Vectors: Reflection of a line in a plane, foot of perpendicular, and parallel planes — TMJC 2025 H2 Math Prelim P1
What this question tests
Question
A drone flies in a straight line towards a rooftop for inspection. The drone's flight path is modelled by the line \[ l:\ \mathbf{r} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix},\quad \lambda \in \mathbb{R}, \] and the rooftop is modelled by the plane \(p: 2x + y - z = 7\).
(a) On Day 1, the drone flies towards the rooftop using the flight path. Upon reaching the rooftop, the drone flies in a new flight path, \(m\). The acute angle between the original flight path and the normal to the rooftop is the same as the acute angle between the new flight path and the same normal to the rooftop. Both flight paths and the normal lie on the same plane. Find the cosine of the angle between the original flight path and the new flight path in an exact non-trigonometrical form.
(b) On Day 2, the drone starts to descend from point \(A\) with coordinates \((2,1,0)\) towards the rooftop using the shortest possible path. Find the exact coordinates of the point where the drone lands on the rooftop.
(c) After landing, the drone glides on the rooftop in another new flight path, \(s\), that is perpendicular to the flight path \(l\). Find a cartesian equation of the flight path \(s\).
A second rooftop is built parallel to the first rooftop \(p\) such that the perpendicular distance between these two rooftops is \(d\) units. Find two possible cartesian equations of the second rooftop in terms of \(d\).
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(a) Cosine of the angle between the original and reflected flight paths
Find the acute angle \(\theta\) between \(l\) and the normal, then use the fact that reflection makes the new path sit at angle \(2\theta\) from the original.
With \(\mathbf{d}_l = (3, -2, 1)\) and \(\mathbf{n} = (2, 1, -1)\):
\[ \cos\theta = \dfrac{|\mathbf{d}_l \cdot \mathbf{n}|}{|\mathbf{d}_l|\,|\mathbf{n}|} = \dfrac{|6 - 2 - 1|}{\sqrt{14}\sqrt{6}} = \dfrac{3}{\sqrt{14}\sqrt{6}}. \]The angle between the original flight path and the new (reflected) flight path is \(2\theta\):
\[ \cos 2\theta = 2\cos^2\theta - 1 = 2\left(\dfrac{3}{\sqrt{14}\sqrt{6}}\right)^2 - 1 = \dfrac{18}{84} - 1 = -\dfrac{11}{14}. \](b) Coordinates where the drone lands on the rooftop
The shortest path is along the normal, so find the foot of perpendicular \(F\) from \(A\) to the plane.
Let \(F\) be the foot of perpendicular from \(A(2,1,0)\) to \(p\). Then \(l_{AF}:\ \mathbf{r} = (2,1,0) + \mu(2,1,-1)\), \(\mu \in \mathbb{R}\). Substituting into \(p\):
\[ \begin{aligned} \begin{pmatrix} 2 + 2\mu \\ 1 + \mu \\ -\mu \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} &= 7 \\ 4 + 4\mu + 1 + \mu + \mu &= 7 \\ 6\mu + 5 &= 7 \\ \mu &= \dfrac{1}{3}. \end{aligned} \]Coordinates of \(F\): \(\left(\tfrac{8}{3},\, \tfrac{4}{3},\, -\tfrac{1}{3}\right)\).
(c) Cartesian equation of flight path \(s\)
A direction perpendicular to both \(l\) and the normal is given by their cross product; the line passes through the landing point \(F\).
\[ \mathbf{d}_s = \mathbf{d}_l \times \mathbf{n} = \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix} \times \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ 5 \\ 7 \end{pmatrix}. \]So \(l_s:\ \mathbf{r} = \begin{pmatrix} 8/3 \\ 4/3 \\ -1/3 \end{pmatrix} + \alpha\begin{pmatrix} 1 \\ 5 \\ 7 \end{pmatrix}\). Converting to cartesian form:
\[ \begin{aligned} x &= \tfrac{8}{3} + \alpha \implies \alpha = \tfrac{3x-8}{3}, \\ y &= \tfrac{4}{3} + 5\alpha \implies \alpha = \tfrac{3y-4}{15}, \\ z &= -\tfrac{1}{3} + 7\alpha \implies \alpha = \tfrac{3z+1}{21}. \end{aligned} \] \[ \therefore\ \dfrac{3x - 8}{3} = \dfrac{3y - 4}{15} = \dfrac{3z + 1}{21}. \]
Second rooftop parallel to \(p\)
A parallel plane shares the normal of \(p\); set its perpendicular distance from \(p\) equal to \(d\) to find the constant.
Let the second rooftop be \(\mathbf{r} \cdot (2,1,-1) = D\). Distance from \(p\):
\[ \left|\dfrac{7 - D}{\sqrt{6}}\right| = d \Rightarrow D = 7 \pm \sqrt{6}\,d. \]The two possible equations are
\[ 2x + y - z = \sqrt{6}\,d + 7 \quad \text{or} \quad 2x + y - z = -\sqrt{6}\,d + 7. \]