Components of Vectors — Question 7

Question 7

A projection is not the traveled route

A hiker starts at O and finishes at P, 4 km east and 3 km north of O. The terrain permits either a direct straight route or any piecewise straight route in the horizontal plane.

1. Find the displacement magnitude and bearing north of east.

2. Compare the direct route with a route 4 km east followed by 3 km north. What quantities are identical and what differs?

3. Find a route with the same displacement and total distance 9 km. Do the given components impose a maximum possible distance?

Question 7 - Solution

A projection is not the traveled route

(a) Use endpoint information.

$$\boxed{D=\sqrt{4^2+3^2}=5\,\mathrm{km}},\quad\theta=\tan^{-1}(3/4)=36.870^\circ$$ $D=\sqrt{4^2+3^2}=5$ km and $\theta=\tan^{-1}(3/4)\approx36.87^\circ$ north of east. These characterize the straight separation O–P regardless of how the hiker travels.

(b) Do not read a path into projection lines.

$$s_{\mathrm{direct}}=5\,\mathrm{km},\quad s_{\mathrm{axes}}=4+3=7\,\mathrm{km},\quad\Delta\vec r=(4,3)\,\mathrm{km}$$ The direct distance traveled is 5 km; the axis-aligned route is 7 km. Both have components $(4,3)$ km and the same 5 km displacement. Dashed projection legs are a method for specifying an arrow, not evidence that the traveler followed those legs.

(c) Add a return excursion.

$$\Delta\vec r=(-1+5,3)=(4,3)\,\mathrm{km},\quad s=1+5+3=9\,\mathrm{km},\quad s_{\mathrm{detour}}=5+2\ell\to\infty$$ Travel 1 km west, then 5 km east, then 3 km north. The net endpoint is still $(4,3)$ km, while the length is $1+5+3=9$ km. Longer out-and-back excursions can increase the distance arbitrarily, so there is no finite upper bound from displacement alone. The straight 5 km route is the minimum.