1. Not finding help here? Sign up for a free 30min 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!

Questions about vector

  1. Aug 26, 2009 #1

    htk

    User Avatar

    1) use the Law of Cosines to find the angle theta between the given vectors. (Assume 0 degree <=theta<=180degrees)v= i+ j, w=2(i - j)

    My teacher told me that me that I have to find

    a. length(v) = a
    b. length(w) = b
    c. length of vector between (1,1), (2,-2) = c. (use the distance formula.) and then use this formula c^2 = a^2 + b^2 - 2ab cos C, and solve for C but I don't know how to find length(v) and length(w).

    the answer on the back of my book is 90 degrees.

    2)three forces with magnitudes 70 pounds, 40 pounds, and 60 pounds act on an object at angles of -30 degrees, 45 degrees, and 135 degrees, respectively with positive x-axis. Find the direction and magnitude of the resultant.

    I know that I need to find v1, v2, and v3. Then find the sum of the vectors, the magnitude of the sum and theta.

    First find the trig form, then convert to rectangular form

    at 70 pounds
    <60.02, -35> I don't know how to get this number.

    at 40 pounds
    <28.28, 28.28>

    at 60 pounds
    <-2.43, 42,43>

    now add the components to find the resultant of the system and use it to find the magnitude and direction angle.

    for this problem, I have no clue what my teacher tries to say.

    3) an airplane's velocity with respect to the air is 580 miles per hour, and it is heading N 60 degrees W. The wind, at the altitude of the lane, is from the southwest and has a velocity of 60 miles per hour. Draw a figure that gives a visual representation of the problem. What is the true direction of the plane, and what is its speed with respect to the ground?.

    u= air speed= 580 = ?
    v= wind speed = 60 = ?
    w = u + v = < -459.9, 332.4> I don't know how to get this number???

    I will have a test on Friday about this kind of problems, but I still don't get it. Can any one please explain it to me step by step? Thank you for your help!
     
  2. jcsd
  3. Aug 26, 2009 #2
    1) remember that vectors are little arrows with a certain magnitude and direction. v= i + j is a little arrow from the origin to point (1,1), while w= 2 i - 2 j is a little arrow from the origin to point (2, -2). So the formula for distance in 2d is sqrt( (x1-x0)^2 + (y1-y0)^2 ), using these two facts you can find the length of the two little "arrow" (or, alternatively, lines). This problem is really easy if you know how to take a dot product, but if you don't, don't worry as it can be done without it.

    2) Use a similar technique to "decompose" vectors into their components. In general if a vector is a i + b j, then it's x component is a, and it's y component is b. However we are given an angle and a magnitude, so we need to decompose it into x and y components. To do this we visualize the vector as a little arrow with that direction given by the angle. We can can then visualize it as the hypotenuse of a right triangle, where x is the length of the horizontal leg and y the vertical leg (use trig functions, cosine and sine respectfully). So do this for those 3 vectors, and then add the components to get the resultant.

    3) Use a similar procedure as 2, breaking each into component form and finding the resultant.
     
  4. Sep 3, 2009 #3
    length of v and w = magnitude.

    a + jb => Find the magnitude = |a+jb| = Sqrt of (a^z2 + b^2).

    Hope it helps. Good Luck.
     
  5. Sep 3, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I am sure NJunJie meant
    "ai+ bj => Find the magnitude = |ai+ bj|= Sqrt of (a^2+ b^2).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Questions about vector
  1. Vector Question (Replies: 5)

  2. Vectors question (Replies: 5)

  3. Vector question (Replies: 2)

Loading...