Question

An air traffic controller observes two airplanes approaching the airport. The displacement from the control tower to plane 1 is given by the vector A which has a magnitude of 220km and points in a direction 32 degree north of west. The displacement from the control tower to plane 2 is given by the vector B, which has a magnitude of 140km and points 65 degree east of North. (a) sketch the vectors A, B and D = A -B. Notice that D is the displacement from plane 2 to plane 1. (b) Find the magnitude and direction of the vector D

Answer 1

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,

`\vec{A}=Acos\theta_1(-\hat{i})+A\sin \theta_1\hat{j}`

That is,

`\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}`

And vector B is given by,

`\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}`

Thus, vector D is given as

`\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}`

The magnitude of the vector D is given by,

`\begin{gathered} D=\sqrt[]{(-313.63)^2+57.65^2} \\ =318.88\text{ km} \end{gathered}`

The direction of vector D is given by,

`\begin{gathered} \theta=\tan ^(-1)((57.65)/(313.63)) \\ =10.41^(\circ) \end{gathered}`

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,