Answer :
Answer:
(a) 6.56 km
(b) [tex]59.68^\circ[/tex] north of west
Explanation:
Given:
[tex]\vec{d}_1= 3.35\ km\ east = 3.35\ km\ \hat{i}\\\vec{d}_2= 9.31\ km\ north = 9.31\ km\ \hat{j}\\\vec{d}_3= 6.66\ km\ west = -6.61\ km\ \hat{i}\\\vec{d}_4= 3.65\ km\ south = -3.65\ km\ \hat{j}\\[/tex]
Let the resultant displacement vector be [tex]\vec{D}[/tex].
As the resultant is the vector sum of all the vectors.
[tex]\therefore \vec{D}=\vec{d}_1+\vec{d}_2+\vec{d}_3+\vec{d}_4\\\Rightarrow \vec{D} =(3.35\ km\ \hat{i})+(9.31\ km\ \hat{j})+(-6.61\ km\ \hat{i})+(-3.65\ km\ \hat{j})\\\Rightarrow \vec{D} =-3.31\ km\ \hat{i}+5.66\ km\ \hat{j}\\\textrm{Magnitude of the resultant displacement vector} = \sqrt{(-3.31)^2+(5.66)^2}\ km= 6.56\ km\\[/tex]
[tex]\textrm{Angle with the positive west} = \theta = \tan^{-1}(\dfrac{5.66}{3.31})= 59.68^\circ[/tex]