Question 15
The vectors representing the position (m) of particles A and B are given by
$$\begin{aligned}\boldsymbol{r}_A&=(4t-9)\hat{i}-2(5-t)\hat{j}-8\hat{k}\\\boldsymbol{r}_B&=(t^2+1)\hat{i}-3\hat{j}+(4-at^2)\hat{k}\end{aligned}$$
respectively, where $t$ is the time of motion for $0\le t\le10$ seconds.
Show that particle A passes through the point $P(5,\ -3,\ -8)$.
[2 marks] b)Given that particle B also passes through point $P$, determine the value of $a$.
[2 marks] c)Determine the vector that represents the displacement of particle B relative to particle A during the given time of motion. Express your answer in simplest form.
[1 mark] d)Use your result from Question 15c) to determine the shortest distance between particles A and B during the given time of motion.
[3 marks]Both particles pass through P, but at different times
Watch when each particle reaches the point, then follow the distance between them over time.
B leaves $P$ just before the particles are closest, while A is still 1.5 s away from it.
Move through time and watch the gap between the particles
With $a=3$, pick a time. The readouts show where each particle is, the displacement of B relative to A, and its magnitude.
The graph stops at 4 s. After that the particles keep moving apart, and by 10 s they are about 297 m apart.
Work through the solution one mark at a time
Try each step yourself before you reveal it.
Set the $\hat{i}$ component of $\boldsymbol{r}_A$ equal to the $x$-coordinate of $P$.
$$4t-9=5\ \Rightarrow\ t=3.5\text{ s}$$
Marker: correctly determines time taken for particle A to reach $P$ by considering the $\hat{i}$ component of $\boldsymbol{r}_A$
Now do the same with the $\hat{j}$ component. It has to give the same time.
$$-2(5-t)=-3\ \Rightarrow\ t=3.5\text{ s}$$
The $\hat{k}$ component is always $-8$, which matches $P$.
The guide also accepts finding $t=3.5$ from one component and then showing $\boldsymbol{r}_A(3.5)=5\hat{i}-3\hat{j}-8\hat{k}$.
Marker: correctly determines time taken for particle A to reach $P$ by considering the $\hat{j}$ component of $\boldsymbol{r}_A$
The $\hat{i}$ component of $\boldsymbol{r}_B$ doesn’t contain $a$, so use it to find when B reaches $P$.
$$t^2+1=5\ \Rightarrow\ t^2=4\ \Rightarrow\ t=2\text{ s}$$
Reject $t=-2$ because $0\le t\le10$. Writing $t=\pm2$ or $t^2=4$ also gets this mark.
Marker: correctly determines time for object B to reach $P$
At $t=2$ the $\hat{k}$ component of $\boldsymbol{r}_B$ must equal $-8$.
$$4-a(2)^2=-8\ \Rightarrow\ 4a=12$$
Marker: determines value of $a$
The displacement of B relative to A is $\boldsymbol{r}_B-\boldsymbol{r}_A$. Use $a=3$ and subtract component by component.
$$\boldsymbol{r}_B-\boldsymbol{r}_A=(t^2+1)\hat{i}-3\hat{j}+(4-3t^2)\hat{k}-\big((4t-9)\hat{i}-2(5-t)\hat{j}-8\hat{k}\big)$$
Watch the $\hat{j}$ component: $-3-(-10+2t)=7-2t$. A column vector is also accepted.
Marker: determines the required displacement vector in simplest form
The distance between the particles is the magnitude of the vector from part c).
$$\left|\boldsymbol{r}_B-\boldsymbol{r}_A\right|=\sqrt{(t^2-4t+10)^2+(7-2t)^2+(12-3t^2)^2}\text{ m}$$
The expanded form $\sqrt{10t^4-8t^3-32t^2-108t+293}$ is also accepted, and so is the square of the distance.
Marker: determines an expression that represents the distance between the particles
Say why you are looking for a minimum.
The particles are closest when the distance between them is a minimum.
You can also say you want the minimum of $\left|\boldsymbol{r}_B-\boldsymbol{r}_A\right|$ or of its square, sketch the graph of the distance against time with the minimum marked, or state that the minimum occurs at $t\approx2.03$ s. This mark can be implied by your next line.
Marker: justifies when shortest distance between the particles occurs
Graph the distance on your calculator for $0\le t\le10$ and find the minimum. It occurs at $t\approx2.03$ s.
Rounding to 6.7 m is also accepted.
Marker: determines shortest distance between particles
Each phrase of the question gave you something
Each part is a standard routine: match components to a point, subtract two position vectors, then minimise a magnitude on your calculator. The parts lead into each other, so a slip in b) or c) carries through, but the guide allows follow-through marks.
If I had to guess where you might lose marks, it would be part b). If you set all three components of $\boldsymbol{r}_B$ equal to $P$ at the time A arrives, $t=3.5$, nothing works. The question only says B passes through $P$, so B gets its own time, which comes from the $\hat{i}$ component.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2024 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2024, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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