Another Quick Vector Question

joe215

1. The problem statement, all variables and given/known data

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

3. 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!

rootX

Quote by joe215

1. The problem statement, all variables and given/known data

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

3. 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

joe215

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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joe215	Another Quick Vector Question Tweet 1. The problem statement, all variables and given/known data Find the vector with the components: Ax=3.2 Ay= -5.15 3. 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!