Answer
$\text{Using Eq. 3-6, the magnitude is }$
$$
|\vec{R}|=\sqrt{(-3.18 \mathrm{m})^{2}+(4.72 \mathrm{m})^{2}}=5.69 \mathrm{m}
$$
Work Step by Step
The vector equation is $\vec{R}=\vec{A}+\vec{B}+\vec{C}+\vec{D}$ . Expressing $\vec{B}$ and $\vec{D}$ in unit-vector
notation, we have $(1.69 \hat{\mathrm{i}}+3.63 \hat{\mathrm{j}}) \mathrm{m}$ and $(-2.87 \hat{\mathrm{i}}+4.10 \hat{\mathrm{j}}) \mathrm{m},$ respectively.
Where the length unit is not displayed in the solution below, the unit meter should be understood.
we have $\text{(Eq. 3-6)}$
$$\vec{R}=(-3.18 \mathrm{m}) \hat{\mathrm{i}}+(4.72 \mathrm{m}) \hat{\mathrm{j}}$$
Using Eq. $\text{3-6}$, the magnitude is
$$
|\vec{R}|=\sqrt{(-3.18 \mathrm{m})^{2}+(4.72 \mathrm{m})^{2}}=5.69 \mathrm{m}
$$
