Vectors: Contextual trajectories; angle; distance — RVHS 2025 H2 Math Prelim Paper 2
What this question tests
Question
The Planetary Defense Coordination Office at NASA observes and tracks Near Earth Objects (NEOs) that could be potentially hazardous in a collision with Earth. Over short periods of time, vectors can be used to model the trajectories of the Earth and NEOs. They use coordinates \((x, y, z)\) with units in millions of kilometers relative to the sun which is at position \((0,0,0)\).
Earth's orbit is contained by the plane \(\Pi\) with equation \(\mathbf{r} \cdot \begin{pmatrix}1\\-1\\5\end{pmatrix} = 0\).
- A satellite moves along a path with equation \(\mathbf{r} = \begin{pmatrix}3\\2\\3\end{pmatrix} + \lambda\begin{pmatrix}-2\\3\\1\end{pmatrix}\), \(\lambda \in \mathbb{R}\). Determine whether the satellite crosses Earth's orbital plane.
Over a short period of time, Earth's motion can be modelled by the equation \[ \mathbf{r} = \begin{pmatrix}15\\20\\-1\end{pmatrix} + \mu\begin{pmatrix}4\\-1\\-1\end{pmatrix},\quad\mu\in\mathbb{R}. \] An asteroid moves along a path with equation \[ \frac{x-10}{2} = y+2 = \frac{0.25-z}{k} \] where \(k\) is a positive constant.
- Determine the possible values of \(k\) if the paths of the Earth and the asteroid do not intersect.
It is now given that \(k = 3\).
- Find the acute angle between the paths of the Earth and the asteroid.
The asteroid is closest to the Earth when they are at points \((12,\,18,\,-3)\) and \((12,\,-1,\,-2.75)\) on their respective paths.
- Find the distance between the Earth and the asteroid at this instant.
- By considering the vector between the asteroid and the Earth at this instant, determine whether the distance found in part (d)(i) is the shortest distance between the two paths.
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(a)
Since \(\begin{pmatrix}1\\-1\\5\end{pmatrix}\cdot\begin{pmatrix}-2\\3\\1\end{pmatrix} = -2-3+5=0\), the satellite's path is parallel to Earth's Orbital plane.
Since \(\begin{pmatrix}3\\2\\3\end{pmatrix}\cdot\begin{pmatrix}1\\-1\\5\end{pmatrix} = 3-2+15=16\neq 0\), the satellite does not sit on Earth's Orbital plane.
Therefore, the satellite does not cross Earth's Orbital Plane.
Alternatively:
Earth's orbital plane: \(\mathbf{r}\cdot\begin{pmatrix}1\\-1\\5\end{pmatrix}=0 \quad\cdots(1)\)
Satellite: \(\mathbf{r} = \begin{pmatrix}3\\2\\3\end{pmatrix}+\lambda\begin{pmatrix}-2\\3\\1\end{pmatrix}\), \(\lambda\in\mathbb{R} \quad\cdots(2)\)
Substitute \((2)\) into \((1)\): \[ \left[\begin{pmatrix}3\\2\\3\end{pmatrix}+\lambda\begin{pmatrix}-2\\3\\1\end{pmatrix}\right]\cdot \begin{pmatrix}1\\-1\\5\end{pmatrix}=0 \] \[ 16 + 0\lambda = 0 \] There is no consistent value of \(\lambda\) for this equation. Therefore, the satellite does not cross Earth's Orbital Plane.
(b)
\(l_E\): \(\mathbf{r} = \begin{pmatrix}15\\20\\-1\end{pmatrix}+\mu\begin{pmatrix}4\\-1\\-1\end{pmatrix}\), \(\mu\in\mathbb{R}\)
\(l_A\): \(\dfrac{x-10}{2} = y+2 = \dfrac{0.25-z}{k}\)
\(\mathbf{r} = \begin{pmatrix}10\\-2\\0.25\end{pmatrix}+t\begin{pmatrix}2\\1\\-k\end{pmatrix}\), \(t\in\mathbb{R}\)
Consider the system of equations: \[ \begin{pmatrix}10+2t\\-2+t\\0.25-tk\end{pmatrix} = \begin{pmatrix}15+4\mu\\20-\mu\\-1-\mu\end{pmatrix} \] \[ 2t - 4\mu = 5, \quad t + \mu = 22, \quad \mu - tk = -1.25 \]
From GC: \(t = 15.5\), \(\mu = 6.5\), \(tk = 7.75\). \[ \implies k = \frac{7.75}{15.5} = 0.5 \]
Therefore, the paths intersect when \(k = 0.5\).
(c) Let \(\theta\) be the acute angle between the paths of the Earth and the asteroid. \[ \cos\theta = \frac{\left|\begin{pmatrix}2\\1\\-3\end{pmatrix}\cdot\begin{pmatrix}4\\-1\\-1\end{pmatrix}\right|} {\left|\begin{pmatrix}2\\1\\-3\end{pmatrix}\right|\left|\begin{pmatrix}4\\-1\\-1\end{pmatrix}\right|} = \frac{|8-1+3|}{\sqrt{4+1+9}\sqrt{16+1+1}} = \frac{10}{\sqrt{14}\sqrt{18}} \]
(d)(i)
\[ \text{Distance} = \left|\begin{pmatrix}12\\18\\-3\end{pmatrix}-\begin{pmatrix}12\\-1\\-2.75\end{pmatrix}\right| = \left|\begin{pmatrix}0\\19\\-0.25\end{pmatrix}\right| = \sqrt{19^2+(-0.25)^2} \](d)(ii) The shortest distance between the paths will be when the vector connecting the two points on the paths is perpendicular to each of the two paths.
Check perpendicularity: \[ \begin{pmatrix}0\\19\\-0.25\end{pmatrix}\cdot\begin{pmatrix}2\\1\\-3\end{pmatrix} = 0 + 19 - 0.75 = 19.75 \neq 0 \quad\cdots(1) \]
OR
\[ \begin{pmatrix}0\\19\\-0.25\end{pmatrix}\cdot\begin{pmatrix}4\\-1\\-1\end{pmatrix} = 0 - 19 + 0.25 = -18.75 \neq 0 \quad\cdots(2) \]Since the vector is not perpendicular to one of the paths, the distance found is not the shortest distance between the paths.
To show that \(\left|\begin{pmatrix}0\\19\\-0.25\end{pmatrix}\right|\) is not the shortest distance, it suffices to show either (1) or (2).