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Magnitude or direction of a vector

  1. Jun 22, 2004 #1
    Can someone please tell me what i'm doing wrong?

    I drew a picture of this triangle, but i'm not coming up with the right magnitude or direction. it says a scout troop is practicing it orienreering skills with map and compass. First they walk due east for 1.2 km. Next, they walk 45 degrees west of north for 2.7km. How far and in what direction must they walk to go directly back to their starting point.

    I'm getting the x component of A to be 1.2 and the y component of A to be 0. I'm getting the x component of B to be 2.7cos45 and the y component of B to be -2.7 sin 45. But I'm not getting the right answer. Please help!!
  2. jcsd
  3. Jun 22, 2004 #2
    Let u = (1.2, 0) and v = (-2.7cos(45), 2.7sin(45)) be the given vectors. The location of the scouts is basically u + v. To get back to the starting point, the scouts need to reverse their location, which is just -(u + v). Dead simple. Maybe you have problems calculating the magnitude and direction of this vector. What did you get as an answer?
  4. Jun 22, 2004 #3
    thank you. i had the negative signs switched. i had it as 2.7cos(45) and -2.7sin(45). Also, for another problem i got the y component right but the x component is wrong. It says if a vector is 20 m long and makes an angle 60 degrees counterclockwise from the y-axis, what are the x and y components. I got -10 as the y component and 17.3 as the x component, but the x component is wrong. Do you know what I'm doing wrong?
  5. Jun 22, 2004 #4
    Your response looks OK to me, although your description of the vector is somewhat ambiguous. What are you supposed to get?
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