Modelling with Vectors
Real-world objects rarely come labelled with vector equations — turning a word problem into points, lines and planes is the actual skill being tested, and every model comes with something worth criticising.
The tool used in almost every modelling question here is the perpendicular distance from a fixed point $A$ to a line $\mathbf{r}=\mathbf{p}+t\mathbf{d}$: write a general point $F=\mathbf{p}+t\mathbf{d}$, solve $\overrightarrow{FA}\cdot\mathbf{d}=0$ for $t$, then compute $|\overrightarrow{FA}|$. See Finding Perpendiculars for the full derivation — this lesson is about recognising when to reach for it inside a word problem.
Modelling a new connection to a fixed line
Since $M$ lies on the hillside plane: $3(16)-2k+6(9)=40 \Rightarrow 102-2k=40 \Rightarrow k=31$, so $M=(16,31,9)$.
The cable has direction $\overrightarrow{PQ}=(2,3,6)$, so a general point on it is $F=(4+2t,\,5+3t,\,2+6t)$.
$\overrightarrow{FM}\cdot(2,3,6)=0 \Rightarrow$ solving gives $t=\dfrac{144}{49}$, so $F\approx(9.88,13.82,19.63)$.
Shaft length $=|\overrightarrow{FM}|=\dfrac{\sqrt{21845}}{7}\approx21.1$ (3 s.f.).
$OP=\sqrt{45}\approx6.71$ and $OQ=\sqrt{164}\approx12.8$ — the new shaft, at $21.1$, is actually longer than both existing accessways, so on this model the company should not build it.
Modelling a detection or capture radius
If an object only travels between two named points $A$ and $B$, its path is the segment $\mathbf{r}(t)=A+t(B-A)$ with $0\leq t\leq1$ — not the whole infinite line. Solving $\overrightarrow{FA}\cdot\mathbf{d}=0$ can return a $t$ outside this range, which means the true closest point on the line lies off the segment entirely. When that happens, the closest point actually reached is whichever endpoint, $A$ or $B$, is nearer — always check the value of $t$ before trusting the answer.
Path: $\mathbf{r}(t)=A+t(B-A)=(1,2,2)+t(6,3,6)$, for $0\leq t\leq1$.
$\mathbf{r}(t)\cdot(6,3,6)=0 \Rightarrow$ solving gives $t=-\dfrac{8}{27}$.
This is outside $[0,1]$ — the closest point on the infinite line lies before the fox even starts walking, so it isn't a real position on its path. The closest the fox actually gets is therefore at $t=0$, i.e. at $A$ itself.
$OA=\sqrt{1+4+4}=3\text{ m}$, which is greater than the $2\text{ m}$ detection range, so the sensor does not detect the fox.
(Ignoring the restriction would have given the point on the infinite line at $t=-\tfrac{8}{27}$, a distance of only $\tfrac{\sqrt{17}}{3}\approx1.37\text{ m}$ from $O$ — comfortably inside the range, and the wrong answer.)
Finding $t$ from $\overrightarrow{FA}\cdot\mathbf{d}=0$ and immediately computing a distance without checking whether $t$ lies in the valid range for the scenario. If the object only moves between two given points, always check $0\leq t\leq1$ before accepting the result — otherwise you've measured the distance to a point the object never actually reaches.
Modelling a path meeting a boundary
Setting the $z$-component to zero: $60-20t=0 \Rightarrow t=3$.
Position at $t=3$: $(5-3,\ 8-6,\ 0)=(2,2,0)$.
On this model, the hook reaches the ground at the point $(2,2,0)$.
Nearly every modelling question ends with a part asking you to suggest a limitation or criticise the model. Always tie the answer to the specific object in the question rather than giving a generic statement. Common, exam-acceptable criticisms include:
- A line assumes the object travels in a perfectly straight path with no change of direction — real paths (an animal, a vehicle, a signal) curve, wobble, or adjust course.
- A line also assumes constant speed/heading — real motion speeds up, slows down, or stops.
- A plane assumes a surface (a hillside, the ground, a wall) is perfectly flat — real surfaces are uneven, curved, or textured.
- Modelling something as a point ignores its actual size and shape — a device has physical dimensions, and a living creature has a body, not a single coordinate.
- A fixed detection/capture radius assumes perfectly uniform range in every direction — real sensors and predators have blind spots, directional sensitivity, or an irregular range.
- The model ignores external factors altogether — wind, current, obstacles, or terrain that would realistically affect the path.
Exam-style questions
Question 1 12 marks
A power company maintains a straight underground cable that passes beneath a hill. The cable is modelled as a straight line and one side of the hill is modelled as a plane. Relative to a substation at the origin $O$, two existing access points on the cable are $P(3,6,1)$ and $Q(7,9,13)$.
(a) Find a vector equation for the line $PQ$, giving your answer in the form $\mathbf{r}=\mathbf{a}+t\mathbf{b}$. (2)
The side of the hill is modelled by the plane $2x-3y+6z=36$. The company wants to build a new access shaft, by the shortest possible route, from the point $M(21,k,5)$ on this side of the hill down to the cable, where $k$ is a constant.
(b) Find the value of $k$, and hence find the length of the shaft. (7)
(c) Given that it is only practical to build the shaft if it is significantly shorter than both existing accessways $OP$ and $OQ$, determine whether the company should build it. (2)
(d) Suggest one limitation of this model. (1)
(a) $\overrightarrow{PQ}=(4,3,12)$, so $\mathbf{r}=(3,6,1)+t(4,3,12)$ (B1, B1)
(b) $M$ on the plane: $2(21)-3k+6(5)=36 \Rightarrow 72-3k=36 \Rightarrow k=12$, so $M=(21,12,5)$ (M1, A1)
General point on $PQ$: $F=(3+4t,\,6+3t,\,1+12t)$; requiring $\overrightarrow{FM}\cdot(4,3,12)=0$ (M1, A1)
Solving gives $t=\dfrac{138}{169}$ (M1, A1)
Shaft length $=\dfrac{10\sqrt{445}}{13}\approx16.2$ (3 s.f.) (A1)
(c) $OP=\sqrt{46}\approx6.78$ and $OQ=\sqrt{299}\approx17.3$. The shaft ($\approx16.2$) is shorter than $OQ$ but not significantly shorter than $OP$ — so the company should not build it. (M1, A1)
(d) e.g. the side of the hill is very unlikely to be a perfectly flat plane in reality. (B1)
Question 2 7 marks
A garden sprinkler activates for anything that comes within $6\text{ m}$ of its fixed position. The sprinkler is modelled as a fixed point at the origin $O$. A cat is modelled as a particle moving in a straight line from $A(-1,4,3)$ to $B(0,2,5)$, where distances are in metres.
(a) Use the model to determine whether or not the sprinkler activates for the cat. (5)
(b) Criticise the model in relation to the cat. (1)
(c) Criticise the model in relation to the sprinkler. (1)
(a) Path: $\mathbf{r}(t)=(-1,4,3)+t(1,-2,2)$, for $0\leq t\leq1$ (B1)
$\mathbf{r}(t)\cdot(1,-2,2)=0 \Rightarrow$ solving gives $t=\dfrac13$, which lies in $[0,1]$ (M1, A1)
Closest point $F=\left(-\tfrac23,\tfrac{10}3,\tfrac{11}3\right)$; $OF=\sqrt{\tfrac{225}{9}}=5\text{ m}$ (M1, A1)
Since $5<6$, the sprinkler does activate for the cat. (A1)
(b) e.g. a cat is very unlikely to walk in a perfectly straight line at constant speed. (B1)
(c) e.g. treating the sprinkler as a single point ignores its physical size and its actual spray pattern/direction. (B1)