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Conceptual explanation of Dot Product

by M.Hamilton
Tags: conceptual, explanation, product
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M.Hamilton
#1
Sep11-05, 01:41 AM
P: 6
I'm a peer leader for a general physics lab and someone asked me to explain what the Dot Product meant conceptually.

I told him it was the projection of A onto B multiplied by the magnitude of B.

He looked even more confused after that; my questions are:

a) Did I explain it correctly?
b) Is there a better way to explain it?

Sincerely,
Merle
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matt grime
#2
Sep11-05, 03:28 AM
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According to you the dot product is a vector, when it is in fact a scalar.

personally i don't go in for 'conceptual' explanations preferring to state its definition and uses, and you can pick any one that suits you.

given two vectors a and b, a.b is the quantity |a||b|cos(t) where t is the angle from a to b in an anticlockwise sense.

roughly it measures the angle between two vectors then, that is a.b/|a||b| is the angle between them.

if b is a unit vector then it is the length of the component of a lying in the direction of b.
arildno
#3
Sep11-05, 06:18 AM
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If you want a hand-wavy "explanation" (I don't know what else a "conceptual" explanation is) you could start with that:
a) The length of a vector is an intrinsic property of it, i.e, it does not depend upon the perspective or coordinate system you choose to use.
and
b) The angle between two vectors is an intrinsic property belonging to those two vectors, it does not depend upon the perspective or coordinate system you choose to use.

The dot product enables you to readily find the angle between two vectors once you know their individual lengths.

For vectors both of unit length, the dot product IS the cosine to the angle between them.

Hurkyl
#4
Sep11-05, 09:56 AM
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Conceptual explanation of Dot Product

Another way to explain it is...

The dot product is simply the (continuous) function satisfying:

the product of a unit vector with itself to be 1.
the product of orthogonal vectors to be 0.
it's distributive.
paperwings
#5
Sep11-05, 10:11 AM
P: 6
It might help if you draw it out, and give an example. (but thats how I learn)
amcavoy
#6
Sep11-05, 10:26 AM
P: 668
Quote Quote by M.Hamilton
I'm a peer leader for a general physics lab and someone asked me to explain what the Dot Product meant conceptually.

I told him it was the projection of A onto B multiplied by the magnitude of B.

He looked even more confused after that; my questions are:

a) Did I explain it correctly?
b) Is there a better way to explain it?

Sincerely,
Merle
Component. Not the projection.
arildno
#7
Sep11-05, 10:53 AM
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Quote Quote by paperwings
It might help if you draw it out, and give an example. (but thats how I learn)
Very good suggestion!
On the "elementary" level, few things are as educational as a good, visual representation!
Dr Avalanchez
#8
Sep12-05, 02:39 AM
P: 18
You could explain it pure algebraically (this holds for higher dimensions):

Starting by the proof of the Cauchy-Schwarz inequality we have:
[tex]\left| {x \cdot y} \right| \leqslant \left\| x \right\|\left\| y \right\|[/tex]. This is of course the same as:
[tex]-\left\| x \right\|\left\| y \right\| \leqslant x \cdot y \leqslant \left\| x \right\|\left\| y \right\|[/tex]. Or:
[tex]-1 \leqslant \frac{x \cdot y}{\left\| x \right\|\left\| y \right\|} \leqslant 1[/tex], So that we know that [tex]\exists!\theta \in [0,\pi][/tex] so that
[tex]\frac{x \cdot y}{\left\| x \right\|\left\| y \right\|} = \cos\theta[/tex].

Only then we define [tex]\theta[/tex] to be the angle between the 2 vectors x and y in n-dimensional euclidean/unitarian space.


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