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New Member
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Jan 23, 2011, 07:11 PM
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vectors
Assume that a and b are two vectors in R^3 with the property that a*x = b*x
for every vector x in R^3. Must a = b? If so, prove it. If not, give a
counterexample.
what is this question asking?
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Senior Member
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Jan 23, 2011, 07:49 PM
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It seems likely that the original question is actually asking about a dot-product ( ) or a cross-product ( ). If so, which is it?
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Senior Member
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Jan 23, 2011, 08:39 PM
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I'll assume that the question is asking about a dot-product. If not, ignore this answer. :)
Let's denote the unit vectors in (three-dimensional real space) as , , and . (Those would be unit vectors in the x, y, and z directions, respectively). Any vector, , in can then be written as
,
where , , and are scalars.
Likewise, we can define two more vectors and as
The question is asking "If , does that automatically mean that ?"
Now recall the definition of the dot product (whose result is a scalar):
and
.
So if for any , , and , does that mean that , , and ?
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New Member
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Jan 26, 2011, 11:49 PM
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I think it is no because it is the addition of all the quantities on the left = the addition of all the quantities on the right, so it is not necessary that a1=b1, a2=b2, a3=b3. If so, how can I make an counterexample?
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Senior Member
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Jan 27, 2011, 02:24 PM
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You'd be right if the equation had to be true for just one specific vector x, but the problem said it has to remain true for all vectors in R3. The only way that can be true is if a and b are equal.
Try it. Make up a really simple vector [a1, a2, a3] (I'd do something simple like [1, 2, 3]). Then make up some simple test vectors [x1, x2, x3] (like [1,1,1]; [1,3,2]; [2,1,1]; [2,3,1]; etc.). For any one x-vector, you should find that you can easily come up with a variety of different b-vectors which make the equation true. For example, if you choose [1,1,1] for your a-vector and then [1,2,3] for your x-vector, the left side of the equation (a1x1 + a2x2 + a3x3) is 6. A b-vector of [1,2,3] gives you the same answer for the right side of the equation, of course, but so does [-1,2,1], [0,0,2], and an infinite number of others.
However, now if you try to find a b-vector which simultaneously makes the equation true for several different x-vectors, you'll find it's impossible unless b = a.
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