91Ó°ÊÓ

The dimensions of the room are${}^{{}^{\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{}\mathbf{\left(}\mathbf{height}\mathbf{\right)}\mathbf{}\mathbf{×}\mathbf{3}\mathbf{.}\mathbf{70}\mathbf{×}\mathbf{4}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{m}}}$

Displacement may be defined as the shortest distance between two points.

The magnitude of the displacement ${}^{\mathbf{d}}$can be expressed in three dimension as:

$\mathit{d}\mathbf{=}\sqrt{{\left(\mathrm{Δ}x\right)}^{\mathbf{2}}\mathbf{+}{\left(\mathrm{Δ}y\right)}^{\mathbf{2}}\mathbf{+}{\left(\mathrm{Δ}z\right)}^{\mathbf{2}}}$ … (i)

Here,${}^{\mathbf{Δ}\mathbf{x}}$,${}^{\mathbf{Δ}\mathbf{y}}$, and ${}^{\mathbf{Δ}\mathbf{z}}$can be consider as width, length, and height respectively.

If a fly is start moving from one corner (the ceiling) and the opposite corner is on the floor, the magnitude of the displacement can be calculated by equation (i) as:

$$\begin{gathered} \mathbf{d}\mathbf{=}\sqrt{{\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{7}\mathbf{}\mathbf{m}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{m}\mathbf{)}}^{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{417}\mathbf{}\mathbf{m} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{≈}\mathbf{6}\mathbf{.}\mathbf{42}\mathbf{}\mathbf{m} \end{gathered}$$

Thus, the magnitude of the displacement is${}^{\mathbf{6}\mathbf{.}\mathbf{42}\mathbf{}\mathbf{m}}$ .

Since, the straight line gives the shortest distance between two points, length of the path can’t be less than the above displacement.

If the fly crawls along the edges of the room, the path length would

$\mathbf{3}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{7}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{3}\mathbf{=}\mathbf{11}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$

Thus, the path length can’t be less than the displacement.

Since, the straight line gives the shortest distance between two points, length of the path can’t be less than the above displacement.

If the fly crawls along the edges of the room, the path length would

$\mathbf{3}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{7}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{3}\mathbf{=}\mathbf{11}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$

Thus, the path length can be greater than the displacement.

Since, the straight line gives the shortest distance between two points, length of the path can’t be less than the above displacement.

If the fly flies along the displacement vector, then the path length can be equal to the displacement.

Taking 3.7 m along x-axis, 4.30 m along y-axis and 3.0 m along z axis, the displacement vector can be written as,

$\mathbf{d}\mathbf{}\mathbf{=}\left(3.7m\right)\mathbf{}\mathbf{i}\mathbf{}\left(4.3m\right)\mathbf{}\hat{\mathbf{j}}\mathbf{}\left(3.0m\right)\mathbf{}\hat{\mathbf{k}}$

If we flatten out the corners of the box, we would get the rectangles as shown in figure.

There can be three different routes to reach to the top, ${}^{{\mathbf{R}}_{\mathbf{1}\mathbf{\text{'}}}}$${}^{{\mathbf{R}}_{\mathbf{2}}}$, and ${}^{{\mathbf{R}}_{\mathbf{3}}}$, thus it can be concluded that ${}^{{\mathbf{R}}_{\mathbf{1}}}$would be smaller than ${}^{{\mathbf{R}}_{\mathbf{2}}}$, and ${}^{{\mathbf{R}}_{\mathbf{3}}}$.

So, the length of the shortest path is,

$$\begin{gathered} {\mathbf{R}}_{\mathbf{1}}\mathbf{=}\sqrt{{\left(6.7\right)}^{\mathbf{2}}\mathbf{+}{\left(4.3\right)}^{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{96}\mathbf{}\mathbf{m} \end{gathered}$$

Thus, the length of the shortest path is${}^{\mathbf{7}\mathbf{.}\mathbf{96}\mathbf{}\mathbf{m}}$ .