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Quick Question

  • #1
I am doing a assingment for my classical mechanics class that requires the proof of:
The dot product of |A dot B| <= (less than or equeal to) |A| |B| .

I did the algebraic proof fine but we are required to do a geometic proof as well. This leaves me with the question what is the geometic view of the Dot Product?

At first I brushed it off thinking that it was a prjection of some sort, but I have no idea what |A||B|Cos(Angle) would be geometirally?

Thanks for the help
Derek
 

Answers and Replies

  • #2
Fermat
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derekmohammed said:
...

At first I brushed it off thinking that it was a prjection of some sort, but I have no idea what |A||B|Cos(Angle) would be geometirally?

Thanks for the help
Derek
|B|Cos(Angle) is the projection of B upon A.

Lat A and B be two vectors coming out from the origin. They have an angle of Φ between them.
If you draw a line from the tip of B to the vector A such that the line is perdindicular to A, and intersects it at P, say, then |B|cosΦ is the distance |OP| which is the projection of B onto A.
 
  • #3
I know that |B|Cos(angle) is the projection on A but what is |A|B|Cos(angle)???
This is the question
 
  • #4
Fermat
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|A||B|Cos(Φ) is simply the product of two scalars and is usually interpreted as two co-linear lines, |A| and |B|Cos(Φ).

Hmmm.

Can you do this ?

Let |A||B|Cos(Φ) be the product of two scalars, S1 and S2, where S1 = |A| and S2 = |B|cos(Φ).
The product of any two scalars is an area: A1 = S1*S2.

So now sketch a little rectangle with sides labelled S1 and S2.

Then |A||B| is also the product of two scalars S1 and S3, where S1 = |A| and S3 = |B|.
The product of these two scalars is A2 = S1*S3.

So now sketch another little rectangle with sides labelled S1 and S3.

Can you do something like this to (geometrically) show that A1 <= A2 ?
 
  • #5
Integral
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|A||B|cos ([itex] \Theta[/itex]) is a scaler and represents the magnitude of the vector, the direction is that of A,
 
  • #6
HallsofIvy
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You want a "geometric" proof of |A dot B|<= |A||B| and you know that, geometrically, A dot B= |A||B|cos(&theta;)?

Okay how large or how small can cos(&theta;) be?
 

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