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Multivariable Calculus - Dot Products

  1. Feb 13, 2008 #1
    Anyone familiar with dot products of two vectors? What does the dot product show, in other words what is the point of doing a dot product?
  2. jcsd
  3. Feb 13, 2008 #2


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    Welcome to PF,

    I'm assuming that this is a homework assignment, so what is the definition of the scalar product? How to you calculate it? How is it related to the angle between the two vectors?
  4. Feb 13, 2008 #3
    I'm still confused. Here's an example. Im given a problem X "dot" X and that would give me x1*x1 + x2*x2 + x3*x3 . What is the point of having this information? What does the dot product mean geometrically?
  5. Feb 13, 2008 #4


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    One can also define the dot product thus,


    Where [itex]\theta[/itex] is the angle between the two vectors and [itex]\left|\underline{v_1}\right|[/itex], [itex]\left|\underline{v_2}\right|[/itex] are the magnitudes of the respective vectors. Therefore, geometrically the scalar product represent the projection of [itex]\underline{v_1}[/itex] on the unit vector in the direction of [itex]\hat{\underline{v_2}} = \underline{v_2}/\left|\underline{v_2}\right|[/itex]. The opposite is also true. For more information see http://mathworld.wolfram.com/DotProduct.html" [Broken]
    Last edited by a moderator: May 3, 2017
  6. Feb 13, 2008 #5
    thank you sir. i think i understand it now. By the way, how did you make those cool symbols like for theta or the dot in between v1 and v2
  7. Feb 13, 2008 #6


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    By using [t e x]symbol notation here[/t e x] or [i t e x]symbol notation here[/i t e x].
  8. Feb 15, 2008 #7
    The dot product has another useful use:

    Work = Force•distance
  9. Feb 15, 2008 #8


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    Well, I certainly wouldn't put it that way. It is common to think of Force as a vector but "distance" is just a number. I would say Work = Force•displacement.

    One important application, of which the work formula above is a specific example, is finding the projection of one vector on another. If [itex]\vec{u}[/itex] and [itex]\vec{v}[/itex] are vectors, then the "projection of [itex]\vec{u}[/itex] on [itex]\vec{v}[/itex] is:

    And, of course, the fact that [itex]\vec{u}\cdot\vec{v}= 0[/itex] if and only if [itex]\vec{v}[/itex] and [itex]\vec{u}[/itex] are perpendicular.

    In general, [itex]\vec{u}\cdot\vec{v}= ||\vec{u}||||\vec{v}|| cos(\theta)[/itex] where [itex]\theta][/itex] is the angle between the vectors. "Formally" extending that to vectors with more than three components allows us to define the angle between two vector in higher dimensions.
  10. Feb 15, 2008 #9
    Oh of course, my bad. The dot product is extremely useful throughout physics and maths, such as in Line Integrals.
  11. Feb 15, 2008 #10
    In my experience, the dot product tells you more about the geometry than the other way around. What I mean is that often times, although you do things in real life one way (defining vectors as being having distance and angles and stuff and then defining your dot products in terms of these), mathematicians will often turn it around and start with the vectors and then define distance and angles in terms of the dot product.

    This is why often times the formulas relating distance and angles to the dot product are sometimes unintuitive. (And why the dot product is a fairly confusing beast at first) In real life, it's useful to think one way. Mathematicians, however, have found that it's sometimes more useful to basically think backwards.

    As an example, using the formula [tex]\underline{v_1}\cdot\underline{v_2}=\left|\underline{v_1}\right|\left|\underline{v_2}\right|\cos\theta[/tex], you can show that 2 vectors at right angles to each other have a dot product of 0. Mathematicians on the other hand will often define right angles as having a dot product of 0.

    Makes some things really nice, lets you easily generalize a lot of things that you find out about vectors... but confuses the heck out engineering and physics students
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