Vectors and finding coordinates question

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The discussion revolves around solving a vector problem to find the coordinates of a treasure located at C = 4A - 3B, where A and B are defined vectors. The initial confusion stemmed from understanding the relationship between the vectors and their coordinates, particularly the direction of vector B given the angle θ = 328°. After clarifying the calculations for the x and y coordinates of vector B, the user successfully determined the coordinates of the treasure, with y = 14.36 km and x = -30.17 km. The conversation highlights the importance of vector addition and the correct application of trigonometric functions to resolve the problem. Overall, the collaborative effort led to a successful resolution of the coordinates.
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Homework Statement


On a treasure map,
A = -5 (km)x + 2 (km)y, B = 4 km, and theta = 328 deg. The treasure is located at C = 4A - 3B. What is the x-coordinate of the treasure?

What is the y-coordinate of the treasure?



Homework Equations


a^2 + b^2 = c^2
Vector addition


The Attempt at a Solution


So to tackle this one I decided that the C couldn't have been making a triangle between the two vectors since C wasn't equal to sqrt(A^2 + B^2). I thought that maybe if I just find 4A in C that would be the x coordinate, and 3B would be the y. I guess that line of logic was wrong, as I didn't get a correct answer.

So can someone help? Please also explain, vectors are confusing to me.
 
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So to tackle this one I decided that the C couldn't have been making a triangle between the two vectors since C wasn't equal to sqrt(A^2 + B^2).
-------
\vec{A}=\vec{B}+\vec{C}

Vector B and C can be of any direction and magnitude.
The angle between B and C can be from zero to 360°.

A2=B2+ C2 means the angle between B and C is 90°.
 
azizlwl said:
So to tackle this one I decided that the C couldn't have been making a triangle between the two vectors since C wasn't equal to sqrt(A^2 + B^2).
-------
\vec{A}=\vec{B}+\vec{C}

Vector B and C can be of any direction and magnitude.
The angle between B and C can be from zero to 360°.

A2=B2+ C2 means the angle between B and C is 90°.

Alright, so the angle is already given to me is 328°. So does that mean that when I'm drawing my vectors, the angle between A and B is that degree since θB = 328°? Furthermore C should be a point, not a line, right? It's 4A-3B (in relation to the two vectors).
 
A = -5 (km)x + 2 (km)y, B = 4 km, and theta = 328 deg. The treasure is located at C = 4A - 3B. What is the x-coordinate of the treasure?
What is the y-coordinate of the treasure?.
......
You have to find the x and y coordinates of 4A -3B.
The x and y coordinate of A is given.

You have find the value of x and y coordinate for vector B.

Normal addition or subtraction can be easily carried out for the vectors in same or opposite direction.
 
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azizlwl said:
A = -5 (km)x + 2 (km)y, B = 4 km, and theta = 328 deg. The treasure is located at C = 4A - 3B. What is the x-coordinate of the treasure?
What is the y-coordinate of the treasure?.
......
You have to find the x and y coordinates of 4A -3B.
The x and y coordinate of A is given.

You have find the value of x and y coordinate for vector B.

Normal addition or subtraction can be easily carried out for the vectors in same or opposite direction.

Alright so B is 4km but I'm not really sure about the direction, is it the θB = 328°? I'm sorry if I'm missing something, I have a hard time understanding vectors.

EDIT- I found the coordinates! It turned out that y=14.36 km. Before that I did some math to find the x and y for B, which ended up being 4sin(-32) = -2.12, and 4cos(-32)= 3.39. Next to find y all I did was 4(2)-3(-2.12). So then (for finding x) 4(-5)-3(3.39) = -30.17. Thank you for all your help! The hints definitely led me to the solution :D
 
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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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