(x,-y,-x,y,x,\dots)$ is generated by rotating a point 90 degrees clockwise on a 2D Cartesian plane.

Cartesian Plane

The Complex Plane

We can also map complex numbers in a 2D grid called the Complex Plane by mapping the Real part on the horizontal axis and the Imaginary part on the vertical axis.

Complex Plane

As shown in the previous sequence, we can say that if we multiply a complex number by i, we can rotate the complex number through the complex plane at 90 degree increments.

Let’s see if this is true. We’ll take an arbitrary point p in the complex plane:

$p=2+i$

and we multiply it by i gives q:

$\begin{array}{rcl}p&=&2+i\\q&=&pi\\&=&(2+i)i\\&=&2i+i^2\\&=&-1+2i\end{array}$

Multiplying q by i gives r:

$\begin{array}{rcl}q&=&-1+2i\\r&=&qi\\&=&(-1+2i)i\\&=&-i+2i^2\\&=&-2-i\end{array}$

And multiplying r by i gives s:

$\begin{array}{rcl}r&=&-2-i\\s&=&ri\\&=&(-2-i)i\\&=&-2i-i^2\\&=&1-2i\end{array}$

And multiplying s by i gives t:

$\begin{array}{rcl}s&=&1-2i\\t&=&si\\&=&(1-2i)i\\&=&i-2i^2\\&=&2+i\end{array}$

Which is exactly what we started with (p). If we plot these complex numbers on the complex plane, we get the following result.

Complex Numbers on the Complex Plane

We can also rotate clock-wise in the complex plane by multiplying the complex number by -i.

Rotors

We can also perform arbitrary rotations in the complex plane by defining a complex number of the form:

$q=\cos\theta+i\sin\theta$

Multiplying any complex number by the rotor q produces the general formula:

$\begin{array}{rcl}p&=&a+bi\\q&=&\cos\theta+i\sin\theta\\pq&=&(a+bi)(\cos\theta+i\sin\theta)\\a^{\prime}+b^{\prime}i&=&a\cos\theta-b\sin\theta+(a\sin\theta+b\cos\theta)i\end{array}$

Which can also be written in matrix form:

$\begin{bmatrix}a^{\prime}&-b^{\prime}\\b^{\prime}&a^{\prime}\end{bmatrix}=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}\begin{bmatrix}a&-b\\b&a\end{bmatrix}$

Which is the method to rotate an arbitrary point in the complex plane counter-clockwise about the origin.

Quaternions

With this knowledge of the complex number system and the complex plane, we can extend this to 3-dimensional space by adding two imaginary numbers to our number system in addition to i.

The general form to express quaternions is

$q=s+xi+yj+zk~~s,x,y,z\in\mathbb{R}$

Were, according to Hamilton’s famous expression:

$i^2=j^2=k^2=ijk=-1$

and

$\begin{array}{ccc}ij=k&jk=i&ki=j\\ji=-k&kj=-i&ik=-j\end{array}$

You may have noticed that the relationship between i, j, and k are very similar to the cross product rules for the unit cartesian vectors:

$\begin{array}{ccc}\mathbf{x}\times\mathbf{y}=\mathbf{z}&\mathbf{y}\times\mathbf{z}=\mathbf{x}&\mathbf{z}\times\mathbf{x}=\mathbf{y}\\\mathbf{y}\times\mathbf{x}=-\mathbf{z}&\mathbf{z}\times\mathbf{y}=-\mathbf{x}&\mathbf{x}\times\mathbf{z}=-\mathbf{y}\end{array}$

Hamilton also recognized that the i, j, and k imaginary numbers could be used to represent three cartesian unit vectors i, j, and k with the same properties of imaginary numbers, such that $\mathbf{i}^2=\mathbf{j}^2=\mathbf{k}^2=-1$.

Visualizing the Properties of ij, jk, ki

The image above visualizes the relationship between the cartesian unit vectors represented by i, j, and k.

Quaternions as an Ordered Pair

We can also represent quaternions as an ordered pair:

$q=[s,\mathbf{v}]~~s\in\mathbb{R},\mathbf{v}\in\mathbb{R}^3$

Where v can also be represented by its individual components:

$q=[s,x\mathbf{i}+y\mathbf{j}+z\mathbf{k}]~~s,x,y,z\in\mathbb{R}$

Using this notation, we can more easily show the similarities between quaternions and complex numbers.

Quaternions can be added and subtracted similar to complex numbers:

$\begin{array}{rcl}q_a&=&[s_a,\mathbf{a}]\\q_b&=&[s_b,\mathbf{b}]\\q_a+q_b&=&[s_a+s_b,\mathbf{a}+\mathbf{b}]\\q_a-q_b&=&[s_a-s_b,\mathbf{a}-\mathbf{b}]\end{array}$

Quaternion Products

We can also express the product of two quaternions:

$\begin{array}{rcl}q_a&=&[s_a,\mathbf{a}]\\q_b&=&[s_b,\mathbf{b}]\\q_{a}q_{b}&=&[s_{a},\mathbf{a}][s_{b},\mathbf{b}]\\&=&(s_{a}+x_{a}i+y_{a}j+z_{a}k)(s_{b}+x_{b}i+y_{b}j+z_{b}k)\\&=&(s_{a}s_{b}-x_{a}x_{b}-y_{a}y_{b}-z_{a}z_{b})\\&&+(s_{a}x_{b}+s_{b}x{a}+y_{a}z_{b}-y_{b}z_{a})i\\&&+(s_{a}y_{b}+s_{b}y_{a}+z_{a}x_{b}-z_{b}x_{a})j\\&&+(s_{a}z_{b}+s_{b}z_{a}+x_{a}y_{b}-x_{b}y_{a})k\end{array}$

Which results in another quaternion. If we replace the imaginary numbers i, j, and k in the previous expression by the ordered pairs (also known as the quaternion units),

$i=[0,\mathbf{i}]~j=[0,\mathbf{j}]~k=[0,\mathbf{k}]$

And substituting back to the original expression together with $[1,\mathbf{0}]=1$ gives:

$\begin{array}{rcl}[s_{a},\mathbf{a}][s_{b},\mathbf{b}]&=&(s_{a}s_{b}-x_{a}x_{b}-y_{a}y_{b}-z_{a}z_{b})[1,\mathbf{0}]\\&&+(s_{a}x_{b}+s_{b}x{a}+y_{a}z_{b}-y_{b}z_{a})[0,\mathbf{i}]\\&&+(s_{a}y_{b}+s_{b}y_{a}+z_{a}x_{b}-z_{b}x_{a})[0,\mathbf{j}]\\&&+(s_{a}z_{b}+s_{b}z_{a}+x_{a}y_{b}-x_{b}y_{a})[0,\mathbf{k}]\end{array}$

And expanding this expression into a sum of ordered pairs gives:

$\begin{array}{rcl}[s_{a},\mathbf{a}][s_{b},\mathbf{b}]&=&[s_{a}s_{b}-x_{a}x_{b}-y_{a}y_{b}-z_{a}z_{b},\mathbf{0}]\\&&+[0,(s_{a}x_{b}+s_{b}x{a}+y_{a}z_{b}-y_{b}z_{a})\mathbf{i}]\\&&+[0,(s_{a}y_{b}+s_{b}y_{a}+z_{a}x_{b}-z_{b}x_{a})\mathbf{j}]\\&&+[0,(s_{a}z_{b}+s_{b}z_{a}+x_{a}y_{b}-x_{b}y_{a})\mathbf{k}]\end{array}$

If we multiply through with the quaternion unit and extract the common vector components, we can rewrite this equation in this way:

$\begin{array}{rcl}[s_{a},\mathbf{a}][s_{b},\mathbf{b}]&=&[s_{a}s_{b}-x_{a}x_{b}-y_{a}y_{b}-z_{a}z_{b},\mathbf{0}]\\&&+[0,s_{a}(x_{b}\mathbf{i}+y_{b}\mathbf{j}+z_{b}\mathbf{k})+s_{b}(x_{a}\mathbf{i}+y_{a}\mathbf{j}+z_{a}\mathbf{k})\\&&+(y_{a}z_{b}-y_{b}z_{a})\mathbf{i}+(z_{a}x_{b}-z_{b}x_{a})\mathbf{j}+(x_{a}y_{b}-x_{b}y_{a})\mathbf{k}]\end{array}$

This equation gives us the sum of two ordered pairs. The first ordered pair is a Real quaternion and the second is a Pure quaternion. These two ordered pairs can be combined into a single ordered pair:

$\begin{array}{rcl}[s_{a},\mathbf{a}][s_{b},\mathbf{b}]&=&[s_{a}s_{b}-x_{a}x_{b}-y_{a}y_{b}-z_{a}z_{b},\\&&s_{a}(x_{b}\mathbf{i}+y_{b}\mathbf{j}+z_{b}\mathbf{k})+s_{b}(x_{a}\mathbf{i}+y_{a}\mathbf{j}+z_{a}\mathbf{k})\\&&+(y_{a}z_{b}-y_{b}z_{a})\mathbf{i}+(z_{a}x_{b}-z_{b}x_{a})\mathbf{j}+(x_{a}y_{b}-x_{b}y_{a})\mathbf{k}]\end{array}$

And if we substitute,

$\begin{array}{rcl}\mathbf{a}&=&x_{a}\mathbf{i}+y_{a}\mathbf{j}+z_{a}\mathbf{k}\\\mathbf{b}&=&x_{b}\mathbf{i}+y_{b}\mathbf{j}+z_{b}\mathbf{k}\\\mathbf{a}\cdot\mathbf{b}&=&x_{a}x_{b}\mathbf{i}^{2}+y_{a}y_{b}\mathbf{j}^2+z_{a}z_{b}\mathbf{k}^{2}\\\mathbf{a}\times\mathbf{b}&=&(y_{a}z_{b}-y_{b}z_{a})\mathbf{i}+(z_{a}x_{b}-z_{b}x_{a})\mathbf{j}+(x_{a}y_{b}-x_{b}y_{a})\mathbf{k}\end{array}$

We get:

$[s_{a},\mathbf{a}][s_{b},\mathbf{b}]=[s_{a}s_{b}-\mathbf{a}\cdot\mathbf{b},s_{a}\mathbf{b}+s_{b}\mathbf{a}+\mathbf{a}\times\mathbf{b}]$

Which is the general equation of a quaternion product.

A Real Quaternion

A Real Quaternion is a quaternion with a vector term of 0:

$q=[s,\mathbf{0}]$

And the product of two Real Quaternions is another Real Quaternion:

$\begin{array}{rcl}q_a&=&[s_a,\mathbf{0}]\\q_b&=&[s_b,\mathbf{0}]\\q_{a}q_{b}&=&[s_a,\mathbf{0}][s_b,\mathbf{0}]\\&=&[s_{a}s_{b},\mathbf{0}]\end{array}$

Which is similar to the product of two complex numbers that contain a zero imaginary term.

$\begin{array}{rcl}z_1&=&a_1+0i\\z_2&=&a_2+0i\\z_{1}z_{2}&=&(a_1+0i)(a_2+0i)\\&=&a_{1}a_{2}\end{array}$

Multiplying a Quaternion by a Scalar

We can also multiply a quaternion by a scalar which should obey the rule:

$\begin{array}{rcl}q&=&[s,\mathbf{v}]\\\lambda{q}&=&\lambda[s,\mathbf{v}]\\&=&[\lambda{s},\lambda\mathbf{v}]\end{array}$

We can confirm this by using the product or Real Quaterions shown above to multiply a quaternion by the scalar as a Real Quaternion:

$\begin{array}{rcl}q&=&[s,\mathbf{v}]\\\lambda&=&[\lambda,\mathbf{0}]\\\lambda{q}&=&[\lambda,\mathbf{0}][s,\mathbf{v}]\\&=&[\lambda{s},\lambda\mathbf{v}]\end{array}$

Pure Quaternions

Similar to Real Quaterions, Hamilton also defined the Pure Quaternion as a quaternion that has a zero scalar term:

$q=[0,\mathbf{v}]$

Or, written in its component parts:

$q=xi+yj+zk$

And we can also take the product of two Pure quaternions:

$\begin{array}{rcl}q_a&=&[0,\mathbf{a}]\\q_b&=&[0,\mathbf{b}]\\q_{a}q_{b}&=&[0,\mathbf{a}][0,\mathbf{b}]\\&=&[-\mathbf{a}\cdot\mathbf{b},\mathbf{a}\times\mathbf{b}]\end{array}$

According to the quaternion product rule shown above.

We can also express quaternions as an addition of the Real and Pure quaternion parts:

$\begin{array}{rcl}q&=&[s,\mathbf{v}]\\&=&[s,\mathbf{0}]+[0,\mathbf{v}]\end{array}$

Unit Quaternion

Given an arbitrary vector v, we can express this vector in both its scalar magnitude and its direction as such:

$\mathbf{v}=v\mathbf{\hat{v}}\text{~where~}v=|\mathbf{v}|\text{,~and~}|\mathbf{\hat{v}}|=1$

And combining this definition with the definition of a pure quaternion gives:

$\begin{array}{rcl}q&=&[0,\mathbf{v}]\\&=&[0,v\mathbf{\hat{v}}]\\&=&v[0,\mathbf{\hat{v}}]\end{array}$

And we can also describe a unit quaternion that has a zero scalar and a unit vector as such:

$\hat{q}=[0,\mathbf{\hat{v}}]$

Binary Form of a Quaternion

We can now combine the definitions of the unit quaternion and the additive form of a quaternion, we can create a representation of quaternions which is similar to the notation used to describe complex numbers:

$\begin{array}{rcl}q&=&[s,\mathbf{v}]\\&=&[s,\mathbf{0}]+[0,\mathbf{v}]\\&=&[s,\mathbf{0}]+v[0,\mathbf{\hat{v}}]\\&=&s+v\hat{q}\end{array}$

This gives us a way to represent the quaternion that is very similar to complex numbers:

$\begin{array}{rcl}z&=&a+bi\\q&=&s+v\hat{q}\end{array}$

Quaternion Conjugate

The quaternion conjugate can be computed by negating the vector part of the quaternion:

$\begin{array}{rcl}q&=&[s,\mathbf{v}]\\q^*&=&[s,-\mathbf{v}]\end{array}$

And the product of a quaternion with its conjugate gives:

$\begin{array}{rcl}qq^*&=&[s,\mathbf{v}][s,-\mathbf{v}]\\&=&[s^2-\mathbf{v}\cdot-\mathbf{v},-s\mathbf{v}+s\mathbf{v}+\mathbf{v}\times-\mathbf{v}]\\&=&[s^2+\mathbf{v}\cdot\mathbf{v},\mathbf{0}]\\&=&[s^2+v^2,\mathbf{0}]\end{array}$

Quaternion Norm

If you recall from the definition of the norm of a complex number:

$\begin{array}{rcl}|z|&=&\sqrt{a^2+b^2}\\zz^*&=&|z|^2\end{array}$

Similarly, the norm (or magnitude) of a quaternion is defined as:

$\begin{array}{rcl}q&=&[s,\mathbf{v}]\\|q|&=&\sqrt{s^2+v^2}\end{array}$

Which allows us to express the norm of a quaternion as:

$qq^*=|q|^2$

Quaternion Normalization

With the definition of a quaternion norm, we can use it to normalize a quaternion. A quaternion is normalized by dividing it by $|q|$:

$q^{\prime}=\frac{q}{\sqrt{s^2+v^2}}$

As an example, let’s normalize the quaternion:

$q=[1,4\mathbf{i}+4\mathbf{j}-4\mathbf{k}]$

First, we must compute the norm of the quaternion:

$\begin{array}{rcl}|q|&=&\sqrt{1^2+4^2+4^2+(-4)^2}\\&=&\sqrt{49}\\&=&7\end{array}$

Then, we must divide the quaternion by the norm of the quaternion to compute the normalized quaternion:

$\begin{array}{rcl}q^{\prime}&=&\cfrac{q}{|q|}\\[1.3em]&=&\cfrac{(1+4\mathbf{i}+4\mathbf{j}-4\mathbf{k})}{7}\\[1.3em]&=&\cfrac{1}{7}+\cfrac{4}{7}\mathbf{i}+\cfrac{4}{7}\mathbf{j}-\cfrac{4}{7}\mathbf{k}\end{array}$

Quaternion Inverse

The inverse of a quaternion is denoted $q^{-1}$. To compute the inverse of a quaternion, we take the conjugate of the quaternion and divide it by the square of the norm:

$q^{-1}=\frac{q^*}{|q|^2}$

To show this, we can take the fact that by definition of the inverse:

$qq^{-1}=[1,\mathbf{0}]=1$

And multiply both sides by the conjugate of the quaternion gives:

$q^{*}qq^{-1}=q^{*}$

And by substitution we get:

$\begin{array}{rcl}|q|^{2}q^{-1}&=&q^{*}\\[0.8em]q^{-1}&=&\cfrac{q^{*}}{|q|^{2}}\end{array}$

And for unit-norm quaternions whose norm is 1, we can write:

$q^{-1}=q^{*}$

Quaternion Dot Product

Similar to vector dot-products, we can also compute the dot product between two quaternions by multiplying the corresponding scalar parts and summing the results:

$\begin{array}{rcl}q_1&=&[s_1,x_1\mathbf{i}+y_1\mathbf{j}+z_1\mathbf{k}]\\q_2&=&[s_2,x_2\mathbf{i}+y_2\mathbf{j}+z_2\mathbf{k}]\\q_1{\cdot}q_2&=&s_{1}s_{2}+x_{1}x_{2}+y_{1}y_{2}+z_{1}z_{2}\end{array}$

We can also use the quaternion dot-product to compute the angular difference between the quaternions:

$\cos\theta=\cfrac{s_{1}s_{2}+x_{1}x_{2}+y_{1}y_{2}+z_{1}z_{2}}{|q_{1}||q_{2}|}$

And for unit-norm quaternions, we can simplify the equation:

$\cos\theta=s_{1}s_{2}+x_{1}x_{2}+y_{1}y_{2}+z_{1}z_{2}$

Rotations

If you recall we defined a special form of the complex number called a Rotor that could be used to rotate a point through the 2D complex plane as:

$q=\cos\theta+i\sin\theta$

Then by its similarities to complex numbers, it should be possible to express a quaternion that can be used to rotate a point in 3D-space as such:

$q=[\cos\theta,\sin\theta\mathbf{v}]$

Let’s test if this theory holds by computing the product of the quaternion q and the vector p. First, we can express p as a Pure quaternion in the form:

$p=[0,\mathbf{p}]$

And q is a unit-norm quaternion in the form:

$q=[s,\lambda\mathbf{\hat{v}}]$

Then,

$\begin{array}{rcl}p^{\prime}&=&qp\\&=&[s,\lambda\mathbf{\hat{v}}][0,\mathbf{p}]\\&=&[-\lambda\mathbf{\hat{v}}\cdot\mathbf{p},s\mathbf{p}+\lambda\mathbf{\hat{v}}\times\mathbf{p}]\end{array}$

We see that the result is a general quaternion with both a scalar and a vector parts.

Let’s first consider the “special” case where p is perpendicular to $\mathbf{\hat{v}}$ in which case, the dot-product term $-\lambda\mathbf{\hat{v}}\cdot\mathbf{p}=0$ and the result becomes the Pure quaternion:

$p^{\prime}=[0,s\mathbf{p}+\lambda\mathbf{\hat{v}}\times\mathbf{p}]$

In this case, to rotate p about $\mathbf{\hat{v}}$ we just substitute $s=\cos\theta$ and $\lambda=\sin\theta$.

$p^{\prime}=[0,\cos\theta\mathbf{p}+\sin\theta\mathbf{\hat{v}}\times\mathbf{p}]$

As an example, let’s rotate a vector p 45° about the z-axis then our quaternion q is:

$\begin{array}{rcl}q&=&[\cos\theta,\sin\theta\mathbf{k}]\\[0.8em]&=&\left[\cfrac{\sqrt{2}}{2},\cfrac{\sqrt{2}}{2}\mathbf{k}\right]\end{array}$

And let’s take a vector p that adheres to the special case that p is perpendicular to k:

$p=[0,2\mathbf{i}]$

Now let’s find the product of qp:

$\begin{array}{rcl}p^{\prime}&=&qp\\[0.8em]&=&\left[\cfrac{\sqrt{2}}{2},\cfrac{\sqrt{2}}{2}\mathbf{k}\right][0,2\mathbf{i}]\\[1.3em]&=&\left[0,2\cfrac{\sqrt{2}}{2}\mathbf{i}+2\cfrac{\sqrt{2}}{2}\mathbf{k}\times\mathbf{i}\right]\\[1.3em]&=&[0,\sqrt{2}\mathbf{i}+\sqrt{2}\mathbf{j}]\end{array}$

Which results in a Pure quaternion that is rotated 45° about the k axis.
We can also confirm that the magnitude of the resulting vector is maintained:

$\begin{array}{rcl}|\mathbf{p}^{\prime}|&=&\sqrt{\sqrt{2}^{2}+\sqrt{2}^{2}}\\&=&2\end{array}$

Which is exactly the result we expected!

We can visualize this by the following image:

Quaternion Rotation (1)

Now let’s consider a quaternion that is not orthogonal to p. If we specify the vector part of our quaternion to 45° offset from p we get:

$\begin{array}{rcl}\mathbf{\hat{v}}&=&\cfrac{\sqrt{2}}{2}\mathbf{i}+\cfrac{\sqrt{2}}{2}\mathbf{k}\\[0.8em]\mathbf{p}&=&2\mathbf{i}\\q&=&[\cos\theta,\sin\theta\mathbf{\hat{v}}]\\p&=&[0,\mathbf{p}]\end{array}$

And multiplying our point p by q we get:

$\begin{array}{rcl}p^{\prime}&=&qp\\&=&[\cos\theta,sin\theta\mathbf{\hat{v}}][0,\mathbf{p}]\\&=&[-\sin\theta\mathbf{\hat{v}}\cdot\mathbf{p},\cos\theta\mathbf{p}+\sin\theta\mathbf{\hat{v}}\times\mathbf{p}]\end{array}$

And substituting $\mathbf{\hat{v}}$, p and $\theta=45^{\circ}$ gives:

$\begin{array}{rcl}p^{\prime}&=&\left[-\cfrac{\sqrt{2}}{2}\left(\cfrac{\sqrt{2}}{2}\mathbf{i}+\cfrac{\sqrt{2}}{2}\mathbf{k}\right)\cdot(2\mathbf{i}),\cfrac{\sqrt{2}}{2}2\mathbf{i}+\cfrac{\sqrt{2}}{2}\left(\cfrac{\sqrt{2}}{2}\mathbf{i}+\cfrac{\sqrt{2}}{2}\mathbf{k}\right)\times2\mathbf{i}\right]\\&=&[-1,\sqrt{2}\mathbf{i}+\mathbf{j}]\end{array}$

Which is no longer a pure quaternion, and it has not been rotated 45° and the vector’s norm is no longer 2 (instead it has been reduced to $\sqrt{3}$).

This result can be visualized by the image.

Quaternion Rotation (2)

Technically, it’s incorrect to represent the quaternion p’ in 3D space because it’s actually a 4D vector! For the sake of simplicity, I will only visualize the vector component of quaternions.

However, all is not lost. Hamilton recognized (but didn’t publish) that if we post-multiply the result of qp by the inverse of q then the result is a pure quaternion and the norm of the vector component is maintained. Let’s see if we can apply this to our example.

First, let’s compute $q^{-1}$:

$\begin{array}{rcl}q&=&\left[\cos\theta,\sin\theta\left(\cfrac{\sqrt{2}}{2}\mathbf{i}+\cfrac{\sqrt{2}}{2}\mathbf{k}\right)\right]\\[1.3em]q^{-1}&=&\left[\cos\theta,-\sin\theta\left(\cfrac{\sqrt{2}}{2}\mathbf{i}+\cfrac{\sqrt{2}}{2}\mathbf{k}\right)\right]\end{array}$

For $\theta=45^{\circ}$ gives:

$\begin{array}{rcl}q^{-1}&=&\left[\cfrac{\sqrt{2}}{2},-\cfrac{\sqrt{2}}{2}\left(\cfrac{\sqrt{2}}{2}\mathbf{i}+\cfrac{\sqrt{2}}{2}\mathbf{k}\right)\right]\\[1.3em]&=&\cfrac{1}{2}\left[\sqrt{2},-\mathbf{i}-\mathbf{k}\right]\end{array}$

And combining the previous value of qp and $q^{-1}$ gives:

$\begin{array}{rcl}qp&=&\left[-1,\sqrt{2}\mathbf{i}+\mathbf{j}\right]\\[0.8em]qpq^{-1}&=&\left[-1,\sqrt{2}\mathbf{i}+\mathbf{j}\right]\cfrac{1}{2}\left[\sqrt{2},-\mathbf{i}-\mathbf{k}\right]\\[1.3em]&=&\cfrac{1}{2}\left[-\sqrt{2}-\left(\sqrt{2}\mathbf{i}+\mathbf{j}\right)\cdot(-\mathbf{i}-\mathbf{k}),\mathbf{i}+\mathbf{k}+\sqrt{2}\left(\sqrt{2}\mathbf{i}+\mathbf{j}\right)-\mathbf{i}+\sqrt{2}\mathbf{j}+\mathbf{k}\right]\\[1.3em]&=&\cfrac{1}{2}\left[-\sqrt{2}+\sqrt{2},\mathbf{i}+\mathbf{k}+2\mathbf{i}+\sqrt{2}\mathbf{j}-\mathbf{i}+\sqrt{2}\mathbf{j}+\mathbf{k}\right]\\[0.8em]&=&\left[0,\mathbf{i}+\sqrt{2}\mathbf{j}+\mathbf{k}\right]\end{array}$

Which is a pure quaternion and the norm of the result is:

$\begin{array}{rcl}|p^{\prime}|&=&\sqrt{1^2+\sqrt{2}^2+1^2}\\&=&\sqrt{4}\\&=&2\end{array}$

which is the same as p so the norm of the vector is maintained.

The image below visualizes the result of the rotation.

Quaternion Rotation (3)

So we can see that the result is a pure quaternion and that the norm of the initial vector is maintained, but the vector has been rotated 90° rather than 45° which is twice as much as desired! So in order to correctly rotate a vector p by an angle $\theta$ about an arbitrary axis $\mathbf{\hat{v}}$, we must consider the half-angle and construct the following quaternion:

$q=\left[\cos\cfrac{1}{2}\theta,\sin\cfrac{1}{2}\theta\mathbf{\hat{v}}\right]$

Which is the general form of a rotation quaternion!

Quaternion Interpolation

One of the most important reasons for using quaternions in computer graphics is that quaternions are very good at representing rotations in space. Quaternions overcome the issues that plague other methods of rotating points in 3D space such as Gimbal lock which is an issue when you represent your rotation with eular angles.

Using quaternions, we can define several methods that represents a rotational interpolation in 3D space. The first method I will examine is called SLERP which is used to smoothly interpolate a point between two orientations. The second method is an extension of SLERP called SQAD which is used to interpolate through a sequence of orientations that define a path.

SLERP

SLERP stands for Spherical Linear Interpolation. SLERP provides a method to smoothly interpolate a point about two orientations.

I will represent the first orientation as $q_1$ and the second orientation as $q_2$. The point that is interpolated will be prepresented by $\mathbf{p}$ and the interpolated point will be represented by $\mathbf{p}^{\prime}$. The interpolation parameter $t$ will interpolate $\mathbf{p}$ from $q_1$ when $t=0$ to $q_2$ when $t=1$.

The standard linear interpolation formula is:

$\mathbf{p}^{\prime}=\mathbf{p_1}+t(\mathbf{p_2}-\mathbf{p_1})$

The general steps to apply this equation are:

• Compute the difference between $\mathbf{p_1}$ and $\mathbf{p_2}$.
• Take the fractional part of that difference.
• Adjust the original value by the fractional difference between the two points.

We can use the same basic principle to interpolate between two quaternion orientations.

Quaternion Difference

The first step dictates that we must compute the difference between $q_1$ and $q_2$. With regards to quaternions, this is equivalent to computing the angular difference between the two quaternions.

$\Deltaq=q_1^{-1}q_2$

Quaternion Exponentiation

The next step is to take the fractional part of that difference. We can compute the fractional part of a quaternion by raising it to a power whose value is in the range 0…1.

The general formula for quaternion exponentiation is:

$q^t=\exp(t\log{q})$

Where the exponential function for quaternions is given by:

$\begin{array}{rcl}\exp(q)&=&\exp\left([0,\theta\mathbf{\hat{v}}]\right)\\&=&[\cos\theta,\sin\theta\mathbf{\hat{v}}]\end{array}$

And the logarithm of a quaternion is given by:

$\begin{array}{rcl}\log{q}&=&\log(\cos\theta{+}\sin\theta\mathbf{\hat{v}})\\&=&\log\left(\exp(\theta\mathbf{\hat{v}})\right)\\&=&\theta\mathbf{\hat{v}}\\&=&[0,\theta\mathbf{\hat{v}}]\end{array}$

For t=0, we have:

$\begin{array}{rcl}q^0&=&\exp(0\log{q})\\&=&\exp([\cos(0),\sin(0)\mathbf{\hat{v}}])\\&=&\exp([1,\mathbf{0}])\\&=&[1,\mathbf{0}]\end{array}$

And for t=1, we have:

$\begin{array}{rcl}q^1&=&\exp(\log{q})\\&=&q\end{array}$

Fractional Difference of Quaternions

And to compute the interpolated angular rotation, we adjust the original orientation $q_1$ and adjust it by the fractional part of the difference between $q_1$ and $q_2$.

$q^{\prime}=q_1\left(q_1^{-1}q_2\right)^t$

Which is the general form of spherical linear interpolation using quaternions. However, this is not the form of the slerp equation that is commonly used in practice.

We can apply a similar formula for performing a spherical interpolation of vectors to quaternions. The general form of a spherical interpolation for vectors is defined as:

$\mathbf{v}_t=\cfrac{\sin(1-t)\theta}{\sin\theta}\mathbf{v}_1+\cfrac{\sin{t\theta}}{\sin\theta}\mathbf{v}_2$

This is visualized in the following image.

Quaternion Interpolation

This formula can be applied unmodified to quaternions:

$q_t=\cfrac{\sin(1-t)\theta}{\sin\theta}q_1+\cfrac{\sin{t\theta}}{\sin\theta}q_2$

And we can obtain the angle $\theta$ by computing the dot-product between $q_1$ and $q_2$.

$\begin{array}{rcl}\cos\theta&=&\cfrac{q_1{\cdot}q_2}{|q_1||q_2|}\\[1.3em]&=&\cfrac{s_{1}s_{2}+x_{1}x_{2}+y_{1}y_{2}+z_{1}z_{2}}{|q_1||q_2|}\\[1.3em]\theta&=&\cos^{-1}\left(\cfrac{s_{1}s_{2}+x_{1}x_{2}+y_{1}y_{2}+z_{1}z_{2}}{|q_1||q_2|}\right)\end{array}$

Considerations

There are two issues with this implementation which must be taken into consideration during implementation.

First, if the quaternion dot-product results in a negative value, then the resulting interpolation will travel the “long-way” around the 4D sphere which is not necessarily what we want. To solve this problem, we can test the result of the dot product and if it is negative, then we can negate one of the orientations. Negating the scalar and the vector part of the quaternion does not change the orientation that it represents but by doing this we guarantee that the rotation will be applied in the “shortest” path.

The other problem arises when the angular difference between $q_1$ and $q_2$ is very small then $\sin\theta\$ becomes 0. If this happens, then we will get an undefined result when we divide by $\sin\theta$. In this case, we can fall-back to using linear interpolation between $q_1$ and $q_2$.

Just as a SLERP can be used to compute an interpolation between two quaternions, a SQUAD (Spherical and Quadrangle) can be used to smoothly interpolate over a path of rotations.

If we have the sequence of quaternions:

$q_1,q_2,q_3,\cdots,q_{n-2},q_{n-1},q_{n}$

And we also define the “helper” quaternion ($s_i$) which we can consider an intermediate control point:

$s_i=\exp\left(-\cfrac{\log\left(q_{i+1}q_i^{-1}\right)+\log\left(q_{i-1}q_i^{-1}\right)}{4}\right)q_i$

Then the orientation along the sub-cuve defined by:

$q_{i-1},q_i,q_{i+1},q_{i+2}$

at time t is given by:

$\mathrm{squad}(q_i,q_{i+1},s_i,s_{i+1},t)=\mathrm{slerp}(\mathrm{slerp}(q_i,q_{i+1},t),\mathrm{slerp}(s_i,s_{i+1},t),2t(1-t))$

Conclusion

Besides being extremely difficult to understand, quaternions provide a few obvious advantages over using matrices or Euler angles for representing rotations.

• Quaternion interpolation using SLERP and SQUAD provide a way to interpolate smoothly between orientations in space.
• Rotation concatenation using quaternions is faster than combining rotations expressed in matrix form.
• For unit-norm quaternions, the inverse of the rotation is taken by subtracting the vector part of the quaternion. Computing the inverse of a rotation matrix is considerably slower if the matrix is not orthonormalized (if it is, then it’s just the transpose of the matrix).
• Converting quaternions to matrices is slightly faster than for Euler angles.
• Quaternions only require 4 numbers (3 if they are normalized. The Real part can be computed at run-time) to represent a rotation where a matrix requires at least 9 values.

However for all of the advantages in favor of using quaternions, there are also a few disadvantages.

• Quaternions can become invalid because of floating-point round-off error however this “error creep” can be resolved by re-normalizing the quaternion.
• And probably the most significant deterrent for using quaternions is that they are very hard to understand. I hope that this issue is resolved after reading this article.

There are several math libraries that implement quaternions and a few of those libraries implement quaternions correctly. In my personal experience, I find GLM (OpenGL Math Library) to be a good math library with a good implementation of quaternions. If you are interested in using quaternions in your own applications, this is the library I would recommend.

I created a small demo that demonstrates how a quaternion is used to rotate an object in space. The demo was created with Unity 3.5.2 which you can download for free and view the demo script files. The zip file also contains a Windows binary executable but Using Unity, you can also generate a Mac application (and Unity 4 introduces Linux builds as well).

Understanding Quaternions.zip

Reference

 Quaternions for Computer Graphics Vince, J (2011). Quaternions for Computer Graphics. 1st. ed. London: Springer. 3D Math Primer for Graphics and Game Development Dunn, F. and Parberry, I. (2002). 3D Math Primer for Graphics and Game Development. 1st. ed. Plano, Texas: Wordware Publishing, Inc. Quaternions – Wikipedia Quaternions and Spatial Rotation – Wikipedia

43 thoughts on “Understanding Quaternions”

• Thanks Lawrence, you comments are welcome.

1. This is a wonderfully clear and concise introduction to quaternions! There are a few typos that could be corrected. “It’s” should be “its” throughout. (The first is a contraction of “it is”, the second is the possessive form of “it”.) Also, “gimble’ should be “gimbal”. And “principal” should be “principle”: both are words but they have different meanings.

I will show this tutorial to all my robotics students.

• Dave,

Thanks for the corrections! Let me know if you find anything else.

2. Well done! I find this website through google and found it very informative and useful. I would like to learn graphics programming for a long time, but it’s quite difficult to get started, especially the mathematics, may you kindly suggest what mathematics I should know and how to learn graphics programming effectively? Thank you very much.

• Gary,

For basic graphics programming you will want to have a good understanding of vector algebra (vector addition & subtraction, dot & cross products) as well as matrix algebra (matrix products, inverse & transpose). Also knowledge of basic lighting equations is important.

I recommend you read the books I’ve referenced in the math articles here: http://3dgep.com/?cat=27 as well as read the the OpenGL articles here http://3dgep.com/?cat=11.

If you’d like to get into shader programming, then you should follow the Cg articles here: http://3dgep.com/?cat=108.

I also plan to make a set of GLSL articles, but I can’t promise when they will be ready.

3. Thank you for writing such a thorough post on quaternions, it is much appreciated!

Unfortunately though, quite a few of the equation listings don’t show properly in neither Firefox nor Chrome on my Win7 computer ( see http://grab.by/hQJm for example). Just wanted to let you know

• Thanks for pointing this out. A few weeks after creating this tutorial, the LaTeX generator I was using went offline and i had to switch to another provider. This provider seems to have a different parser than the one I was using when I created the page on quaternions in the first place. I switched again and some of the formulas that were not showing up before are fixed now but others are broken. I really need to install my own CGI handler for LaTex on my own server.

• I find this to be a great tutorial but I’m having the same problem Trond had with every single equation, but the way it looks it seems to be a great tutorial.

• Luke,

I have moved the script that is used to generate the formulas to a new server. There have been some hiccups with this script but I hope to have them all fixed now.

• Henry, when I created this article I was using a different source for the LaTeX generation. That source has since become unavailable so I had to switch to another source which apparently handles LaTeX differently. I need to go through each equation which isn’t rendering correctly to find out why this is happening but this takes time… It will be fixed in the future.

4. This is the only coherent introduction to quaternions I have found on the web. It is failry complex and I will have to read through this several times. I am assuming you are drawing heavily from Vince, as Dunn is pretty sketchy. Thanks for pulling this together–I feel like I have found a startting point!

5. Excellent tutorial/article on quaternions, it helped me get a good insight, thanks for the effort and this website is great!

6. While familiar with the use of rotational matrices all my previous attempts to get a hold of quaternions have failed.

What worked for me with your explanation is that you explained it by analogy to rotation in the complex plane which I already understood: the next step was then easy for me.

Many thanks for your work and making it available.

7. Great tutorial, loving it so far!

I’ve spot one minor error I think though:
> The interpolation parameter t will interpolate P from q1 when t = 0 to q1 when t = 1.

Shouldn’t this be:
> The interpolation parameter t will interpolate P from q1 when t = 0 to q2 when t = 1.

• Shammah,

Thanks for pointing this out!

8. WOW! Very nice article. I actually implemented a simple quaternion library in C++ while reading it :). Very well presented too.

9. Please it look like very importent&impresive post, but you aware that we can’t read it becouse lack of picture? equations?
Can you attach or sent a full pdf format to read it please?

• Mica,

Sorry if the formulas did not show up for you. The images for the formulas are generated on another sever than where the article itself is hosted. If the server that generates the images is inaccessible, then the formulas may not show up. This is annoying but if the images don’t show up, then check back in a few minutes and perhaps it is fixed at that time. I strive for 100% up-time, but some downtime is unavoidable.

10. This was fantastic! Thank you so much for writing it up.

11. This is a great explanation but I got really stuck at why the product of two quaternions is minus the dot product. We have:
SaSb – XaXb – YaYb – ZaZb, then we substitute in that a.b = XaXbi^2 etc. Does this i^2 (i squared) not resolve to -1 as per Hamilton? That would make a.b equal to -XaXb – YaYb – ZaZb already and it wouldn’t need further negation. I’m obviously missing something in the notation, perhaps that i^2 here is the cosine of 0 and therefore equals 1. Confused!

• Gaz,

You are right about the fact that $\mathbf{i}^2=\mathbf{j}^2=\mathbf{k}^2=-1$ as is stated in the section titled Quaternions.

Maybe if I write the product rule like this:

$\begin{array}{rcl}q_a&=&[s_a,\mathbf{a}]\\q_b&=&[s_b,\mathbf{b}]\\q_{a}q_{b}&=&[s_{a},\mathbf{a}][s_{b},\mathbf{b}]\\&=&(s_{a}+x_{a}i+y_{a}j+z_{a}k)(s_{b}+x_{b}i+y_{b}j+z_{b}k)\\&=&(s_{a}s_{b}+x_{a}x_{b}i^{2}+y_{a}y_{b}j^{2}+z_{a}z_{b}k^{2})\\&&+(s_{a}x_{b}+s_{b}x{a}+y_{a}z_{b}-y_{b}z_{a})i\\&&+(s_{a}y_{b}+s_{b}y_{a}+z_{a}x_{b}-z_{b}x_{a})j\\&&+(s_{a}z_{b}+s_{b}z_{a}+x_{a}y_{b}-x_{b}y_{a})k\end{array}$

And we can write the dot product of the vector parts of the two quaternions as:

$\begin{array}{rcl}\mathbf{a}\cdot\mathbf{b}&=&x_{a}x_{b}\mathbf{i}^{2}+y_{a}y_{b}\mathbf{j}^2+z_{a}z_{b}\mathbf{k}^{2}\\&=&(-1)x_{a}x_{b}+(-1)y_{a}y_{b}+(-1)z_{a}z_{b}\\&=&-(x_{a}x_{b}+y_{a}y_{b}+z_{a}z_{b})\end{array}$

Which we can substitute back into the original equation:

$\begin{array}{rcl}[s_{a},\mathbf{a}][s_{b},\mathbf{b}]&=&(s_{a}s_{b}-\mathbf{a}\cdot\mathbf{b})\\&&+(s_{a}x_{b}+s_{b}x{a}+y_{a}z_{b}-y_{b}z_{a})i\\&&+(s_{a}y_{b}+s_{b}y_{a}+z_{a}x_{b}-z_{b}x_{a})j\\&&+(s_{a}z_{b}+s_{b}z_{a}+x_{a}y_{b}-x_{b}y_{a})k\end{array}$

So we are not adding any extra negatives, we are just factoring out the $\mathbf{i}^2=\mathbf{j}^2=\mathbf{k}^2=-1$ from the dot-product to get the real part of the quaternion product.

Does this make it more clear?

12. Where it says:
“The set of complex numbers (represented by the symbol ) is the sum of a real number and an imaginary number and has the form:”, the following formula is wrong. It should say: z = a + bi; a, b \in R; i^2 = -1, but it does not.

• Oskar,

Yes, you are right. Thanks for pointing this out. After switching to a different LaTeX generator some formulas didn’t render correctly anymore. I hope I have fixed all of the places (in this page) where formulas were not rendered correctly anymore but if you seen any more then please let me know!

13. … and in the following formulas, about addition, subtraction and multiplication, you should not use an implication arrow, you should use an equals sign.

• Oskar,

Yes, again you are correct. I can’t remember why I had an implication arrow there but it’s fixed now.

14. Thanks, this was really useful. I finally understood quaternions. I’ll use them in 3D reconstruction from multiple photos.

15. Thanks for the article. Found an error in the formulas after: “Complex numbers can also be multiplied by applying normal algebraic rules.”

In row three there stand a2a2 where it should be a1a2.

Bye

• Henning,

Thanks for pointing this out. I have fixed the article now.

16. beautiful article.
I suggest you try mathjax.js for equation generation. Its client-based. no need for host or anything.

17. i have a doubt. when u calculate the norms int the example, u said, instead of 2 its is root(3). but how is it?? sqrt(1 + 2 + 1) = srt(4) = 2 ???

• Akash,

In the example we are rotating the vector $\mathbf{p}=2\mathbf{i}$ 45° by the quaternion $q=[\cos\theta,\sin\theta\mathbf{k}]$ but in order to perform this operation, we must express $\mathbf{p}$ as a pure quaternion in the form $p=[0,\mathbf{p}]$. (notice how vectors are expressed use bold-face characters and quaternions are expressed as normal (not bold) characters).

If the quaternion $q$ correctly rotated the vector $\mathbf{p}$ then the result should also be a pure quaternion (with no scalar part) and the magnitude of the vector part should be the same as the original vector (because a rotation should not scale the original vector) however this example shows that this is not the case.

The magnitude of the original vector is

$\begin{array}{rcl}|\mathbf{p}|&=&\sqrt{2^{2}}\\&=&2\end{array}$

But the magnitude of the vector part of the resulting quaternion is:

$\begin{array}{rcl}|\mathbf{p}\prime|&=&\sqrt{\sqrt{2}^{2}+1^{2}}\\&=&\sqrt{2+1}\\&=&\sqrt{3}\end{array}$

Which is not the same magnitude as the original vector.

If I wanted to compute the magnitude of the resulting quaternion then I would need to consider the quaternion’s scalar part according to the formula described in the section titled Quaternion Norm. But since I’m only interested in rotating a vector by a quaternion I only want to consider the result of the vector part (and thus discard the scalar part when I compute the magnitude of the resulting vector).

18. and I have a question.whats that suppose to mean?

if a+bi is a point,isn`t that a Matrix multipy a point equals to a point more correct?

• cubase01,

Sorry, the matrix representation of a complex number is not explained in this article. I will try to briefly explain it here.

We can represent a complex number as the matrix $\mathbf{C}$ which is the sum of two other matrices representing the real $\mathbf{R}$ and the imaginary $\mathbf{I}$ parts (note that bold, upper-case characters represent matrices):

$\mathbf{C}=\mathbf{R}+\mathbf{I}$

Which can also be expressed in the more familiar form of the complex number:

$\mathbf{C}=a\mathbf{\hat{R}}+b\mathbf{\hat{I}}~~a,b\in\mathbb{R}$

where $\mathbf{\hat{R}}{\equiv}1$ and $\mathbf{\hat{I}}{\equiv}i$.

The matrix equivalent of 1 is the 2 x 2 identity matrix:

$\mathbf{\hat{R}}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$

And as was mentioned in the section titled Powers of i the imaginary component $i$ can be treated as a 90° counter-clockwise rotation in the complex plane which can also be represented by a rotation matrix:

$\begin{array}{rcl}\mathbf{\hat{I}}&=&\begin{bmatrix}\cos{90^{\circ}}&-\sin{90^{\circ}}\\\sin{90^{\circ}}&\cos{90^{\circ}}\end{bmatrix}\\&=&\begin{bmatrix}0&-1\\1&0\end{bmatrix}\end{array}$

Now we can express the complex number in matrix form:

$\begin{array}{rcl}\mathbf{C}&=&a\mathbf{\hat{R}}+b\mathbf{\hat{I}}\\&=&a\begin{bmatrix}1&0\\0&1\end{bmatrix}+b\begin{bmatrix}0&-1\\1&0\end{bmatrix}\\&=&\begin{bmatrix}a&-b\\b&a\end{bmatrix}\end{array}$

So what you see in the section titled Rotors is the matrix form of a complex number and the $a$ and $b$ are the real and imaginary parts of a complex number and rotating a complex number (represented in matrix form) by the 2×2 counter-clockwise rotation matrix produces another complex number (represented in matrix form).

It is also interesting to note that if we express the equation $i^2=-1$ in matrix form we get:

$\begin{bmatrix}0&-1\\1&0\end{bmatrix}\begin{bmatrix}0&-1\\1&0\end{bmatrix}=-1\begin{bmatrix}1&0\\0&1\end{bmatrix}$

which verifies that the square of the imaginary number is -1.

I hope this helps.

• Carlo,

Thanks for pointing this out. It should actually be:

$q_1{\cdot}q_2$

I recently switched LaTeX engines. The new one seems to be a bit more picky about separating parts of the equation for correct rendering.

19. Thank you ! THANK YOU ! this is an awesome website you made, and while the subjects are technical, they remain accessible… thanks again !

20. Thank you sooo much for this explanation! I searched for good ones in the web for nearly a month and this is far the best and most detailed I read. Now I can implement Quaternion rotation without shaming me for not knowing the stuff I am using…

21. Many thanks for this extraordinary introduction into quaternions.
I really appreciate this work!
Thank you.