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Scalar/Dot Product

  1. May 22, 2014 #1
    I'm taking "physics for engineers" right now - the condensed 4 hour summer course over 7 weeks.

    I'm doing fine in the class. I feel confident about the ideas and concepts we've covered so far, sure enough, but I'm having a hard time grasping the concept (geometrically at least) of the scalar/dot product.

    Certainly ABcos(ø) is simple enough to calculate. And I understand I'm getting a scalar, rather than a vector, so there isn't going to be an actual geometric representation of the number calculated... but... what is it?

    For example if we have two vectors, A & B, and we say that A runs along the x-axis in the positive direction and has a magnitude of 4, and we have a vector B that runs 77° from A with a magnitude of 5 (that is to say B is in the first quadrant, between the x and y axis, but closer to the y axis.) then we have the following:

    [4][5]cos[77] = 4.5

    What is the 4.5 representative of? A magnitude of a non-existent vector? The force one vector applies to another?
     
  2. jcsd
  3. May 22, 2014 #2

    jedishrfu

    Staff: Mentor

    its the projection of vector A on vector B or alternatively the projection of B on A.

    http://en.wikipedia.org/wiki/Dot_product

    So say B was a unit vector normal to some surface and vector A represented some sort of flow crossing this surface then the projection of A on B would be the effective flow thru the surface. If A were perpendicular to B then it would be tangent to the surface and the projection of A on B would be zero meaning there is no flow across the surface.
     
  4. May 23, 2014 #3
    I've read that projection part before. Maybe it will clear up later; I was hoping for some alternate - perhaps watered down - explanation of it in non-technical terms. "The projection of vector A on to vector B" doesn't mean anything to me at this point... what is it projecting? It's magnitude? Can you give a real world example?

    I feel like I'm probably over complicating this, but I want to get an intuitive sense of what's going on.
     
  5. May 23, 2014 #4
    Projecting: imagine a length of wood sticking up out of the ground at some angle. You stand over the wood looking straight down. You would see a shortened view of the wood - the projection of the wood on the ground. The wood represents a vector. Treating the wood as the hypotenuse of a right angled triangle with one side (the projection we want) on the ground and the other perpendicular to the ground this gives the calculation of the size of the projection on the ground.
     
  6. May 23, 2014 #5
    Another view.
    You have a video projector and it is projecting the image of a straight line onto a screen. Usually you set the screen to be at right angles to the direction of projection (it is square on). But this time you move the projector so the beam is not straight on to the screen but strikes it at an angle other than a right angle. The image of the straight line would be foreshortened - again giving the projection.
    The straight line represents the vector in magnitude and direction. Its image is the projection.
     
  7. May 23, 2014 #6

    BiGyElLoWhAt

    User Avatar
    Gold Member

    Another wat to calculate the dot product is by multiplying the components.

    ##<a,b,c> \cdot <d,e,f> = ad +be +cf ##

    This is how I like to think about it, because using this method you can actually see what you're doing. Your taking the portion of vector abc and finding out (directionally) what it has in common with vector def and multiplying those "chunks" together.

    Probably you'll most often see this with work in your physics class, if this is an early physics class.

    What you're doing here:
    ##f \cdot \Delta r##
    is you're finding out which portion of the work is in the same direction as the displacement and multiplying the 2 "chunks".
     
  8. May 23, 2014 #7
    So, given my example, with the resultant scalar product of 4.5, this would be vector A projected on the vector B. I can say that, I can calculate it... but what property is it projecting? There is nothing coming from this calculation except a number (a scalar) as far as I can tell. It doesn't appear to have a magnitude or direction... so... is 4.5 the "length" of Vector A if it were uhmm... projected on to B?

    Excuse my ignorance.
     
  9. May 23, 2014 #8
    Yes - it is the length. Remember a vector is determined by two values, in the case of a line its length(or size) and its direction. So the dot product gives the size of the projection.
     
  10. May 23, 2014 #9

    jtbell

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    Staff: Mentor

    The first place where physics students usually actually use the dot product is in calculating the quantity that we call "work." Have you started that yet?

    When you push or pull on an object with a constant force, exerting it in the same direction that the object is moving, the work is simply force times distance. For example, pulling on a sled with a rope, with the sled directly behind you as you walk forwards.

    When you exert the force at an angle (when you walk off to one side so the rope is at an angle to the sled's path), then only the component of the force parallel to the sled's motion "contributes" to the work. The component perpendicular to the motion doesn't do any work. So we use the dot product for calculating work: ##W = \vec F \cdot \vec d##.
     
  11. May 23, 2014 #10

    BiGyElLoWhAt

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    Gold Member

    Let ##\vec{A}## lie along the x-axis and have magnitude 4 and ##\vec{B}## have magnitude 5 and direction 77 degrees.

    ##\vec{A} = <4,0,0> ##
    ##\vec{B} = <5cos(77), 5sin(77), 0> ##
    ##\vec{A} \cdot \vec{B}##
    ##= <4,0,0> \cdot <5cos(77), 5sin(77), 0> ##
    ##= 4 \cdot 5cos(77) + 5sin(77) \cdot 0 + 0 \cdot 0##
    ##= (4)(5)cos(77)##

    In the picture what you're doing when calculating the dot product, is finding the projection of B onto A, and then multiplying the magnitude of the projection by the magnitude of A. Does this make sense?
     

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  12. May 23, 2014 #11

    jedishrfu

    Staff: Mentor

    Projecting is just the math way of describing the fact that the A vector can be decomposed into two vectors (P) and (Q) one vector (P) which is parallel to B and one vector (Q) which perpendicular to B.

    In other words:
    Code (Text):

    vector A = vector P + vector Q

        with vector P || vector B

        and with vector Q _|_ vector B
     
    So if the (B) vector was a unit vector that is its length is 1 then:
    Code (Text):

    The projection of A on B is the length of the (P) vector
     
     
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