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Magnitude and vector help

  1. Sep 15, 2009 #1
    1. The problem statement, all variables and given/known data

    What are
    (a) the x component,
    (b) the y component, and
    (c) the z component of
    r= a-b +c if
    a = 6.5i + 4.2j - 6.7k,
    b = -6.3i + 1.5j + 1.4k, and
    c = 8.2i + 4.3j + 5.9k.
    (d) Calculate the angle between r and the positive z axis.
    (e) What is the component of a along the direction of b?
    (f) What is the magnitude of the component of a perpendicular to the direction of b but in the plane of a and b and ?


    2. Relevant equations
    a . b = abcosθ
    c = absinθ
    tanθ= (aj/ai)

    3. The attempt at a solution
    a) 21
    b) 7
    c) -2.2

    21i + 7j - 2.2k

    d)√(441+ 49 + 4.84) = 22.24

    θ=sin-1 = 7/22.24
    θ = 18.34 /*Incorrect im pretty sure ;/ */
    I took y/r r as hypot and y as opposite

    e)Well I know how to find the ay = asinθ etc but the z I do not know.
    So I tried
    direction of b -i,+j,+k
    thus a = -6.5i + 4.2j + 6.7k but then I believe I have three components. So im clueless.
    f) close the book
    a X b produces third vector c

    i j k
    6.5i + 4.2j - 6.7k
    -6.3i +1.5j + 1.4k

    = 15.93i - 33.11j + 36.21k
    = magnitude 51.58
     
  2. jcsd
  3. Sep 15, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    Re: Vectors

    d) You have to find the angle between -z-axis and r, calculate cosθ = z/r, then find (180 - θ ) to find the angle with + z-axis.
     
  4. Sep 16, 2009 #3
    Re: Vectors

    so cosθ = 2.2/22.24
    θ = 84.3
    180 - θ = 95.6 degrees?

    If - 2.2k was positive wouldnt it be the same thing as the + z-axis because cos is same for +/-? Will I always do (180 - θ)?

    I had done e) wrong
    (e) What is the component of a along the direction of b?
    Its basically
    a.b = abcosθ
    a.b / b = component along direction of b

    (f) What is the magnitude of the component of a perpendicular to the direction of b but in the plane of a and b and ?
    What is it talking about :cry:

    I thought it had something to with
    aXb= c = absinθ

    But then c is the direction perpendicular to the plan of a and b not in it.
    So is a = a; b = c; c=a;
    aXc = b = acsinθ
    But I dont have the c :\\\\\\\\\\\\
    :confused:
     
  5. Sep 16, 2009 #4

    rl.bhat

    User Avatar
    Homework Helper

    Re: Vectors

    e) component of a along b is acosθ = abcosθ/b = a.b/b
     
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