Given,
The magnitude of vector A, A=220 km
The direction of vector A, θ₁=32° north of west
The magnitude of vector B, B=140 Km
The direction of vector B, θ₂=65° east of north
(a)
(b)
The vector A is given by,
[tex]\vec{A}=Acos\theta_1(-\hat{i})+A\sin \theta_1\hat{j}[/tex]
That is,
[tex]\begin{gathered} \vec{A}=-220\cos 32^{\circ}\hat{i}+220\sin 32^{\circ}\hat{j} \\ =-186.57\hat{i}+116.52\hat{j} \end{gathered}[/tex]
And vector B is given by,
[tex]\begin{gathered} \vec{B}=B\sin \theta_2\hat{i}+B\cos \theta_2\hat{j} \\ =140\sin 65^{\circ}\hat{i}+140\cos 65^{\circ}\hat{j}_{} \\ =126.88\hat{i}+59.17\hat{j} \end{gathered}[/tex]
Thus, vector D is given as
[tex]\begin{gathered} \vec{D}=\vec{A}-\vec{B} \\ =(-186.75-126.88)\hat{i}+(116.52-59.17)\hat{j} \\ =-313.63\hat{i}+57.35\hat{j} \end{gathered}[/tex]
The magnitude of the vector D is given by,
[tex]\begin{gathered} D=\sqrt[]{(-313.63)^2+57.65^2} \\ =318.88\text{ km} \end{gathered}[/tex]
The direction of vector D is given by,
[tex]\begin{gathered} \theta=\tan ^{-1}(\frac{57.65}{313.63}) \\ =10.41^{\circ} \end{gathered}[/tex]
Therefore the magnitude of vector D is 318.88 km and the direction of the vector D is 10.41° north of the west.
The direction of the vector D is represented as,
