Multivariable Calculus - Dot Products

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Discussion Overview

The discussion revolves around the concept of dot products in multivariable calculus, exploring its definition, geometric interpretation, and applications in physics and mathematics. Participants share their understanding and confusion regarding the significance and calculation of dot products.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants inquire about the definition and purpose of the dot product, questioning its geometric meaning.
  • One participant presents the formula for the dot product in terms of vector magnitudes and the cosine of the angle between them, suggesting it represents a projection.
  • Another participant mentions the application of the dot product in calculating work, while another corrects this to emphasize displacement instead of distance.
  • Some participants discuss the relationship between the dot product and the concept of perpendicular vectors, noting that a dot product of zero indicates orthogonality.
  • A participant reflects on the geometric implications of the dot product, suggesting that mathematicians often define angles and distances based on the dot product, which can be counterintuitive for students in physics and engineering.

Areas of Agreement / Disagreement

Participants express varying interpretations of the dot product and its applications, with some agreeing on its significance in geometry and physics, while others contest specific definitions and applications, indicating that the discussion remains unresolved.

Contextual Notes

There are unresolved questions regarding the definitions of terms like "distance" versus "displacement" in the context of work, as well as the implications of defining geometric concepts based on the dot product.

engstudent363
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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?
 
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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?
 
I'm still confused. Here's an example. I am 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?
 
One can also define the dot product thus,

[tex]\underline{v_1}\cdot\underline{v_2}=\left|\underline{v_1}\right|\left|\underline{v_2}\right|\cos\theta[/tex]

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"
 
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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
 
engstudent363 said:
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
By using [t e x]symbol notation here[/t e x] or [i t e x]symbol notation here[/i t e x].
 
The dot product has another useful use:

Work = Force•distance
 
flebbyman said:
The dot product has another useful use:

Work = Force•distance
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:
[tex]\frac{\vec{u}\cdot\vec{v}}{||\vec{v}||}[/tex].

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.
 
HallsofIvy said:
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.

Oh of course, my bad. The dot product is extremely useful throughout physics and maths, such as in Line Integrals.
 
  • #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, let's 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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