Racquel Sanderson

April 21, 2026 · 11:26 am

Find the Equation of a Plane Given Three Points

Find the Cartesian equation of the plane containing the points $A(1, 2, 3), B(-3, 0, 2)$ and $C(2, -4, -1)$

Find $\overrightarrow{AB}$ and $\overrightarrow{AC}$
$\overrightarrow{AB}=\begin{pmatrix}-2\\0\\2\end{pmatrix}-\begin{pmatrix}1\\2\\3\end{pmatrix}=\begin{pmatrix}-3\\-2\\-1\end{pmatrix}$
$\overrightarrow{AC}=\begin{pmatrix}2\\-4\\-1\end{pmatrix}-\begin{pmatrix}1\\2\\3\end{pmatrix}=\begin{pmatrix}1\\-6\\-4\end{pmatrix}$

Find the cross product $\overrightarrow{AB}\times \overrightarrow{AC}$
$\begin{pmatrix}-3\\-2\\-1\end{pmatrix}\times\begin{pmatrix}1\\-6\\-4\end{pmatrix}=\begin{pmatrix}2\\-13\\20\end{pmatrix}$
This is the normal, $overrightarrow{n}$ , to the plane.
We know $A$ is on the plane

$\begin{equation*}\overrightarrow{n}\cdot (\overrightarrow{r}-\overrightarrow{OA})=0\end{equation}$

Hence $\overrightarrow{n}\cdot \overrightarrow{r}=\overrightarrow{n}\cdot \overrightarrow{OA}$
$\overrightarrow{n}\cdot \overrightarrow{r}=\begin{pmatrix}2\\-13\\20\end{pmatrix} \cdot \begin{pmatrix}1\\2\\3\end{pmatrix}=38$
Therefore the Cartesian equation of the plane is

$\begin{equation*}2x-13y+20z=38\end{equation}$

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Filed under Algebra, Cross Product, Vectors, Year 12 Specialist Mathematics

April 6, 2026 · 9:23 am

The Secret Life of Numbers – Kate Kitagawa and Timothy Revell

The first place I went to buy this was sold out! Seems somewhat amazing – perhaps they only had one copy. I found it at my favourite book store – Subiaco Bookshop.

Here’s the blurb …

Mathematics shapes almost everything we do. But despite its reputation as the study of fundamental truths, the stories we have been told about it are wrong. In The Secret Lives of Numbers, historian Kate Kitagawa and journalist Timothy Revell introduce readers to the mathematical boundary-smashers who have been erased by history because of their race, gender or nationality.

From the brilliant Arabic scholars of the ninth-century House of Wisdom, and the pioneering African American mathematicians of the twentieth century, to the ”lady computers” around the world who revolutionised our knowledge of the night sky, we meet these fascinating trailblazers and see how they contributed to our global knowledge today.

This revisionist, completely accessible and radically inclusive history of mathematics is as entertaining as it is important.

This has a lovely style and is very easy to read. As part of my maths degree, I studied some history, but it was very western and I enjoyed the global approach in this book. I do think it is accessible and anyone with an interest in maths or history could read it.

A review

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Tagged as book review, kate kitagawa, the secret life of numbers, timothy revell

March 21, 2026 · 7:47 am

Complex Loci

Sketch on an Argand diagram

$\begin{equation*}\lvert{z-6}\rvert-\lvert{z+6}\rvert=3\end{equation}$

where $z\in \mathbb{C}$

Let $z=x+yi$

$\begin{equation*}\lvert{x+yi-6}\rvert-\lvert{x+yi+6\rvert=3\end{equation}$

$\begin{equation*}\lvert{x-6+yi}\rvert-\lvert{x+6+yi\rvert=3\end{equation}$

$\begin{equation*}\sqrt{(x-6)^2+y^2}-\sqrt{(x+6)^2+y^2}=3\end{equation}$

$\begin{equation*}\sqrt{(x-6)^2+y^2}=3+\sqrt{(x+6)^2+y^2}\end{equation}$

Square both sides of the equation

$\begin{equation*}{(x-6)^2+y^2=9+6\sqrt{(x+6)^2+y^2}+(x+6)^2+y^2\end{equation}$

$\begin{equation*}x^2-12x+36+y^2-9-x^2-12x-36-y^2=6\sqrt{(x+6)^2+y^2}\end{equation}$

(1) $\begin{equation*}-24x-9=6\sqrt{(x+6)^2+y^2}\end{equation*}$

From equation $1$ we know $-24x-9\ge0$

Hence $x\le\frac{-3}{8}$ , which means we only have the left section of the hyperbola.

$\begin{equation*}-8x-3=2\sqrt{(x+6)^2+y^2}\end{equation}$

Square both sides of the equation

$\begin{equation*}(-8x-3)^2=4((x+6)^2+y^2)\end{equation}$

$\begin{equation*}64x^2+48x+9=4x^2+48x+144+y^2\end{equation}$

$\begin{equation*}60x^2-y^2=135\end{equation}$

$\begin{equation*}\frac{4x^2}{9}-\frac{y^2}{135}=1\end{equation}$

(2) $\begin{equation*}\frac{x^2}{\frac{9}{4}}-\frac{y^2}{135}=1\end{equation*}$

Remember, we have the left part of the hyperbola.

The $x-$ intercept $=-\sqrt{\frac{9}{4}}=-\frac{3}{2}$ and the asymptotes are $y=\pm \frac{\sqrt{135}}{\frac{3}{2}}x$

$y=\pm 2\sqrt{15}x$

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Filed under Algebra, Complex Numbers, Simplifying fractions, Sketching Complex Regions, Solving Equations, Year 12 Specialist Mathematics

Tagged as complex loci, hyperbola

March 11, 2026 · 5:34 am

Function Composition

If $g(f(x))=\frac{x}{x+1}$ and $f(x)=\frac{1}{1-2x}$ , find $g(x)$ .

$\begin{equation*}g(f(x))=g(\frac{1}{1-2x})=\frac{x}{x+1}\end{equation}$

Sometimes you can do this type of question by inspection, but this one is a bit harder. I am going to use a variable substitution.

Let $u=\frac{1}{1-2x}$

$\begin{equation*}u=\frac{1}{1-2x}\end{equation}$

$\begin{equation*}1-2x=\frac{1}{u}\end{equation}$

$\begin{equation*}1-\frac{1}{u}=2x\end{equation}$

$\begin{equation*}\frac{u-1}{u}=2x\end{equation}$

$\begin{equation*}x=\frac{u-1}{2u}\end{equation}$

Therefore

$\begin{equation*}g(u)=\frac{\frac{u-1}{2u}}{\frac{u-1}{2u}+1}\end{equation}$

$\begin{equation*}g(u)=\frac{\frac{u-1}{2u}}{\frac{u-1+2u}{2u}}\end{equation}$

$\begin{equation*}g(u)=\frac{u-1}{3u-1}\end{equation}$

Therefore

$\begin{equation*}g(x)=\frac{x-1}{3x-1}\end{equation}$

Let’s test it

$g(x)=\frac{x-1}{3x-1}$ and $f(x)=\frac{1}{1-2x}$

$\begin{equation*}g(f(x))=\frac{\frac{1}{1-2x}-1}{\frac{3}{1-2x}-1}\end{equation}$

$\begin{equation*}g(f(x))=\frac{\frac{1-(1-2x)}{1-2x}}{\frac{3-(1-2x)}{1-2x}}\end{equation}$

$\begin{equation*}g(f(x))=\frac{2x}{2+2x}\end{equation}$

$\begin{equation*}g(f(x))=\frac{x}{x+1}\end{equation}$

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Filed under Algebra, Composition, Functions, Simplifying fractions, Year 12 Specialist Mathematics

Tagged as function composition

February 25, 2026 · 1:22 pm

Complex Locus Question

My Year 12 Specialist students are working on complex loci again. The following type of question always creates confusion.

$arg(z-w_1)$ is the angle the vector from $w_1$ to $z$ makes with the positive $x-$ axis, likewise for $arg(z-w_2)$ .

I am going to plot a possible $z$ and try to see the geometry that works.

We want $arg(z-w_1)-arg(z-w_2)=\frac{\pi}{6}$

I am going to take advantage of some triangle geometry

Using the External Angle Theorem, we know $\alpha=\beta+\theta$

$\begin{equation*}arg(z-w_1)-arg(z-w_2)=(\alpha+\frac{\pi}{2})-(\beta+\frac{\pi}{2}0=\alpha-\beta=\theta\end{equation}$

Therefore $\theta=\frac{\pi}{6}$

So we want all of the $z$ values that have an angle of $\frac{\pi}{6}$

Now we are going to use some circle geometry -The angle at the circumference subtended by the same arc are congruent. So we need to find a circle that has those three points ( $z, w_1$ and $w_2$ ) on the circumference.

Hence the locus is

Now we need to find the radius and centre of the circle.

Using another circle theorem, the angle at the centre is twice the angle at the circumference.

The triangle must be equilateral (it is isosceles with a vertex angle of $\frac{\pi}{3}$ )

Hence the radius is 2.

$h=\sqrt{2^2-1^2}=\sqrt{3}$

Hence the centre is $(-\sqrt{3}+1, 0)$

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Filed under Circle Theorems, Complex Numbers, Geometry, Interesting Mathematics, Pythagoras, Sketching Complex Regions, Year 12 Specialist Mathematics

Tagged as complex loci

February 11, 2026 · 3:28 am

Complex Numbers and Trig Idenities

My Year 12 Specialist Students are using complex numbers to prove trigonometric identities.

Things like

$\begin{equation*}sin(5\theta)=16sin^5(\theta) -20sin^3(\theta)+5sin(\theta) \end{equation}$

Method 2 might be a little bit easier depending upon how your brain works.

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Filed under Algebra, Binomial Expansion Theorem, Complex Numbers, Identities, Trig Identities, Trigonometry, Year 12 Specialist Mathematics

Tagged as complex numbers, de moivre's theorem, trigonometric identities

February 1, 2026 · 8:42 am

Drawing Threefold Symmetry with Geogebra

I am an embroiderer and I have been experimenting with geometric drawings. Like the diagram below.

I have been using Geogebra – I use the desktop app.

(1) Start with a circle (mine has a radius of 5)

(2) Place 6 points equidistant around the circle (I placed a point on the circle and then used the angle of a given size tool – 60 degrees)

I have hidden the angle markers to make everything a bit easier to see.

(3) Draw circles (with the same radius as the original) on every second point.

(4) Find the distance between two centre points of the new circles

(5) Using the centres of the three new circles, draw circles where the radius is the distance you found in (4).

(6) Mark the points of intersection.

(7) Hide the big circles

(8) Draw arcs

Finished – change the colour (if you want) and remove labels, etc.

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Filed under Drawing, Geometry, Interesting Mathematics

Tagged as geometry, threefold symmetry

January 22, 2026 · 9:27 am

Complex Loci Question

A sketch of the locus of a complex number $z$ is shown above, determine the maximum value of $arg(z)$ correct to two decimal places where $0\le z \le 2\pi$

Draw tangent lines from the origin to the circle.

Remember tangent lines are perpendicular to the radii

The maximum argument is this angle

I am going to find the angle in two sections

From the diagram the radius of the circle is $2$ and the centre is $(4, 3)$ . Hence the distance from the origin to the centre is $5$ .

$\begin{equation*}sin(\theta_1)=\frac{2}{5}\end{equation}$

$\begin{equation*}\theta_1=0.412\end{equation}$

$\begin{equation*}sin(\theta_2)=\frac{3}{5}\end{equation}$

$\begin{equation*}\theta_2=0.644\end{equation}$

Hence maximum $arg(z)=1.06$

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Filed under Complex Numbers, Right Trigonometry, Sketching Complex Regions, Trigonometry, Year 12 Specialist Mathematics

Tagged as argument of a complex number, complex locus, right trig

January 8, 2026 · 9:25 am

Love & Math – Edward Frenkel

A friend left this behind when they moved countries (knowing that I liked maths). This was my second attempt at reading it.

Here’s the blurb …

What if you had to take an art class in which you were only taught how to paint a fence? What if you were never shown the paintings of van Gogh and Picasso, weren’t even told they existed? Alas, this is how math is taught, and so for most of us it becomes the intellectual equivalent of watching paint dry.

In Love and Math , renowned mathematician Edward Frenkel reveals a side of math we’ve never seen, suffused with all the beauty and elegance of a work of art. In this heartfelt and passionate book, Frenkel shows that mathematics, far from occupying a specialist niche, goes to the heart of all matter, uniting us across cultures, time, and space.

Love and Math tells two intertwined stories: of the wonders of mathematics and of one young man’s journey learning and living it. Having braved a discriminatory educational system to become one of the twenty-first century’s leading mathematicians, Frenkel now works on one of the biggest ideas to come out of math in the last 50 years: the Langlands Program. Considered by many to be a Grand Unified Theory of mathematics, the Langlands Program enables researchers to translate findings from one field to another so that they can solve problems, such as Fermat’s last theorem, that had seemed intractable before.

At its core, Love and Math is a story about accessing a new way of thinking, which can enrich our lives and empower us to better understand the world and our place in it. It is an invitation to discover the magic hidden universe of mathematics.

I believe the aim of this book was to show non-maths people the beauty and joy of mathematics. However, I think it would be hard going if you didn’t have a mathematics background. This is high level mathematics (at one stage he mentions that there are probably about 12 people in the world who understand what he is discussing). Having said that, I liked it. I particularly liked the personal aspects of the narrative – his life experiences, and the people he met and with whom he worked.

A review

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Tagged as edward frenkel, frenkel edward, Love and math

January 2, 2026 · 7:08 am

Geometry Puzzle (finding a fraction of an area)

Geometry Puzzles in Felt Tip: A Compilation of puzzles from 2018 – Catriona Shearer

Band $1$ and $3$ have the same area.

We want to find the area of the shaded segment.

As the dots are equally spaced, the sector’s angle is $\frac{\pi}{2} = (\frac{2\pi}{12}\times 3)$

Remember the area of a segment is $A=\frac{1}{2}r^2(\theta-sin(\theta))$ where the angle measurement is in radians.

(1) $\begin{equation*}A=\frac{1}{2}r^2(\frac{\pi}{2}-sin(\frac{\pi}{2}))=\frac{1}{2}r^2(\frac{\pi}{2}-1))=\frac{\pi r^2}{4}-\frac{r^2}{2}\end{equation*}$