Vectors: Lines, planes, intersection, angle between lines and shortest distance — RVHS 2025 H2 Math Prelim P2
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 kilometres 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) 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.
(b) Determine the possible values of \(k\) if the paths of the Earth and the asteroid do not intersect.
It is now given that \(k = 3\).
(c) 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.
(d)(i) Find the distance between the Earth and the asteroid at this instant.
(d)(ii) 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) Determine whether the satellite crosses Earth's orbital plane
If the line's direction is perpendicular to the plane's normal the line is parallel to the plane; then check whether the starting point lies on the plane.
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 lie on Earth's orbital plane.
Therefore, the satellite does not cross Earth's orbital plane.
Alternatively, substitute the line into the plane equation: \[ \left[\begin{pmatrix}3\\2\\3\end{pmatrix}+\lambda\begin{pmatrix}-2\\3\\1\end{pmatrix}\right]\cdot\begin{pmatrix}1\\-1\\5\end{pmatrix}=0 \] \[ 16 + 0\lambda = 0 \] There is no consistent value of \(\lambda\), so the satellite does not cross Earth's orbital plane.
(b) Possible values of \(k\) for non-intersection
Convert the asteroid's symmetric equation to vector form, solve the resulting system, and find the value of \(k\) that makes the lines meet — the lines fail to intersect for all other positive \(k\).
\(l_E\): \(\mathbf{r} = \begin{pmatrix}15\\20\\-1\end{pmatrix}+\mu\begin{pmatrix}4\\-1\\-1\end{pmatrix}\), \(\mu\in\mathbb{R}\)
\(l_A\): \(\dfrac{x-10}{2} = y+2 = \dfrac{0.25-z}{k}\) gives \(\mathbf{r} = \begin{pmatrix}10\\-2\\0.25\end{pmatrix}+t\begin{pmatrix}2\\1\\-k\end{pmatrix}\), \(t\in\mathbb{R}\)
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: \(t = 15.5\), \(\mu = 6.5\), \(tk = 7.75\). \[ \implies k = \frac{7.75}{15.5} = 0.5 \]
The paths intersect only when \(k = 0.5\). Hence the paths do not intersect for \(k\neq 0.5\), \(k\in\mathbb{R}^+\).
(c) Acute angle between the paths (with \(k = 3\))
Use the scalar product of the two direction vectors, taking the modulus to obtain the acute angle.
Let \(\theta\) be the acute angle between the paths of the Earth and the asteroid. \[ \cos\theta = \frac{\left|\begin{pmatrix}2\\1\\-3\end{pmatrix}\cdot\begin{pmatrix}4\\-1\\-1\end{pmatrix}\right|}{\left|\begin{pmatrix}2\\1\\-3\end{pmatrix}\right|\left|\begin{pmatrix}4\\-1\\-1\end{pmatrix}\right|} = \frac{|8-1+3|}{\sqrt{4+1+9}\,\sqrt{16+1+1}} = \frac{10}{\sqrt{14}\,\sqrt{18}} \] \[ \theta = 51.0^\circ \quad(\text{to 1 d.p.}) \]
(d)(i) Distance between Earth and asteroid at the closest instant
Find the magnitude of the vector joining the two given points.
\[ \text{Distance} = \left|\begin{pmatrix}12\\18\\-3\end{pmatrix}-\begin{pmatrix}12\\-1\\-2.75\end{pmatrix}\right| = \left|\begin{pmatrix}0\\19\\-0.25\end{pmatrix}\right| = \sqrt{19^2+(-0.25)^2} = 19.0 \text{ million km} \]
(d)(ii) Is this the shortest distance between the paths?
The shortest distance occurs when the connecting vector is perpendicular to both direction vectors; show it fails the test for at least one path.
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 connecting vector is not perpendicular to (at least) one of the paths, the distance found is not the shortest distance between the paths. It suffices to show either (1) or (2).