Solving for Vector Magnitude and Direction: Ax=3.2, Ay=-5.15 - Quick Question

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The vector with components Ax=3.2 and Ay=-5.15 has a calculated magnitude of 6.06. The initial direction was found to be 58.7 degrees, but this is incorrect since the vector lies in the fourth quadrant. Using the tangent function, the correct angle is calculated as -57.99 degrees, indicating a negative angle due to the vector's position. The discussion highlights the importance of quadrant consideration when determining direction. The final angle should be interpreted correctly in the context of the vector's quadrant.
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Homework Statement



Find the vector with the components: Ax=3.2 Ay= -5.15

The Attempt at a Solution



-5.15^2 + 3.2^2= 36.76
sq root of 36.76= 6.06
So, the magnitude is 6.06. Now for the direction...

CosD=3.2/6.06
ArcCos(0.52)=58.7 degrees
So, the direction is 58.7

Am I right? Thanks so much!
 
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joe215 said:

Homework Statement



Find the vector with the components: Ax=3.2 Ay= -5.15

The Attempt at a Solution



-5.15^2 + 3.2^2= 36.76
sq root of 36.76= 6.06
So, the magnitude is 6.06. Now for the direction...

CosD=3.2/6.06
ArcCos(0.52)=58.7 degrees
So, the direction is 58.7

Am I right? Thanks so much!

with y negative and x positive, the vector falls in fourth quadrant. But 58.7 is in first..

other than that, everything else seems good.

try using tan theta = y/x .. it's easier to use
 
So does that mean...

Tan theta=-5.15/3.2
ArcTan(-1.6)= -57.99 degrees

How can I have a negative angle?
 
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