1. Limited time only! Sign up for a free 30min personal 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!

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

    User Avatar
    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

    User Avatar
    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

    User Avatar
    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.
    $$\vec{2i} \cdot -\vec{i} = |2\vec{i}| |-\vec{i}| \cos(180°) = (2)(1)(-1) = -2$$
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Multiplying vectors -- where does cos(theta) go?
Loading...