Answer
We orient $\hat{\mathrm{i}}$ eastward, $\hat{\mathrm{j}}$ northward, and $\hat{\mathrm{k}}$ upward. The displacement for the return portion is $$\vec{d}=-(1300 \mathrm{\ m}) \hat{\mathrm{i}}-(2200 \mathrm{\ m}) \hat{\mathrm{j}}$$ and the magnitude is $$d^{\prime}=\sqrt{(-1300 \mathrm{m})^{2}+(-2200 \mathrm{m})^{2}}=2.56 \times 10^{3} \mathrm{\ m}$$ The net displacement is $(zero)$ since his final position matches his initial position.
Work Step by Step
We orient $\hat{\mathrm{i}}$ eastward, $\hat{\mathrm{j}}$ northward, and $\hat{\mathrm{k}}$ upward. The displacement for the return portion is $$\vec{d}=-(1300 \mathrm{\ m}) \hat{\mathrm{i}}-(2200 \mathrm{\ m}) \hat{\mathrm{j}}$$ and the magnitude is $$d^{\prime}=\sqrt{(-1300 \mathrm{m})^{2}+(-2200 \mathrm{m})^{2}}=2.56 \times 10^{3} \mathrm{\ m}$$ The net displacement is $(zero)$ since his final position matches his initial position.
