- #1

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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

- Thread starter M.Hamilton
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- #1

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I told him it was the projection of

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

- #2

matt grime

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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.

- #3

arildno

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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.

- #4

Hurkyl

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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.

- #5

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It might help if you draw it out, and give an example. (but thats how I learn)

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M.Hamilton said:

I told him it was the projection ofAontoBmultiplied by the magnitude ofB.

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

- #7

arildno

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Very good suggestion!paperwings said:It might help if you draw it out, and give an example. (but thats how I learn)

On the "elementary" level, few things are as educational as a good, visual representation!

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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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