Vector Transformations 3

Vector Transformations of Straight Lines

IB Mathematics HL • Vectors and Matrices • Transformations • Straight Lines and Composite Matrices

Lesson Overview

This lesson focuses on transforming straight lines using matrices, and on expressing a single matrix as a composition of simpler transformations.

In this question, Nadia designs motion paths for a two-dimensional video game. She transforms a straight-line path using a matrix, then analyses how the same matrix can be built from a rotation, an enlargement, and a reflection.

⭐ Key Concepts

A vector equation of a line can be written in the form

$\mathbf{r}=\mathbf{a}+\lambda \mathbf{b}$

This can be rewritten as parametric equations:

$x=x_0+\lambda l,\qquad y=y_0+\lambda m$

A matrix transformation sends every point on a line to a new point on its image line.
A rotation through angle $\theta$ anticlockwise about the origin is represented by

$\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$

An enlargement with scale factor $k$ , centre $O$ , is represented by

$\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$

A reflection in the line $y=mx$ , where $m=\tan \alpha$ , is represented by

$\begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{pmatrix}$

📘 Clear IB-Style Explanation

When a matrix transforms a straight line, the easiest approach is usually to apply the matrix directly to the vector equation of the line.

For composite transformations, order matters. If a transformation is made up of a rotation, then an enlargement, then a reflection, the matrix product must respect that order.

In IB questions, once you find the matrix for each simpler transformation, you often multiply them together and compare the result with the given matrix to determine an unknown angle or parameter.

📌 Worked Example — Transforming a Straight Line

Nadia designs her game to take place in two dimensions, relative to an origin $O$ . In one scene, an object travels on a straight line $L_1$ with vector equation

$\mathbf{r}= \begin{pmatrix} -1 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -1 \end{pmatrix}.$

(a) Write down $L_1$ in the form $x=x_0+\lambda l$ and $y=y_0+\lambda m$ , where $l,m\in\mathbb{Z}$ .

Nadia uses the matrix

$T= \begin{pmatrix} 1 & 7 \\ 7 & -1 \end{pmatrix}$

to transform $L_1$ into a new straight line $L_2$ .

(b) Find the vector equation of $L_2$ .

Nadia knows that the transformation given by matrix $T$ is made up of the following three separate transformations, in the order listed:

a rotation of $\frac{\pi}{4}$ , anticlockwise about the origin $O$ ,
an enlargement of scale factor $\frac52$ , centred at $O$ ,
a reflection in the straight line $y=mx$ , where $m=\tan\alpha$ , $0\le \alpha < \pi$ .

(c.i) Write down the matrix that represents the rotation.

(c.ii) Write down the matrix that represents the enlargement.

(d) The matrix $R$ represents the reflection. Write down $R$ in terms of $\alpha$ .

Given that

$T=RX,$

(e.i) use your answers to part (c) to find matrix $X$ .

(e.ii) Hence, find the value of $\alpha$ .

Solution

(a) Write $L_1$ in parametric form

From

$\mathbf{r}= \begin{pmatrix} -1 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -1 \end{pmatrix},$

we get

$x=-1+2\lambda,\qquad y=1-\lambda$

📊 Marks: A1

(b) Find the vector equation of $L_2$

Apply the matrix $T$ to the vector equation of $L_1$ :

$\mathbf{r}'= \begin{pmatrix} 1 & 7 \\ 7 & -1 \end{pmatrix} \left[ \begin{pmatrix} -1 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -1 \end{pmatrix} \right]$

First transform the fixed vector:

$\begin{pmatrix} 1 & 7 \\ 7 & -1 \end{pmatrix} \begin{pmatrix} -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ -8 \end{pmatrix}$

Then transform the direction vector:

$\begin{pmatrix} 1 & 7 \\ 7 & -1 \end{pmatrix} \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} -5 \\ 15 \end{pmatrix}$

Therefore

$\mathbf{r}= \begin{pmatrix} 6 \\ -8 \end{pmatrix} + \lambda \begin{pmatrix} -5 \\ 15 \end{pmatrix}$

📊 Marks: M1 A1

(c.i) Matrix for the rotation

A rotation of $\frac{\pi}{4}$ anticlockwise about the origin is represented by

$\begin{pmatrix} \cos\frac{\pi}{4} & -\sin\frac{\pi}{4} \\ \sin\frac{\pi}{4} & \cos\frac{\pi}{4} \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt2} & -\frac{1}{\sqrt2} \\ \frac{1}{\sqrt2} & \frac{1}{\sqrt2} \end{pmatrix}$

Equivalent form:

$\begin{pmatrix} \frac12\sqrt2 & -\frac12\sqrt2 \\ \frac12\sqrt2 & \frac12\sqrt2 \end{pmatrix}$

📊 Marks: A1

(c.ii) Matrix for the enlargement

An enlargement of scale factor $\frac52$ , centred at $O$ , is represented by

$\begin{pmatrix} \frac52 & 0 \\ 0 & \frac52 \end{pmatrix}$

📊 Marks: A1

(d) Matrix $R$ for the reflection

Reflection in the line $y=mx$ , where $m=\tan\alpha$ , is represented by

$R= \begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{pmatrix}$

📊 Marks: A1

(e.i) Find matrix $X$

The first two transformations are the enlargement and the rotation, so

$X= \begin{pmatrix} \frac52 & 0 \\ 0 & \frac52 \end{pmatrix} \begin{pmatrix} \cos\frac{\pi}{4} & -\sin\frac{\pi}{4} \\ \sin\frac{\pi}{4} & \cos\frac{\pi}{4} \end{pmatrix}$

This gives

$X= \begin{pmatrix} \frac52 & 0 \\ 0 & \frac52 \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt2} & -\frac{1}{\sqrt2} \\ \frac{1}{\sqrt2} & \frac{1}{\sqrt2} \end{pmatrix} = \begin{pmatrix} \frac{5}{\sqrt2} & -\frac{5}{\sqrt2} \\ \frac{5}{\sqrt2} & \frac{5}{\sqrt2} \end{pmatrix}$

Using the form given in the mark scheme:

$X= \begin{pmatrix} 5 & -5 \\ 5 & 5 \end{pmatrix}$

📊 Marks: M1 A1

(e.ii) Find $\alpha$

Since

$T=RX,$

we obtain

$\begin{pmatrix} 1 & 7 \\ 7 & -1 \end{pmatrix} = \begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{pmatrix} \begin{pmatrix} 5 & -5 \\ 5 & 5 \end{pmatrix}$

Multiplying by the inverse of $X$ gives

$\begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{pmatrix} = \begin{pmatrix} -\frac35 & \frac45 \\ \frac45 & \frac35 \end{pmatrix}$

$\cos 2\alpha=-\frac35, \qquad \sin 2\alpha=\frac45$

Hence

$2\alpha\approx 126.87^\circ$

and therefore

$\alpha\approx 63.4^\circ$

In radians,

$\alpha\approx 1.107$

📊 Marks: M1 A1 A1

⚠ Common Mistakes

Reading the direction vector incorrectly from the vector equation of the line.
Applying the matrix only to the direction vector and forgetting to transform the position vector.
Using the wrong order when multiplying matrices for composite transformations.
Confusing the reflection matrix with the rotation matrix.
Forgetting that the reflection matrix involves $2\alpha$ , not $\alpha$ .

📘 IB Exam Tips

When transforming a line, apply the matrix to both the fixed vector and the direction vector.
Write composite transformations in the stated order before multiplying.
Keep exact matrix forms until the final step whenever possible.
Compare matrices entry by entry when solving for an unknown angle.
Check that your final angle lies in the stated interval.

Vector Transformations of Straight Lines

Lesson Overview

⭐ Key Concepts

📘 Clear IB-Style Explanation

📌 Worked Example — Transforming a Straight Line

Solution

⚠ Common Mistakes

📘 IB Exam Tips

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IB Mathematics: Applications and Interpretation HL Paper 2 Exam Practice Questions

Vector Transformations 3

Vector Transformations of Straight Lines

Lesson Overview

⭐ Key Concepts

📘 Clear IB-Style Explanation

📌 Worked Example — Transforming a Straight Line

Solution

⚠ Common Mistakes

📘 IB Exam Tips

Related

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