Adding Vectors by Components — Question 1

Question 1

Add signed components before taking a length

Three displacements in one east–north frame are $\vec A=(5\hat\imath+2\hat\jmath)$ m, $\vec B=(-3\hat\imath+4\hat\jmath)$ m, and $\vec C=(\hat\imath-7\hat\jmath)$ m.

1. Find $\vec R=\vec A+\vec B+\vec C$, its magnitude and standard direction.

2. Find the distance traveled if the three displacements are consecutive straight legs. Compare it with $|\vec R|$.

3. Would reordering the legs change either result? Would reversing only the last leg preserve them?

Question 1 - Solution

Add signed components before taking a length

(a) Keep horizontal and vertical sums separate.

$$\boxed{\vec R=(5-3+1,2+4-7)=(3,-1)\,\mathrm m},\quad R=\sqrt{10}\,\mathrm m,\quad\theta=341.565^\circ$$ $R_x=5-3+1=3$ m and $R_y=2+4-7=-1$ m. Thus $\vec R=(3\hat\imath-\hat\jmath)$ m, $R=\sqrt{10}\approx3.162$ m, and $\theta=360^\circ-\tan^{-1}(1/3)\approx341.57^\circ$.

(b) Add segment lengths for the route.

$$s=\sqrt{29}+5+\sqrt{50}=17.456\,\mathrm m>R$$ The distance is $\sqrt{29}+5+\sqrt{50}\approx17.456$ m. It exceeds the displacement magnitude because the legs partly cancel. Taking a magnitude after the vector sum is a different operation from summing the magnitudes.

(c) Track what changes.

$$\sum\vec d_{\pi(i)}=\sum\vec d_i,\quad\vec R_{C\to-C}=\vec R-2\vec C=(1,13)\,\mathrm m,\quad R'=\sqrt{170}\,\mathrm m$$ Reordering preserves the vector sum and the sum of the three lengths, though it changes the intermediate route. Replacing C by $-\vec C$ gives $(1,13)$ m and magnitude $\sqrt{170}$ m. The route length is unchanged, but the endpoint and net displacement change.