# Vector problem redux

Discussion in 'Math' started by Mark44, Apr 10, 2008.

1. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
Someone posted a problem about vectors in this folder a couple of years ago. A member posted a reply, but the member didn't realize that the problem involved vectors. To be fair, the OP didn't provide much context to make that clear.

Anyway, here it is again...

Given that a$\times$b = a$\times$c and a$\bullet$b = a$\bullet$c, where a, b, and c are vectors in R$^{3}$, show that b = c.

The hint given in the original problem is to "cross" both sides of the first equation with a.

Mark

2. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
519
Proofs.

the man wants proofs now!

In general yes, it follows from the transitive property of vectors in the definition of an inner product space.
but
if a is the zero vector then the statement does not hold.

3. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
OK, let's assume that a is a nonzero vector. (The OP made no such assumption, and it didn't occur to me).

I don't get that this follows from the transitive property. The transitive property of what operation? Cross product? Dot product? For any readers who are rusty on the transitive property, for a relation R over a set of vectors, if a R b and b R c, then a R c.

Maybe I'm being obtuse here, but I don't see how this applies.
Mark

4. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
519
OK I've been away for a few days but back now.

Mark you are quite correct I was too quick, the transitive property does not help here.

So here is my solution - To consider the new vector (B-C).

Now

AxB = AxC and A.B = A.C

Thus AxB - AxC = 0 (zero vector) and A.B - A.C = 0

i.e. Ax(B-C) = 0 (zero vector) and A.(B-C) = 0

Thus mod{Ax(B-C)} = mod{zero vector} and mod{A.(B-C)} = 0

i.e. mod{A} x mod{(B-C)} x sin$\theta$ = 0 :: the x sign now means ordinary multiplication of numbers

and

mod{A} x mod{(B-C)} x cos$\theta$ = 0

Since A $\neq$ 0 and both sin$\theta$ and cos$\theta$ cannot simultaneously be zero

mod{(B-C)} = 0

(B-C) = 0 (zero vector)

B = C

as required.

This question does highlight the fact that both cross and dot products must be zero for the equality to hold. It is easy to construct counter examples when only one holds.
I don't understand the hint. How can you cross an equation?

5. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
Studiot,
Looks good!

Regarding the hint, it suggested crossing each side of the equation, not the equation itself.
Mark

6. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
519
This question is actually quite instructive as it elicits some important points about vectors and vector products.

Firstly, just as we can find more than one pair of numbers r,s that satisfy the equation
r times s =12 (6x2; 3x4; etc)
We can find more than one pair of vectors that satisfy the dot or cross product of two vectors.

Secondly when equating to zero we need to be careful with objects other than numbers. It is not true to say that for two vectors A,B if

AxB or A.B = 0 then

either A or B = 0

It may be true, but not necessarily.

Here are some concrete examples

A = (1,1,1)
B = (7,8,9)
C = (2,3,4)

B-C = (5,5,5)

D = (5,5,5)
E = (1,-2,1)

AxB = {(9-8),(7-9),(8-7)} = (1,-2,1)
AxC = {(4-3),(2-4),(3-2)} = (1,-2,1)
Ax(B-C) = {(5-5),(5-5),(5-5)} = (0,0,0)

AxB = AxC but B $\neq$C

Because
A.B = (7+8+9) = 21
A.C = (2+3+4) = 9

A.B $\neq$ A.C

Also Consider
D.E = (5,-10,5) = 0

neither D nor E = 0

7. ### Ratch New Member

Mar 20, 2007
1,068
4
Taking the hint:

Ax(AxB) = Ax(AxC) =(A.B)A - (A.A)B = (A.C)A - (A.A)C

since A.B = A.C; then -(A.A)B = -(A.A)C

therefore B = C for the two given relationships

Ratch

8. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
519
If that is what the hint means then I'm afraid the proof fails.

It fails because it has to work for all A.
In particular it has to work for A=Zero Vector.

If A = zero vector all you have proven is the consistancy of the statement

0 = 0

Otherwise 'proofs' such as the following would be possible.

0 x 3 = 0 x 7 (true)

Therefore

3 = 7 (false)

Last edited: May 16, 2008
9. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
All that is required is that we stipulate (again--see the third post of this thread) that a not be the zero vector.

10. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
519
Ratch,
Looking back I see that we excluded A as zero vector so it's only fair to allow your manipulation to do the same.

So long as your formula is allowable, it is a good piece of manipulation.

Out of intrest do you know the proof of your formula? It is a fairly standard formula.

11. ### Ratch New Member

Mar 20, 2007
1,068
4
studiot,

Which is self evident.

The equation balances.

Yes, false because in order to reduce the above equation, division by zero is necessary. That is not allowed in algebraic manipulations. Ratch

12. ### Ratch New Member

Mar 20, 2007
1,068
4
studiot,

It is a straight forward derivation consisting of first finding the cross product in parenthesis, and next applying the cross product formula again for the last vector. There are lots of algebraic manipulations to carry out to reduce the terms. It is called the vector cross product, and I am sure any book on vector analysis will show the method. Ratch