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Vectors and finding coordinates question

  • Thread starter Freemark
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
1,065
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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).
-------
[itex]\vec{A}=\vec{B}+\vec{C}[/itex]

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°.
 
  • #3
14
0
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).
-------
[itex]\vec{A}=\vec{B}+\vec{C}[/itex]

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).
 
  • #4
1,065
10
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.
 
Last edited:
  • #5
14
0
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
 
Last edited:

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