IBDP MAI : Topic 3 Geometry and trigonometry - AHL 3.11 Vector equation of a line in two and three dimensions AI HL Paper 3
Luca is a video game creator. He crafts his games to occur in a two-dimensional space, relative to an origin O. In one of his projects, an item moves along a straight path \(L_1\) with vector equation
\[ \mathbf{r} = \begin{pmatrix} -1 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -1 \end{pmatrix}, \quad \lambda \in \mathbb{Z}. \]
(a) Express \(L_1\) in the form \(x = x_0 + \lambda l\) and \(y = y_0 + \lambda m\), where \(l, m \in \mathbb{Z}\). [1]
Luca employs the matrix \[ T = \begin{pmatrix} 1 & 7 \\ 7 & -1 \end{pmatrix} \] to transform \(L_1\) into a new straight path \(L_2\). The item will then travel along \(L_2\).
(b) Find the vector equation of \(L_2\). [4]
Luca understands that the transformation provided by matrix \(T\) consists of the following three distinct transformations (in the order listed):
- A rotation of \(\frac{\pi}{4}\), anticlockwise around the origin \(O\)
- An enlargement with a scale factor of \(5\sqrt{2}\), centered at \(O\)
- A reflection over the straight line \(y = mx\), where \(m = \tan\alpha\), \(0 \leq \alpha < \pi\)
(c) Write down the matrix that represents
- the rotation,
- the enlargement.[2]
(d) The matrix \(R\) represents the reflection. Write down \(R\) in terms of \(\alpha\). [1]
(e) Given that \(T = RX\),
- use your answers to part (c) to find matrix \(X\).
- hence, find the value of \(\alpha\).[6]
▶️ Answer/Explanation
(a)
Extracting the components from the vector equation to express the line parametrically.
From \(\mathbf{r} = \begin{pmatrix} -1 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -1 \end{pmatrix}\), the x-component is \(-1 + 2\lambda\) and the y-component is \(1 – \lambda\).
Thus, \( x = -1 + 2\lambda \), \( y = 1 – \lambda \).
[1 mark]
(b)
Applying the transformation matrix \( T \) to the position vector of \( L_1 \) to derive the new line \( L_2 \).
Compute \( T \begin{pmatrix} -1 + 2\lambda \\ 1 – \lambda \end{pmatrix} \):
\( \begin{pmatrix} 1 & 7 \\ 7 & -1 \end{pmatrix} \begin{pmatrix} -1 + 2\lambda \\ 1 – \lambda \end{pmatrix} = \begin{pmatrix} 1 \times (-1 + 2\lambda) + 7 \times (1 – \lambda) \\ 7 \times (-1 + 2\lambda) + (-1) \times (1 – \lambda) \end{pmatrix} \)
\( = \begin{pmatrix} (-1 + 2\lambda) + (7 – 7\lambda) \\ (-7 + 14\lambda) + (-1 + \lambda) \end{pmatrix} = \begin{pmatrix} -1 + 2\lambda + 7 – 7\lambda \\ -7 + 14\lambda – 1 + \lambda \end{pmatrix} \)
\( = \begin{pmatrix} 6 – 5\lambda \\ -8 + 15\lambda \end{pmatrix} \)
Formulating the vector equation using the resulting coordinates, ensuring the direction vector is correctly adjusted.
So, the vector equation is \( r = \begin{pmatrix} 6 \\ -8 \end{pmatrix} + \lambda \begin{pmatrix} -5 \\ 15 \end{pmatrix} \).
[4 marks]
(c)
(i) Using the standard rotation matrix for an anticlockwise rotation of \(\frac{\pi}{4}\), where \(\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\) and \(\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\).
The rotation matrix is \( \begin{pmatrix} \cos(\frac{\pi}{4}) & -\sin(\frac{\pi}{4}) \\ \sin(\frac{\pi}{4}) & \cos(\frac{\pi}{4}) \end{pmatrix} \).
Substituting, \( \begin{pmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{pmatrix} \), or approximately \( \begin{pmatrix} 0.707 & -0.707 \\ 0.707 & 0.707 \end{pmatrix} \).
[1 mark]
(ii) Constructing the enlargement matrix with the given scale factor applied uniformly to both axes.
The enlargement matrix with scale factor \(5\sqrt{2}\) is \( \begin{pmatrix} 5\sqrt{2} & 0 \\ 0 & 5\sqrt{2} \end{pmatrix} \).
[1 mark]
(d)
Recalling the reflection matrix formula over a line \( y = mx \), expressed in terms of the angle \(\alpha\).
The reflection matrix is \( \begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{pmatrix} \).
[1 mark]
(e)
(i) Performing matrix multiplication of the rotation and enlargement matrices in the specified order to determine \( X \).
Multiply \( \begin{pmatrix} 5\sqrt{2} & 0 \\ 0 & 5\sqrt{2} \end{pmatrix} \) by \( \begin{pmatrix} \cos(\frac{\pi}{4}) & -\sin(\frac{\pi}{4}) \\ \sin(\frac{\pi}{4}) & \cos(\frac{\pi}{4}) \end{pmatrix} \):
\( \begin{pmatrix} 5\sqrt{2} \times \frac{\sqrt{2}}{2} + 0 \times \frac{\sqrt{2}}{2} & 5\sqrt{2} \times (-\frac{\sqrt{2}}{2}) + 0 \times \frac{\sqrt{2}}{2} \\ 0 \times \frac{\sqrt{2}}{2} + 5\sqrt{2} \times \frac{\sqrt{2}}{2} & 0 \times (-\frac{\sqrt{2}}{2}) + 5\sqrt{2} \times \frac{\sqrt{2}}{2} \end{pmatrix} \)
\( = \begin{pmatrix} 5 \times 1 & -5 \times 1 \\ 5 \times 1 & 5 \times 1 \end{pmatrix} = \begin{pmatrix} 5 & -5 \\ 5 & 5 \end{pmatrix} \).
So, \( X = \begin{pmatrix} 5 & -5 \\ 5 & 5 \end{pmatrix} \).
[3 marks]
(ii) substituting T, R and X
\[ \begin{pmatrix} 1 & 7 \\ 7 & -1 \end{pmatrix} = \begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{pmatrix} \begin{pmatrix} 5 & -5 \\ 5 & 5 \end{pmatrix} \]
multiplying by inverse (in any order)
\[ \begin{pmatrix} 1 & 7 \\ 7 & -1 \end{pmatrix} \begin{pmatrix} 5 & -5 \\ 5 & 5 \end{pmatrix}^{-1} = \begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{pmatrix} \]
\[ \begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{pmatrix} = \begin{pmatrix} -\frac{3}{5} & \frac{4}{5} \\ \frac{4}{5} & \frac{3}{5} \end{pmatrix} \]
\(\cos 2\alpha = -\frac{3}{5}\) AND \(\sin 2\alpha = \frac{4}{5}\)
\(\alpha = 1.11 \; (\approx 1.107148\ldots) \quad \text{OR} \quad 63.4^\circ \; (63.4349\ldots^\circ)\)
[Total 14 marks]