Question

Displacement vector points due east and has a magnitude of 3.35 km. Displacement vector points due north and has a magnitude of 9.31 km. Displacement vector points due west and has a magnitude of 6.66 km. Displacement vector points due south and has a magnitude of 3.65 km. Find (a) the magnitude of the resultant vector + + + , and (b) its direction as a positive angle relative to due west.

Answer 1

Answer:

(a) 6.56 km

(b) $59.68^\circ$ north of west

Explanation:

Given:

$\vec{d}_1= 3.35\ km\ east = 3.35\ km\ \hat{i}\\\vec{d}_2= 9.31\ km\ north = 9.31\ km\ \hat{j}\\\vec{d}_3= 6.66\ km\ west = -6.61\ km\ \hat{i}\\\vec{d}_4= 3.65\ km\ south = -3.65\ km\ \hat{j}$

Let the resultant displacement vector be $\vec{D}$.

As the resultant is the vector sum of all the vectors.

$\therefore \vec{D}=\vec{d}_1+\vec{d}_2+\vec{d}_3+\vec{d}_4\\\Rightarrow \vec{D} =(3.35\ km\ \hat{i})+(9.31\ km\ \hat{j})+(-6.61\ km\ \hat{i})+(-3.65\ km\ \hat{j})\\\Rightarrow \vec{D} =-3.31\ km\ \hat{i}+5.66\ km\ \hat{j}\\\textrm{Magnitude of the resultant displacement vector} = \sqrt{(-3.31)^2+(5.66)^2}\ km= 6.56\ km$

$\textrm{Angle with the positive west} = \theta = \tan^{-1}(\dfrac{5.66}{3.31})= 59.68^\circ$