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 magnitude of 6.06 and a direction of 58.7 degrees when calculated using cosine. However, since the vector lies in the fourth quadrant, the correct angle should be adjusted to reflect this, resulting in an angle of -57.99 degrees when using the tangent function. The discussion highlights the importance of quadrant consideration in vector direction calculations.

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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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