# Direction of a vector after I have already found magnitude

1. May 24, 2012

### wbetting

1. The problem statement, all variables and given/known data
The x component of vector is -16.2 m and the y component is +54.6 m. (a) What is the magnitude of ? (b) What is the angle (in radians) between the direction of and the positive direction of x?

2. Relevant equations
magnitude of vector is sqrt of a^2+b^2
direction of vector is tan theta= ay/ax

3. The attempt at a solution
I got A part correct by finding the magnitude using sqrt of a^2+b^2= 56.95m but i can't seem to get b part. i did inverse tan of 56.4m/-16.2m and got -1.28 radians and its wrong. HELP!!!

2. May 24, 2012

### Parzival

You did 56.4/-16.2.
Your given info, however states that the y component is +54.6 m.

So:

3. May 25, 2012

### wbetting

That was a typo. I have been calculating inverse tan of 54.6/-16.2 and get -1.28 which is wrong according to the automatic grader

4. May 25, 2012

### Staff: Mentor

The inverse tangent function cannot "tell" whether the negative sign is to be associated with the numerator or denominator in its argument; $\frac{-y}{x}$ is indistinguishable from $\frac{y}{-x}$. It will return a value that is in the range -90 ≤ θ ≤ +90 degrees.

You need to apply some smarts regarding the signs of the given components in order to place the resulting vector in the correct quadrant; You may have to add or subtract 180° ($\pi$ radians) from the value it gives you in order to "shift" the quadrant.

Alternatively, if your calculator has an atan2(y,x) function, then it'll handle the signs automatically. It may also have a "rectangular to polar" conversion capability which will both find the magnitude and the proper angle for you in one step.

5. May 25, 2012

### wbetting

oh ok so since -16.5 x and 54.6 y puts it in the 2nd quadrant the angle has to be between 90 and 180. -73.47 is in the 4th quadrant so by adding 180 to that and converting it to radians i get 1.855 radians which is correct!