Here we present three solutions to the one problem. The vector solution is probably the slickest. The geometry solution here can be simplified by being less rigorous, and the coordinate geometry solution might be made easier by using the $-b\pm\sqrt{\cdots}$ formula.

### LCHL 2007, Q. 2(a)

Ship B is travelling west at 24 km/h. Ship A is travelling north at 32 km/h.
At a certain instant ship B is 8 km north-east of ship A.

(i) Find the velocity of ship A relative to ship B.

(ii) Calculate the length of time, to the nearest minute, for which the ships
are less than or equal to 8 km apart.

Solution to (i): We have that $\overrightarrow{V_A}=32\mathbf{j}$ and $\overrightarrow{V_B}=-24\mathbf{i}$ and so $\overrightarrow{V_{AB}}=\overrightarrow{V_A}-\overrightarrow{V_B}=32\mathbf{j} -(-24\mathbf{i})=24\mathbf{i}+32\mathbf{j}$.

### Vector Approach to (ii)

First of all we draw a picture. As we are talking relative to ship B we will put B at the origin. If ship B is 8 km north-east of ship A then ship A is 8 km south-west of ship B. Where $\overrightarrow{R_0}$ is the initial displacement of ship A relative to ship B, the displacement of ship A relative to ship B, as a function of time, is given by $\overrightarrow{R_{AB}}(t)=\overrightarrow{R_0}+t\cdot \overrightarrow{V_{AB}}$.

Using some trigonometry — vertical and horizontal components of $\overrightarrow{R_0}$ — we have $\displaystyle\overrightarrow{R_0}=-\frac{8}{\sqrt{2}}\mathbf{i}-\frac{8}{\sqrt{2}}\mathbf{j}$,

and so $\displaystyle \overrightarrow{R_{AB}(t)}=\left(24t-\frac{8}{\sqrt{2}}\right)\mathbf{i}+\left(32t-\frac{8}{\sqrt{2}}\right)\mathbf{j}$.

We are interested in finding out for how long this displacement of ship A from ship B is less than eight: $\displaystyle\left|\overrightarrow{R_{AB}(t)}\right|\leq 8$ $\displaystyle \Rightarrow \sqrt{\left(24t-\frac{8}{\sqrt{2}}\right)^2+\left(32t-\frac{8}{\sqrt{2}}\right)^2}\leq 8$ $\displaystyle \Rightarrow \left(24t-\frac{8}{\sqrt{2}}\right)^2+\left(32t-\frac{8}{\sqrt{2}}\right)^2\leq 64$ $\displaystyle \Rightarrow 576t^2-\frac{384}{\sqrt{2}}t+\frac{64}{2}+1024t^2-\frac{512}{\sqrt{2}}t+\frac{64}{2}\leq 64$ $\displaystyle \Rightarrow 1600t^2-\frac{896}{\sqrt{2}}t\leq 0$.

This $\displaystyle 1600t^2-\frac{896}{\sqrt{2}}t$ is a $\bigcup$-quadratic and so is negative between the roots, where $\displaystyle 1600t^2-\frac{896}{\sqrt{2}}t=0$ $\displaystyle \underset{\div_{64}}{\Rightarrow} 25t^2-\frac{14}{\sqrt{2}}t=0$ $\displaystyle \Rightarrow \underbrace{t}_{\text{start when A is distance eight away}}\cdot\left(25t-\frac{14}{\sqrt{2}}\right)=0$ $\displaystyle \Rightarrow 25t-\frac{14}{\sqrt{2}}=0$ $\displaystyle \Rightarrow 25t=\frac{14}{\sqrt{2}}$ $\displaystyle \Rightarrow t=\frac{14}{25\sqrt{2}}\approx 0.3960\text{ hours}=0.3960\,(60\text{mins})\approx 24 \text{ mins}$.

### Coordinate Geometry Approach to (ii)

We have: Relative to B, ship A travels along a line —and the set of points a distance eight from $B$ comprise a circle, so we have: The line through $[AP]$ has slope equal to $\frac{32}{24}=\frac{4}{3}$ (why) and has as a point the initial position of A relative to B, $\displaystyle (x_1,y_1)=\left(-\frac{8}{\sqrt{2}},-\frac{8}{\sqrt{2}}\right)$

and so the equation of the line through $[AP]$ is: $\displaystyle y-\left(-\frac{8}{\sqrt{2}}\right)=\frac{4}{3}\left(x-\left(-\frac{8}{\sqrt{2}}\right)\right)$ $\displaystyle \Rightarrow y=\frac43 x+\frac{8}{3\sqrt{2}}$.

The equation of the circle is $x^2+y^2=8^2$. These curves — the line and the circle — intersect at $A$ and $P$ — and we already know where $A$ is.

To find the intersection, substitute $\displaystyle y=\frac43 x+\frac{8}{3\sqrt{2}}$ into $x^2+y^2=8^2$: $\displaystyle x^2+\left(\frac43 x+\frac{8}{3\sqrt{2}}\right)^2=8^2$ $\displaystyle \Rightarrow x^2+\frac{16}{9}x^2+\frac{64}{9\sqrt{2}}x+\frac{32}{9}=64$ $\displaystyle \underset{\times 9\sqrt{2}}{\Rightarrow} 9\sqrt{2}x^2+16\sqrt{2}x^2+64x+32\sqrt{2}=576\sqrt{2}$ $\displaystyle \Rightarrow 25\sqrt{2}x^2+64x-544\sqrt{2}=0$.

The sensible thing here is to use the $-b\pm\sqrt{\cdots}$ formula but this actually has nice factors. Multiply $a\cdot c=25\sqrt{2}\cdot(-544\sqrt{2})=-27200$. Consider $272\times (-100)=-27200$. The sum is 172 — too big — but try $200\times(-136)=-27200$: bingo $200+(-136)=64$.

So we can rewrite $25\sqrt{2}x^2+200x-136x-544\sqrt{2}=0$ $\displaystyle \Rightarrow 25x(\sqrt{2}x+9)-\frac{136}{\sqrt{2}}(\sqrt{2}x+8)=0$ $\displaystyle \Rightarrow (\sqrt{2}x+8)\left(25x-\frac{136}{\sqrt{2}}\right)=0$

Setting the first factor equal to zero gives $(x_1,y_1)=(-8/\sqrt{2},-8/\sqrt{2})$. The other gives $\displaystyle 25x-\frac{136}{\sqrt{2}}=0$ $\displaystyle \underset{+_{136/\sqrt{2}}}{\Rightarrow} 25x=\frac{136}{\sqrt{2}}$ $\displaystyle \underset{\div_{25}}{\Rightarrow} x_2=\frac{136}{25\sqrt{2}}$.

We can find the corresponding $y$ coordinate by substituting back into $\displaystyle y=\frac43 x+\frac{8}{3\sqrt{2}}$: $\displaystyle y_2=\frac43 \left(\frac{136}{25\sqrt{2}}\right)+\frac{8}{3\sqrt{2}}=\frac{124\sqrt{2}}{25}$.

Now the distance between these points is found using the distance formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ $\displaystyle =\sqrt{\left(\frac{136}{25\sqrt{2}}-\left(-\frac{8}{\sqrt{2}}\right)\right)^2+\left(\frac{124\sqrt{2}}{25}-\left(-\frac{8}{\sqrt{2}}\right)\right)^2}=\frac{56\sqrt{2}}{5}$.

The speed of ship A relative to B is $\displaystyle |\overrightarrow{V_{AB}}|=\sqrt{24^2+32^2}=40$,

and so using $\displaystyle\text{time}=\frac{\text{distance}}{\text{speed}}=\frac{\frac{56\sqrt{2}}{5}}{40}\approx 0.3960\text{ hours}=0.3960\,(60\text{mins})\approx 24 \text{ mins}$.

### Geometric/Trigonometric Approach to (ii)

There are probably a number of geometric/trigonometric approaches. One approach is to drop a perpendicular from $B$ to $[AP]$ — probably the easiest way. Here is another approach.

We have: We have $\overrightarrow{V_{AB}}=24\mathbf{i}+32\mathbf{j}$ — and the set of points a distance eight from $B$ comprise a circle. Therefore we consider the following — ship A is headed to $P$ (relative to ship $B$: Now produce the line segment $[PB]$ and the line segment $[AQ]$. Note that $\tan(\angle PAQ)=\frac{32}{24}=\frac43$ and $\tan(\angle BAQ)=1$. Define $\theta:=\angle BAP$.

(It is possible to get an approximate value for $\theta$ using $\theta=\tan^{-1}(4/3)-45^\circ$ — and when I saw that the answer was to be approximate (nearest minute) — I thought perhaps I should have done this instead of what follows.) Using $\displaystyle \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$, $\displaystyle \tan(\theta)=\tan(\angle PAQ-\angle BAQ)=\frac{\tan(\angle PAQ)-\tan(\angle BAQ)}{1+\tan(\angle PAQ)\cdot \tan(\angle BAQ)}$ $\displaystyle =\frac{\frac43-1}{1+\frac43\times 1}=\frac{1}{7}$.

Using the following model triangle: we have $\displaystyle \cos\theta=\frac{7}{\sqrt{50}}$. Now apply the cosine rule to $\Delta ABP$:  $a^2=b^2+c^2-2bc\cos A$ $\displaystyle \Rightarrow 8^2=b^2+8^2-2b(8)\frac{7}{\sqrt{50}}$ $\displaystyle \Rightarrow b^2-\frac{112}{\sqrt{50}}b=0$ $\displaystyle \Rightarrow b\cdot \left(b-\frac{112}{\sqrt{50}}\right)=0$;

we know that $b$ is non-zero and so $\displaystyle b=\frac{112}{\sqrt{50}}$.

The speed of ship A relative to B is $\displaystyle |\overrightarrow{V_{AB}}|=\sqrt{24^2+32^2}=40$,

and so using $\displaystyle\text{time}=\frac{\text{distance}}{\text{speed}}=\frac{\frac{112}{\sqrt{50}}}{40}\approx 0.3960\text{ hours}=0.3960\,(60\text{mins})\approx 24 \text{ mins}$.