Answer:
- magnitude: 23.37 km
- direction: 217.35°
Step-by-step explanation:
You want the magnitude and direction of the sum of the displacements 26 km on a bearing of 175° and 18 km on a bearing of 294°.
Vectors
A vector is a quantity that has magnitude and direction. Here, the direction is specified as a bearing angle, so the angle is measured clockwise from north (the +y axis). In terms of the way we usually measure angles, CCW from the +x axis, the angles relate to each other by ...
conventional angle = 90° - bearing angle
The (x, y) components of a vector r∠θ can be found using ...
(x, y) = r(cos(θ), sin(θ))
As long as the angle θ is consistently measured, you can use either bearing angles or conventional angles for θ. In the calculator attachment, we have shown bearing angles being used.
Addition
In a diagram, vectors are added nose-to-tail, just like segments on a number line. The attachment shows vector b moved to b' so its tail is at the nose of vector 'a'. Its head then represents the sum of the two vectors.
The addition of vectors "by hand" is done by adding corresponding components.
The attached calculator display shows the components of the vectors as ...
26∠175° = (-25.901, 2.266)
18∠294° = (7.321, -16.444)
The sum of these vectors gives the total displacement:
(-25.901+7.321, 2.266-16.444) = (-18.580, -14.178)
To convert this back to magnitude and direction, we use the relations ...
r = √(x² +y²)
θ = arctan(y/x) . . . . . . . . . with attention to quadrant
This means the magnitude of the resultant displacement is ...
r = √((-18.580)² +(-14.178)²) ≈ 23.371
θ = arctan(-14.178/-18.580) ≈ 217.35° . . . . . a 3rd-quadrant angle
The resultant is a displacement of 23.37 km on a bearing of 217.35°.
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Additional comment
If you use (x, y) = r(cos(θ), sin(θ)) with bearing angles, the resulting (x, y) coordinates are (displacement north, displacement east). You will notice this is the reverse of the usual interpretation of the x-y coordinates. This is seen in the way the vectors are drawn in the second attachment. For example, above, we have said the coordinates of 'a' are (-25.9, 2.2). You will notice it is plotted on the graph with its head at (2.2, -25.9).
If this is confusing, you can translate all of the angles to conventional angles, then convert the result back to a bearing angle when you're done. We like to use the built-in calculator ability with as little extra manipulation as possible.
Angles are converted from the ±90° (or ±180°) result of the arctangent function to the 0–360° result required for bearing angles by recognizing negative angles can be added to 360° to make them positive, and by paying attention to the signs of the x- and y-coordinates. For negative x, the angles are in quadrants 2 and 3, so will be opposite of the quadrant 1 and 4 angles returned by the arctangent function. In this problem the arctangent gives 37.34°, but the signs of the components tell us the bearing is actually 180° more than that.
Some calculators, and all spreadsheets, have an ATAN2(x, y) function that takes signs into account in the angle it returns. (This will be the conventional interpretation of the angle. If you want a bearing angle result, you need to use x=north, y=east coordinates.)
Takeaways:
- vectors are converted to/from magnitude and direction using
(x, y) = r(cos(θ), sin(θ))
r = √(x²+y²); θ = arctan(y/x) . . . with attention to signs - vectors are added by adding their components
And, (they don't tell you this in school) you can use bearing angles directly in the calculations as long as you do it consistently, and recognize the rectangular coordinates you get are (north, east) coordinates.
