Final answer:
The plane will end up flying 1.62 degrees off course (east of north) due to the wind. The plane's speed relative to the ground will be approximately 750.71 km/hr.
Step-by-step explanation:
To solve this problem, we will use vector addition, where the velocity of the airplane and the velocity of the wind combine to give the resultant velocity or the plane's speed relative to the ground. The airplane's velocity is 750 km/h north (we'll assume this is along the positive y-axis) and the wind's velocity is 30 km/h southeast. To resolve this southeast wind into its component vectors, we have to split it into east (x-axis) and south (negative y-axis) components. Since the southeast is at a 45-degree angle from these axes, the components of the wind will be equal in magnitude.
Therefore, each component of the wind's velocity will be 30 km/hr divided by the square root of 2, which is approximately 21.21 km/hr both south and east. Subtract the south component from the airplane's north velocity and add the east component to get the plane's ground speed and direction. Using the Pythagorean theorem, the magnitude of the velocity relative to the ground will be the square root of the sum of the squares of the north and east components.
Let Vg be the ground speed of the airplane and θ be the angle at which the airplane is off course. Therefore, Vg = √(750^2 + 21.21^2) and θ = αtan(21.21 / 750). After calculating, Vg ≈ 750.71 km/hr, and θ is approximately 1.62° east of north.
The plane will end up flying 1.62° off course. The plane's speed relative to the ground will be 750.71 km/hr.