# An airplane has an airspeed of 500 kilometers per hour bearing N45°E. The wind velocity is 60 kilometers per hour in the direction N30°W. Find the resultant vector representing the path of the plane relative to the ground?

Nov 9, 2015

#### Answer:

Use unit vectors to solve the addition of two vectors.

#### Explanation:

Airplane : $500 \left[\cos \left(45\right) i + \sin \left(45\right) j\right] = 500 \left[\frac{\sqrt{2}}{2} i + \frac{\sqrt{2}}{2} j\right]$

Wind : $60 \left[\cos \left(120\right) i + \sin \left(120\right) j\right] = 60 \left[- \frac{1}{2} i + \frac{\sqrt{3}}{2} j\right]$

Next, sum the two vectors:

Sum $= 323.5534 i + 405.5149 j$

Angle of resultant vector: $T a {n}^{-} 1 \left(\frac{405.5149}{323.5534}\right) \approx {51.414}^{o}$
or $\approx N {38.6}^{o} E$

Magnitude of resultant vector: $\sqrt{{323.5534}^{2} + {405.5149}^{2}} \approx 518.7766$ km/hr

Here is a sketch of the plane vector $\left(A B\right)$ and the wind vector $\left(A C\right)$. The resultant vector is $A D$

hope that helped! 