A helicopter flies 62 miles east, then turns and flies at an angle of \( 45^{\circ} \) north of east for 199 miles. What is the magnitude of the helicopter's resultant vector? Draw a vector diagram. \[ |\vec{R}|=[\text { ? ] miles } \]

To find the magnitude of the helicopter's resultant vector, we can break down its path into components. The first leg, 62 miles east, contributes \( \vec{A} = 62 \hat{i} \) miles. The second leg can be broken into its east (\(\hat{i}\)) and north (\(\hat{j}\)) components. The helicopter flies 199 miles at \( 45^\circ \) north of east, giving both components as: \[ \vec{B}_x = 199 \cos(45^\circ) = 199 \cdot \frac{\sqrt{2}}{2} \approx 140.14 \hat{i} \text{ miles} \] \[ \vec{B}_y = 199 \sin(45^\circ) = 199 \cdot \frac{\sqrt{2}}{2} \approx 140.14 \hat{j} \text{ miles} \] Now, we can sum the components: \[ \vec{R}_x = 62 + 140.14 = 202.14 \hat{i} \text{ miles} \] \[ \vec{R}_y = 140.14 \hat{j} \text{ miles} \] The magnitude of the resultant vector is calculated using the Pythagorean theorem: \[ |\vec{R}| = \sqrt{(202.14)^2 + (140.14)^2} \approx \sqrt{40856.32 + 19638.67} \approx \sqrt{60494.99} \approx 245.95 \text{ miles} \] So, the magnitude of the helicopter's resultant vector is approximately \( 245.95 \) miles. As for the vector diagram, you can illustrate it by drawing: 1. A horizontal line (to the right) for the first leg (62 miles). 2. An angled line (at \( 45^\circ \) from the end of the first line) representing the second leg (199 miles). 3. The resultant vector would be a straight line from the starting point to the end of the second line, completing the triangle formed by these two segments. With that, you've got a visual representation of how the helicopter flew!