Physics Forums (http://www.physicsforums.com/index.php)
-   Calculus (http://www.physicsforums.com/forumdisplay.php?f=109)
-   -   Help with simple dot product proof (http://www.physicsforums.com/showthread.php?t=6658)

 Juntao Oct3-03 09:12 AM

help with simple dot product proof

Here's what I got to prove where '.' is dot.

A.B=A.C Then B=C True or false? If true, prove it in general terms, if false, provide a counter-example.

Ok, I just need some body to comment on my little proof here, and any guidelines to make it more thorough or whatnot.
I know that the dot product is commutative,
A.(B+C)=A.B +A.C, but not sure if it really needs to be in my proof or not.

Proof
------
Say A.B=N and A.C=N (where N is a scalar number)
so if N=N
Then A.B=A.C
If I cancel the A's, I get B=C.

Is that a good way to approach that, or is there a better way of expressing it?

 Arden1528 Oct3-03 11:04 AM

You know that (A^-1)(A)=1 or the identity. Then
(A^-1)(A)(B)=(A^-1)(A)C)
with this we can multiply both sides and get
1(B)=1(C) or B=C

The raeson that (A^-1)(A)=1 is because (A^-1) is the inverse for A.

 gnome Oct3-03 11:12 AM

What if A is the zero vector? Then A.B=A.C no matter what B and C are.

And even if A <> 0 if you break A, B and C down into components, I think you will find that you can come up with other situations where A.(B-C) must equal 0 even though you know nothing about the values of B and C individually.

Try it.

 HallsofIvy Oct3-03 11:48 AM

No.

In the first place, there is no "A-1" when you are talking about dot product. There is, start with, no "identity" since
A.I= A would not make sense. A is a vector and the dot product of two vectors is a number, not a vector.

You are not really using either commutative or distributive laws:
you are using cancellation which is exactly what you are asked about: Is is true that when A.B= A.C, B MUST equal C. You cannot use what you are asked to prove.

Here is a hint. Choose two vectors at right angles. Call them A and B. Now choose a third vector at right angles to A. Call it C.

 mathman Oct3-03 09:22 PM

a.b=a.c
a.(b-c)=0
Therefore a is perpendicular to b-c. This does not imply b=c.

Example (3 space):
a=(1,0,0)
b=(x,u,v)
c=(x,s,t)
where x,u,v,s,t may assume any values.

 Juntao Oct5-03 10:26 AM

I thought this problem was going to be easy, but I keep on getting confused each time I come back here. Let's see if I get this straight mathman.

Let's say that A and B are perpendicular to each other. Now another vector, C, is perpendicular to A and B.

So A.B=0 and A.C=0, but this doesn't imply that B and C HAVE to equal each other?

And one more thing.
Example (3 space):
a=(1,0,0)
b=(x,u,v)
c=(x,s,t)

Just some clarification. Does x for vecter b and c have to be the same number?

 HallsofIvy Oct5-03 03:51 PM

Take A= (1,0,0), B= (0,1,0), and C= (0,0,1). It can't get any simpler than that.

You also say:
"And one more thing.
Example (3 space):
a=(1,0,0)
b=(x,u,v)
c=(x,s,t)

Just some clarification. Does x for vecter b and c have to be the same number?"

I have absolutely no idea. Generally speaking we do NOT use the same letter to represent two different numbers, but what was the context?

 All times are GMT -5. The time now is 06:44 AM.