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Multiplying vectors -- where does cos(theta) go?

  1. Sep 8, 2015 #1
    For example if I have to find out work done.Force and displacement are given to be (2i^+3j^+4k^)and
    (-i^-2j^) respectively.
    till now i have been thinking that work done=Force multiplied by displacement cos theta but in my book in solution of the problem above work done is calculated by simply multiplying
    (2i^+3j^+4k^)and
    (-i^-2j^).
    where did cos theta go?when vectors are given in terms of i^,j^, k^ we don't write costheta,that's how it works,right?if yes,reason? I know displacement cos theta implies component of displacement along force.
     
    Last edited: Sep 8, 2015
  2. jcsd
  3. Sep 8, 2015 #2

    Titan97

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    Gold Member

    That is not a simple multiplication. Its called dot product. Work is defined as:
    $$W=\vec{F}\cdot\vec{s}$$

    When force and displacement are given in vector forms, you can also find the angle between them by using$$\cos\theta=\frac{\vec{F}\cdot\vec{s}}{|F|.|s|}$$

    You can read it here : https://en.m.wikipedia.org/wiki/Dot_product
     
  4. Sep 8, 2015 #3
    I know that!but should not magnitude of work done be F S COS theta
     
  5. Sep 8, 2015 #4

    Titan97

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    Gold Member

    Its is.
    $$W=|F|.|s|.\cos\theta$$
    But you have to find cosθ.
    This will be the long way to calculate work though.
     
  6. Sep 8, 2015 #5
    Your unit vectors i, j, k are mutually orthogonal to one another (at right angles), so cos(θ) is zero for any i≠j terms. You need to look up the dot product in component form for the question above.
     
  7. Sep 8, 2015 #6
    And what about i ,-i terms?will it be cos 180 i.e -1?
     
  8. Sep 8, 2015 #7
    Yes,that's correct.
     
  9. Sep 8, 2015 #8

    jbriggs444

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    Science Advisor

    The -i component can be understood as (value -1, direction i). With that understanding, when you multiply 2i^ by -i^ the relevant angle is the angle between i^ and i^. That is zero degrees. The result is 2 * -1 * cos(0).

    Understood that way, 180 degrees never enters in. All the cosines are equal to 1.
     
  10. Sep 8, 2015 #9

    Mark44

    Staff: Mentor

    Assuming that both force and displacement are vector constants (i.e., the force doesn't vary and acts along a straight line path), the work done is given by this dot product:
    $$W = \vec{F} \cdot \vec{S}$$
    Let's assume that ##\vec{F} = <f_1, f_2, f_3>## and ##\vec{S} = < s_1, s_2, s_3>##. Personally, I like this notation better than ##f_1 \vec{i} + f_2 \vec{j} + f_3 \vec{k}## as it is easier to write.

    There are two ways to calculate the dot product shown above.
    Coordinate form
    ##W = f_1 s_1 + f_2 s_2 + f_3 s_3##

    Coordinate-free form
    ##W = |\vec{F}| |\vec{S}| cos(\theta)##, where ##\theta## is the angle between the two vectors, and |F| and |S| are the magnitudes of the two vectors. For vectors in R3, the magnitude is the square root of the sum of the squares of the three components. For example, ##|\vec{F} = \sqrt{f_1^2 + f_2^2 + f_3^2}##.
     
  11. Sep 8, 2015 #10

    Mark44

    Staff: Mentor

    Just as valid would be the interpretation that -i has a value of 1 in the direction of -i (i.e., to the left).
    Using my interpretation, we get the same answer via a different route.
    $$\vec{2i} \cdot -\vec{i} = |2\vec{i}| |-\vec{i}| \cos(180°) = (2)(1)(-1) = -2$$
     
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