Adding Vectors by Components — Question 6

Question 6

Close a path in three dimensions

Three consecutive displacements are $\vec A=(2,-1,3)$ m, $\vec B=(-4,5,1)$ m, and $\vec C=(1,-2,-6)$ m in a common $(x,y,z)$ frame.

1. Find the three-leg resultant and its length.

2. Find a straight closing leg and the total four-leg traveled distance.

3. A planar map drops the z components. Find the apparent three-leg resultant length and explain what information it misses.

Question 6 - Solution

Close a path in three dimensions

(a) Add each coordinate column.

$$\boxed{\vec R=(2-4+1,-1+5-2,3+1-6)=(-1,2,-2)\,\mathrm m},\quad R=3\,\mathrm m$$ $\vec R=(-1,2,-2)$ m and $R=\sqrt{1+4+4}=3$ m. The component method works in space exactly as in the plane, with an additional independent sum.

(b) Reverse the resultant to close.

$$\vec D=-\vec R=(1,-2,2)\,\mathrm m,\quad s=\sqrt{14}+\sqrt{42}+\sqrt{41}+3=19.626\,\mathrm m$$ The closing leg is $\vec D=(1,-2,2)$ m, length 3 m. The total traveled distance is $\sqrt{14}+\sqrt{42}+\sqrt{41}+3\approx19.626$ m. The four-leg net displacement is zero despite the positive traveled length.

(c) Distinguish projection from full displacement.

$$R_{xy}=\sqrt{(-1)^2+2^2}=\sqrt5\,\mathrm m,\quad R^2=R_{xy}^2+R_z^2=5+4=9\,\mathrm{m^2}$$ The map gives $\vec R_{xy}=(-1,2)$ m and apparent length $\sqrt5\approx2.236$ m. It hides the $-2$ m vertical component. A closing leg chosen only from the map would leave a vertical error unless that missing component were also corrected.