Answer:
The airplane is approximately 159.892 miles away from the airport.
Step-by-step explanation:
Let consider the following convention, north is represented by +y axis and east is represented in +x axis. Then, we represented each displacement by vectors:
i) An airplane flies 225 miles south
[tex]\vec r_{A} = - 225\,\hat{j}\,\,\,[mi][/tex] (Eq. 1)
ii) Then 175 miles northwest
[tex]\vec r_{B} = 175\cdot \cos 135^{\circ}\,\hat{i} + 175\cdot \sin 135^{\circ}\,\hat{j}\,\,\,[mi][/tex] (Eq. 2)
The resulting vector position is the sum of the two vectors above:
[tex]\vec r = \vec r_{A}+\vec r_{B}[/tex] (Eq. 3)
[tex]\vec r = -225\,\hat{j}+(175\cdot \cos 135^{\circ}\,\hat{i}+175\cdot \sin 135^{\circ}\,\hat{j})[/tex] [tex][mi][/tex]
[tex]\vec r = 175\cdot \cos 135^{\circ}\,\hat{i}+(-225+175\cdot \sin 135^{\circ})\,\hat{j}[/tex] [tex][mi][/tex]
[tex]\vec r = -123.744\,\hat{i} -101.256\,\hat{j}[/tex]
Lastly, we find the distance of the airplane from the airport by Pythagorean Theorem:
[tex]\|\vec r\| = \sqrt{(-123.744\,mi)^{2}+(-101.256\,mi)^{2}}[/tex] (Eq. 4)
[tex]\|\vec r\| \approx 159.892\,mi[/tex]
The airplane is approximately 159.892 miles away from the airport.
