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Another Quick Vector Question

  • Thread starter joe215
  • Start date
  • #1
26
0

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!
 

Answers and Replies

  • #2
378
2

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
 
  • #3
26
0
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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