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Resnick Halliday Krane Unit Vectors question

  1. Aug 4, 2012 #1
    The start of a new school year in Ap physics C, and our favorite engineering professors causing students distress! Maybe the champions of the textbook can help me out... = )

    1. The problem statement, all variables and given/known data

    1. Two vectors are given by
    a = 4i hat - 3j hat + k hat
    b = -i hat +j hat + 4k hat.

    Find
    a) a + b
    b) a-b
    C) vector c such that a -b +c = 0

    2.Given two vectors
    a = 4i hat - 3j hat
    b = 6i hat + 8j hat

    Find the magnitudes and directions (with the +x axis) of

    a)a
    b)b
    c)a+b
    d)b-a
    e)a-b




    2. Relevant equations
    components equations
    etc



    3. The attempt at a solution


    on question 1, ive gotten parts a and b, but running into difficulty proving a-b + c=0.
    i took a-b from part b and replaced it into the equation.

    C)
    a - b = 3i hat + 2j hat

    3i hat + 2j hat + c = 0

    3i hat + 2j hat + (azk hat + bz k hat)

    3i hat + 2j hat + (1 + 4)k hat

    3i hat + 2j hat + 5k hat = 0

    i solved for vector c as 5k hat, but i am not sure if this satisfies the solution, as k hat runs into the z direction.



    on question 2,

    e) a - b

    a = 4i hat -3j hat
    b= 6i hat +8j hat

    a-b=-2i hat - 11j hat

    -2i hat = cx = -2
    -11j hat = cy = -11

    [itex]\sqrt{125}[/itex] = 11.180

    tan [itex]\Phi[/itex] = (-11/-2) = 79.650 degrees

    vector c = 11.180 at 79.650 degrees with +x-axis.

    this is the answer i had gotten, but the book answer decides to choose 260 degrees. i have looked at my positive and negative symbols for my angle measure, but cannot meet the same number.

    Any help is much appreciated :)
     
  2. jcsd
  3. Aug 4, 2012 #2

    Doc Al

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    Staff: Mentor

    Redo that subtraction.

    Hint: What quadrant should your answer be in?
     
  4. Aug 4, 2012 #3
    For Question 1 C consider it to be a normal algebraic equation. If it were asking you for a number, z, such that if x = 3, y = 5, x - y + z = 0 What would you do? You seem to be thinking that the vector c has to be the z component of this new a-b vector, but it isn't! It is its own unique vector [itex] \vec{c} = c_{x}\hat{i} + c_{y}\hat{j} + c_{z}\hat{k} [/itex]
     
  5. Aug 4, 2012 #4

    cepheid

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    Staff Emeritus
    Science Advisor
    Gold Member

    Keep in mind that to add two vectors that are expressed in component form, you add them component-wise. In other words, the x-component of the resultant is the sum of the x-components of all the individual vectors, and likewise for y and z. Bearing that in mind, I think the best approach for part C is to use a vector c = (cx, cy, cz) whose components are unknown (note: boldface denotes vectors). This way you have three unknowns, but you also have three equations (which come from the component-wise summation) so you can solve for them all.
     
  6. Aug 4, 2012 #5
    Great!! I got both of them. for question 1 it looks like i just got a bit lazy. question 2 i did indeed needed to refer to the quadrant. For my future reference, i should always check which quadrant i am going to end up in at the end.

    oh yea another small question. What does "with + x axis" mean? does it mean with respect to the x axis?

    Thanks alot!!
     
  7. Aug 4, 2012 #6

    Doc Al

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    Staff: Mentor

    Yes. (As opposed to the negative x axis.)
     
  8. Aug 4, 2012 #7
    A handy rule to remember is: When taking an inverse tangent on a calculator, if the answer is in the 2nd or 3rd quadrant, add 180 degrees to the answer that the calculator gives you.
     
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