Select the correct answer. A jet plane is traveling with a velocity of 720kilometers/hour due north. There is a crosswind of 40 kilometers/hour from the east. Find the resultant velocity and the direction of the jet plane. A. 720kilometers/hour in the direction of 3.2^circ west of north B. 720kilometers/hour due north C. 721kilometers/hour in the direction of 3.2^circ west of north D. 520kilometers/hour in the direction of 45^circ east of north

To find the resultant velocity and direction of the jet plane, we can use vector addition. The velocity of the jet plane is represented by a vector pointing due north, while the crosswind is represented by a vector pointing east.<br /><br />Using the Pythagorean theorem, we can find the magnitude of the resultant velocity:<br /><br />$|v_{resultant}| = \sqrt{(720)^2)^2} \approx 720.4 \text{ kilometers/hour}$<br /><br />To find the the resultant velocity, we can use trigonometry. Specifically, we can use the tangent function to find the angle between the resultant velocity and the north direction:<br /><br />$\tan(\theta) = \frac{40}{720} \implies \theta \approx 3.2^{\circ}$<br /><br />Therefore, the correct answer is:<br /><br />C. $721kilometers/hour$ in the direction of $3.2^{\circ }$ west of north