1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Direction of a vector after I have already found magnitude

  1. May 24, 2012 #1
    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. jcsd
  3. May 24, 2012 #2
    You did 56.4/-16.2.
    Your given info, however states that the y component is +54.6 m.

    inverse tan (54.6/-16.2) is your answer.
  4. May 25, 2012 #3
    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
  5. May 25, 2012 #4


    User Avatar

    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.
  6. May 25, 2012 #5
    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook