2024 H2 Maths Paper 1 Q12 — Air Traffic Vectors

The scene, to scalePart (a)(i)

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Position along the aircraft's path,

\lambda

0.00

Aircraft path Drone path 4 km range Measured distance Shortest distance

Part (a)(i)

4 marks

Find the coordinates of the aircraft at its closest point to the air traffic controllers.

Approach

The controllers are not at the origin. The origin is the base of the tower; the controllers sit 0.1 km above it, at $(0,\,0,\,0.1)$ . Everything in part (a) is measured from there, and a solution that forgets the 0.1 loses the marks even though the method is right.

Why perpendicular means closest: any other point on the path makes a right-angled triangle in which the perpendicular is a leg and the new segment is the hypotenuse — and the hypotenuse is always the longer of the two.

A general point $A$ on the aircraft's path is

\mathbf{r}=\begin{pmatrix}4\\3\\1\end{pmatrix}+\lambda\begin{pmatrix}1\\-1\\2\end{pmatrix}\quad\Longrightarrow\quad A\,(4+\lambda,\ 3-\lambda,\ 1+2\lambda).

Writing $C$ for the controllers' position, the vector from the controllers to the aircraft is

\overrightarrow{CA}=\begin{pmatrix}4+\lambda\\3-\lambda\\1+2\lambda\end{pmatrix}-\begin{pmatrix}0\\0\\0.1\end{pmatrix}=\begin{pmatrix}4+\lambda\\3-\lambda\\0.9+2\lambda\end{pmatrix}.

At the closest point this vector is perpendicular to the direction of the path, so their scalar product is zero:

\begin{aligned}\begin{pmatrix}4+\lambda\\3-\lambda\\0.9+2\lambda\end{pmatrix}\cdot\begin{pmatrix}1\\-1\\2\end{pmatrix}&=0\\[2pt](4+\lambda)-(3-\lambda)+2(0.9+2\lambda)&=0\\6\lambda+2.8&=0\\\lambda&=-\frac{7}{15}\end{aligned}

Substituting $\lambda=-\frac{7}{15}$ back into the path:

x=4-\frac{7}{15}=\frac{53}{15},\qquad y=3+\frac{7}{15}=\frac{52}{15},\qquad z=1-\frac{14}{15}=\frac{1}{15}

To check your answer:

At $\lambda=-\frac{7}{15}$ the vector is $\left(\frac{53}{15},\ \frac{52}{15},\ -\frac{1}{30}\right)$ , and $\frac{53}{15}-\frac{52}{15}+2\left(-\frac{1}{30}\right)=\frac{1}{15}-\frac{1}{15}=0$ , so it really is perpendicular to the direction vector.

$\therefore$ the aircraft is closest to the controllers at $\left(\frac{53}{15},\ \frac{52}{15},\ \frac{1}{15}\right)$ .

Note

That last coordinate is $\frac{1}{15}$ km, about 67 m. The aircraft is nearly at ground level when it passes closest — turn the model to a low angle and you can see it skim in over the tower's own height.

Part (a)(ii)

2 marks

Hence determine whether the controllers will see this aircraft, given they can only see aircraft within 4 km.

Thought process

"Hence" is doing real work here. Part (i) found the point where the distance is smallest. If even the smallest distance is more than 4 km, then no point of the whole path is within range — one calculation settles the entire flight.

The distance from the controllers to that closest point is

\begin{aligned}d=\left|\overrightarrow{CA}\right|&=\sqrt{\left(\frac{53}{15}\right)^{2}+\left(\frac{52}{15}\right)^{2}+\left(\frac{1}{15}-\frac{1}{10}\right)^{2}}\\[2pt]&=\sqrt{\frac{2809}{225}+\frac{2704}{225}+\frac{1}{900}}\\[2pt]&=\sqrt{\frac{22053}{900}}=\frac{\sqrt{22053}}{30}\\[2pt]&=4.9501\ldots\end{aligned}

To check your answer:

$\frac{22053}{900}=24.5033\ldots$ and $4.9501^{2}=24.5035\ldots$ , agreeing to 4 decimal places.

Since $d\approx4.95\ \text{km}\gt4\ \text{km}$ , the aircraft never comes within 4 km of the controllers.

$\therefore$ the air traffic controllers will not see this aircraft.

Distance from the controllers as the aircraft flies. The curve dips to 4.95 km and climbs again — it never crosses the 4 km line.

Part (b)

3 marks

Determine the possible values of the positive constant $k$ if the two paths do not intersect.

The trap in the first line

$\dfrac{2x-3}{3}$ is not $\dfrac{x-3}{3}$ , and it is not $\dfrac{x-1.5}{3}$ either. The coefficient of $x$ must be 1 before you can read off a direction vector. Setting the whole expression equal to a parameter avoids the question entirely, which is why it is the safer route.

Setting the common value of the drone's Cartesian equation to $\mu$ :

\frac{2x-3}{3}=y+1=\frac{z-k}{-1}=\mu\quad\Longrightarrow\quad x=\frac{3}{2}+\frac{3}{2}\mu,\quad y=-1+\mu,\quad z=k-\mu

So the drone's path in vector form is

\mathbf{r}=\begin{pmatrix}\tfrac{3}{2}\\-1\\k\end{pmatrix}+\mu\begin{pmatrix}\tfrac{3}{2}\\1\\-1\end{pmatrix},\qquad\text{direction }\begin{pmatrix}3\\2\\-2\end{pmatrix}.

If the paths intersect, there are values of $\lambda$ and $\mu$ satisfying all three coordinates at once:

\begin{aligned}4+\lambda&=\tfrac{3}{2}+\tfrac{3}{2}\mu&&\cdots(1)\\3-\lambda&=-1+\mu&&\cdots(2)\\1+2\lambda&=k-\mu&&\cdots(3)\end{aligned}

Equations (1) and (2) involve only $\lambda$ and $\mu$ , so solve those first:

\begin{aligned}-\tfrac{5}{2}+\tfrac{3}{2}\mu&=4-\mu\\\tfrac{5}{2}\mu&=\tfrac{13}{2}\\\mu&=2.6,\qquad\lambda=4-2.6=1.4\end{aligned}

Equation (3) then fixes the only value of $k$ that lets the paths meet:

1+2(1.4)=k-2.6\quad\Longrightarrow\quad k=6.4

To check your answer:

Putting $\mu=2.6$ into (1) gives $\lambda=-2.5+1.5(2.6)=1.4$ , matching (2). With $k=6.4$ , both paths pass through $(5.4,\ 1.6,\ 3.8)$ .

$\therefore$ the paths intersect only when $k=6.4$ , so they do not intersect for $k\gt0,\ k\neq6.4$ .

Note — what is happening geometrically

The two directions are not parallel, so for every other $k$ the paths are skew: one passes over the other with a gap of $\dfrac{|5k-32|}{\sqrt{93}}$ km. Raising $k$ lifts the whole drone path, and the gap closes to zero exactly at $k=6.4$ before opening again.

The gap between the two paths against

k

. It touches zero at one value only — which is why "do not intersect" excludes a single number rather than an interval.

Part (c)

2 marks

Find the acute angle between the paths of the aircraft and the drone.

The angle between two paths is the angle between their direction vectors:

\begin{pmatrix}1\\-1\\2\end{pmatrix}\cdot\begin{pmatrix}3\\2\\-2\end{pmatrix}=3-2-4=-3

\left|\begin{pmatrix}1\\-1\\2\end{pmatrix}\right|=\sqrt{6},\qquad\left|\begin{pmatrix}3\\2\\-2\end{pmatrix}\right|=\sqrt{17}

Since the angle asked for is acute, take the magnitude of the scalar product:

\cos\theta=\frac{|-3|}{\sqrt{6}\,\sqrt{17}}=\frac{3}{\sqrt{102}}\quad\Longrightarrow\quad\theta=\cos^{-1}\!\left(\frac{3}{\sqrt{102}}\right)

To check your answer:

$\sqrt{102}\approx10.0995$ , so $\cos\theta\approx0.2970$ and $\theta\approx72.7^\circ$ — acute, as required.

$\therefore$ the acute angle between the paths is $\cos^{-1}\!\left(\frac{3}{\sqrt{102}}\right)=72.7^\circ$ (1 d.p.).

Note

Dropping the modulus signs gives $107.3^\circ$ . That is the angle between the direction vectors as drawn, and it is obtuse — the two answers add to $180^\circ$ . The paths themselves cross at both angles; the question names the one it wants.

The scalar product is negative, so the two direction vectors as given point away from each other and meet at 107.3°. Reversing one of them — which changes nothing about the path it describes — gives the acute angle the question asks for.

Part (d)(i)

2 marks

With $k=2$ , find the distance between the aircraft at $(4,3,1)$ and the drone at $(4.2,\,0.8,\,0.2)$ .

The vector from the aircraft to the drone at that instant is

\overrightarrow{AB}=\begin{pmatrix}4.2\\0.8\\0.2\end{pmatrix}-\begin{pmatrix}4\\3\\1\end{pmatrix}=\begin{pmatrix}0.2\\-2.2\\-0.8\end{pmatrix}

so the distance between them is

\left|\overrightarrow{AB}\right|=\sqrt{0.2^{2}+(-2.2)^{2}+(-0.8)^{2}}=\sqrt{0.04+4.84+0.64}=\sqrt{5.52}=\frac{\sqrt{138}}{5}

To check your answer:

$\left(\frac{\sqrt{138}}{5}\right)^{2}=\frac{138}{25}=5.52$ , and $\frac{\sqrt{138}}{5}=2.3494\ldots$ Both given points do lie on their paths: $(4,3,1)$ is the aircraft at $\lambda=0$ , and $(4.2,\,0.8,\,0.2)$ is the drone at $\mu=1.8$ when $k=2$ .

$\therefore$ the distance between them at that instant is $\frac{\sqrt{138}}{5}=2.35$ km (3 s.f.).

Part (d)(ii)

2 marks

By considering the vector between them, determine whether that distance is the shortest distance between the two paths.

Approach

The shortest distance between two paths is measured along a line perpendicular to both of them. So there is nothing to calculate beyond a scalar product: test the joining vector against a direction vector, and a single non-zero answer settles it.

The same argument as in part (a), now with two paths instead of one: only the segment meeting both at right angles measures the true gap. Any other joining segment — the radar reading among them — is a hypotenuse built on it.

Testing $\overrightarrow{AB}$ against the aircraft's direction vector:

\begin{pmatrix}0.2\\-2.2\\-0.8\end{pmatrix}\cdot\begin{pmatrix}1\\-1\\2\end{pmatrix}=0.2+2.2-1.6=0.8

To check your answer:

The drone's direction fails too: $\begin{pmatrix}0.2\\-2.2\\-0.8\end{pmatrix}\cdot\begin{pmatrix}3\\2\\-2\end{pmatrix}=0.6-4.4+1.6=-2.2$ . Either non-zero value is enough to answer the question; showing one is what earns the marks.

Since $\overrightarrow{AB}\cdot\begin{pmatrix}1\\-1\\2\end{pmatrix}=0.8\neq0$ , the joining vector is not perpendicular to the aircraft's path.

$\therefore$ the distance found in (d)(i) is not the shortest distance between the two paths.

Note

Using the gap formula from part (b) with $k=2$ , the true shortest distance is $\dfrac{|5(2)-32|}{\sqrt{93}}=\dfrac{22}{\sqrt{93}}\approx2.28$ km — shorter than the 2.35 km measured on radar, exactly as "not the shortest" requires. The model draws both segments so you can see the radar reading is taken at a slight slant.

What this question was really testing

Reading the position that matters. The controllers are 0.1 km above the origin. One line of setup, four marks riding on it.
Perpendicularity as a tool, not a fact. Closest point, shortest distance, "is this the shortest?" — three different parts, one scalar product set to zero.
Converting a Cartesian equation honestly. Make the coefficient of $x$ equal to 1, or set the whole thing to a parameter. Never read a direction vector off $\frac{2x-3}{3}$ by eye.
Answering the question that was asked. "Hence", "acute", "positive constant" — each word closes off a whole line of work you would otherwise do.

Two aircraft paths, one control tower, and the 4 km question

Part (a)(i)

Part (a)(ii)

Part (b)

Part (c)

Part (d)(i)

Part (d)(ii)

What this question was really testing

This is how we mark, and how we teach.