Sure, let's break it down step by step.
1. We first need to calculate the magnitude of each vector. The magnitude of a vector in two dimensions i.e., vector D=axi+byj, can be found using the Pythagorean theorem, which states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This can be written as:
magnitude = sqrt(a² + b²)
For vector a, D=-20.0 mi + 30.0 mj, the magnitude will be sqrt((-20.0)² + (30.0)²) ≈ 36.05551275463989 mi
For vector b, D= 20.0 mi -10.0 mj, the magnitude will be sqrt((20.0)² + (-10.0)²) ≈ 22.360679774997898 mi
2. Now, let's figure out the direction of each vector. The angle θ between vector a and the positive direction of the x axis can be found by invoking the formula:
θ = atan2(b, a)
and the compass bearing can be found by taking:
360 - θ (if θ is positive)
or
360 + θ (if θ is negative)
Applying it to our vectors and converting the radian to the degree, we have:
For vector a,
θa = atan2(30.0, -20.0) ≈ 236.31˚
For vector b,
θb = atan2(-10.0, 20.0) ≈ 26.57˚
3. Given that the positive x-direction is E, and the positive y-direction is N, we have:
a. Vector D=-20.0 mi + 30.0 mj has a magnitude of approximately 36.06 mi, and its direction, the angle from the positive x-axis, is approximately 236.31˚, which falls in the third quadrant, that is the SW direction.
b. Vector D= 20.0 mi -10.0 mj has a magnitude of approximately 22.36 mi, and its direction, the angle from the positive x-axis, is approximately 26.57˚, which falls in the fourth quadrant, that is the SE direction.
I hope this clarifies your question! Let me know if you need further explanation on any of these concepts.