1. ## solve with vectors

At 12 noon the position vectors r and the velocity vectors v of two ships A and B are

rA = (2i+j)km vA = (3i+j)kmh^-1
rB = (-i-4j)km vB = (11i+3j)kmh^-1

(a) Show that at time t h after noon the position vector of B relative to A is given by:
[(8t-3)i+(2t-5)j]km
(b) Show that the distance d km between the vessels, at this time, is given by
d^2=68[t^2 - t+1/2]
(c) Hence show that the ships are nearest together at 12.30 p.m and find the distance between them at this time.

2. Originally Posted by outlaw
At 12 noon the position vectors r and the velocity vectors v of two ships A and B are

rA = (2i+j)km vA = (3i+j)kmh^-1
rB = (-i-4j)km vB = (11i+3j)kmh^-1

(a) Show that at time t h after noon the position vector of B relative to A is given by:
[(8t-3)i+(2t-5)j]km
(b) Show that the distance d km between the vessels, at this time, is given by
d^2=68[t^2 - t+1/2]
(c) Hence show that the ships are nearest together at 12.30 p.m and find the distance between them at this time.
(a) $r_{BA}(t) = t.v_{BA} + r_{BA}(0)$

(b) $d^2 = |r_{BA}(t)|^2$

(c) Differentiate $d^2$ and equate it to zero to find 't' and subsequently d. Or you can observe that $t^2 - t + \frac12 = (t - \frac12)^2 + \frac14$. Clearly the minimum for this function is at $t = \frac12$, which is 30 minutes......

3. Hello, outlaw!

At 12 noon the position vectors $\vec{r}$ and the velocity vectors $\vec{v}$ of two ships A and B are:

. . . $\begin{array}{ccccccc}\vec{r}_A &= & 2{\bf i} + {\bf j}\text{ km} & & \vec{v}_A &=& 3i + j\text{ km/hr} \\
\vec{r}_B &= & \text{-}{\bf i} -4{\bf j} \text{ km} & & \vec{v}_B &=& 11{\bf i} + 3{\bf j}\text{ km/hr}
\end{array}$

(a) Show that $t$ hours after noon, the position vector of $B$ relative to $A$ is given by:

. . . $\overrightarrow{AB} \;=\;(8t-3){\bf i}+ (2t-5){\bf j}\;\text{km}$
In $t$ hours, the position vectors are:

. . $\begin{array}{ccccc}\vec{r}_A &=& (2{\bf i} + {\bf j}) + (3{\bf i}+{\bf j})t &=& (2+3t){\bf i} + (1 + t){\bf j} \\
\vec{r}_B &=& (\text{-}{\bf i}-4{\bf j}) + (11{\bf i} + 3{\bf j})t &=& (11t-1){\bf i} + (3t-4){\bf j} \end{array}$

Then: . $\overrightarrow{AB} \;=\;\bigg[(11t - 1){\bf i} - (2+3t){\bf j}\bigg] + \bigg[(3t-4){\bf i}-(1+t){\bf j}\bigg]$

Therefore: . $\boxed{\overrightarrow{AB} \;=\;(8t-3){\bf i} + (2t-5){\bf j}}$

(b) Show that the distance $d$ between the vessels at time $t$
. . . is given by: . $d^2 \;=\;68\left(t^2 - t+\frac{1}{2}\right)$
$d^2 \;=\;|\overrightarrow{AB}|^2\;=\;(8t-3)^2 + (2t-5)^2\;=\;68t^2-68t + 34$

Therefore: . $\boxed{d^2\;=\;68\left(t^2 - t + \frac{1}{2}\right)}$

(c) Hence show that the ships are nearest together at 12:30 pm
and find the distance between them at this time.
We have . $d^2 \;=\;68t^2 - 68t + 34$

Note that $d$ is at a minimum when $d^2$ is a minimum.

The graph for $d^2$ is an up-opening parabola.
. . Its minimum occurs at its vertex.
The vertex is at: . $t \:=\:\frac{\text{-}b}{2a}$

We have: . $a = 68,\;b = \text{-}68$
. . Hence, the vertex is at: . $t \:=\:\frac{\text{-}(\text{-}68)}{2(68)} \:=\:\frac{1}{2}$

Since the distance is a minimum when $t \:=\:\frac{1}{2}\text{ hour} \:=\:30\text{ minutes}$,

. . the ships are at their nearest at $\boxed{12:30\text{ pm}}$

When $t = \frac{1}{2}\!:\;\;d^2 \;=\;68\left(\frac{1}{2}\right)^2 - 68\left(\frac{1}{2}\right) + 34 \;=\;17$

. . Therefore, their minimum distance is . $d \;=\;\boxed{\sqrt{17}\text{ km}}$