Vectors and finding coordinates question

AI Thread Summary
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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