# Two people meet in the purple room on the fourth floor of a building. On departure, one person travels West 16 feet, South 9 feet, and Down 9 feet. The other person travels North 16 feet, East 8 feet, and Up 9 feet. How far apart are the two people? Round

Jun 1, 2018

I get $\text{40.5 ft}$.

If you want me to round to the nearest integer, it would actually be about $\text{40 ft}$. Without rounding, it was $40.4598 \cdots$ $\text{ft}$.

Interestingly enough, the distances each person walked were $\left(1\right)$ $\text{20.45 ft}$ and $\left(2\right)$ $\text{20.02 ft}$, respectively, which add up to be just a LITTLE longer than the total distance between them. So, the angle between the two travel paths must be nearly ${180}^{\circ}$.

(In fact, it is ${177.43}^{\circ}$.)

Well, let's draw this out in 3D space to see what is asked for. I assume up/down refers to altitude ($z$ axis). West is $- x$ direction, north is $+ y$ direction, etc.

Suppose person $1$ was the $\textcolor{b l u e}{\text{blue}}$ arrow and person $2$ was the $\textcolor{red}{\text{red}}$ arrow. The distance between them is the $\textcolor{g r e e n}{\text{green}}$ line.

The point that they land upon is a vector.

$\textcolor{b l u e}{{\vec{v}}_{1}} = \left\langle- 16 , - 9 , - 9\right\rangle$

$\textcolor{red}{{\vec{v}}_{2}} = \left\langle+ 16 , + 8 , + 9\right\rangle$

Do not simply add the vectors together as they are:

$\textcolor{b l u e}{{\vec{v}}_{1}} + \textcolor{red}{{\vec{v}}_{2}} = \left\langle0 , - 1 , 0\right\rangle$

What we actually need to do is reverse the vector ${\vec{v}}_{1}$ so that both point in the same direction. That way, the components don't cancel out in the $x$ and $z$ directions, and we do get a distance farther than $\text{1 ft}$, which would be physically NOT what is represented on the diagram.

Thus, what we should do is add $- {\vec{v}}_{1}$ instead:

$\textcolor{g r e e n}{{\vec{v}}_{3}} = - \textcolor{b l u e}{{\vec{v}}_{1}} + \textcolor{red}{{\vec{v}}_{2}}$

$= - \left\langle- 16 , - 9 , - 9\right\rangle + \left\langle+ 16 , + 8 , + 9\right\rangle$

$= \left\langle+ 16 , + 9 , + 9\right\rangle + \left\langle+ 16 , + 8 , + 9\right\rangle$

$= \left\langle+ 32 , + 17 , + 18\right\rangle$

Lastly, we need to find the length of this vector to determine the distance:

$\textcolor{g r e e n}{| {\vec{v}}_{3} |} = \sqrt{{x}^{2} + {y}^{2} + {z}^{2}}$

$= \sqrt{{\left(\text{32 ft")^2 + ("17 ft")^2 + ("18 ft}\right)}^{2}}$

$=$ $\textcolor{g r e e n}{\text{40.5 ft}}$