Adding the vectors ijk notation

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The discussion focuses on adding three displacement vectors A, B, and C in ijk notation and determining the resultant vector D. The components of D are calculated as i = 2 and j = -2, leading to a magnitude of approximately 2.83 m and an angle of -45°. There is confusion regarding the resultant vector's angle and its relationship to vector A, with concerns about the quadrant placement in the XY plane. The importance of correctly interpreting the signs in arctan calculations is emphasized to avoid quadrant errors. Accurate graphing of the vectors is also discussed, highlighting that the resultant vector's angle does not necessarily match that of vector A.
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Homework Statement


Add the following 3 displacement vectors and give the answer in ijk notation and in magnitude and direction format.
A = 3i - 3j m
B = i - 4j m
C = -2i + 5j m.

Then I have to do a scetch of the resultant vector...which i'll call D.


Homework Equations





The Attempt at a Solution


So I know to get vector D's i hat and j hat components...i just add the i's and j's of the vectors that make it up so

i = 3 + 1 - 2 = 2
j = -3 - 4 + 5 = -2

right?

so then to find the magnitude of D i do

√(2)² + (-2)²

which comes out to √8 which is roughly 2.83 m here

Then to find the angle i do the inverse tan of (-2/2) or -1...which comes out to -45°

Now to graph...should I find the magnitude and angle of each vector?

A = √(3)² + (-3)² = √18 = 4.24 m at -45°
B = √(1)² + (-4)² = √17 = 4.12 m at -76°
C= √(-2)² + (5)² = √29 = 5.39 m at -68°

Do I graph them just like this? Because how can the resultant vector be at the SAME angle at vector A? It would be a smaller version of vector A because the magnitude of A is >D and the angle is the same.

Can anyone point out where I'm misunderstanding this?
 
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The angle of the resultant vector is not 45°. In which quadrant of the XY plane is it?

The problem with arctan is that it confuses (-a)/b with a/(-b), and (-a)/(-b) with a/b. So you have to keep an eye on the signs to determine which quadrant it is, and what the angle really is.
 
You don't need to work out the magnitude and angle of the vectors.
For example, for A = (3i - 3j) m, the x-component is 3m and the y-component is -3m.
So, just draw a line from the origin to the point (3m, -3m)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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