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?

It seems likely that the original question is actually asking about a dot-product (a \cdot x = b \cdot x) or a cross-product (a \times x = b \times x). If so, which is it?

I'll assume that the question is asking about a dot-product. If not, ignore this answer. :)

Let's denote the unit vectors in \Re ^3 (three-dimensional real space) as \hat{\mathbf {i}}, \hat{\mathbf {j}}, and \hat{\mathbf {k}}. (Those would be unit vectors in the x, y, and z directions, respectively). Any vector, \hat{\mathbf {x}}, in \Re ^3 can then be written as

\hat{\mathbf {x}}= x_1 \hat{\mathbf {i}} + x_2 \hat{\mathbf {j}} + x_3 \hat{\mathbf {k}},

where x_1, x_2, and x_3 are scalars.

Likewise, we can define two more vectors \hat{\mathbf {a}} and \hat{\mathbf {b}} as

\hat{\mathbf {a}}= a_1 \hat{\mathbf {i}} + a_2 \hat{\mathbf {j}} + a_3 \hat{\mathbf {k}}

\hat{\mathbf {b}}= b_1 \hat{\mathbf {i}} + b_2 \hat{\mathbf {j}} + b_3 \hat{\mathbf {k}}

The question is asking "If \hat{\mathbf {a}} \cdot \hat{\mathbf {x}}=\hat{\mathbf {b}} \cdot \hat{\mathbf {x}}, does that automatically mean that \hat{\mathbf {a}}=\hat{\mathbf {b}}?"

Now recall the definition of the dot product (whose result is a scalar):

\hat{\mathbf {a}} \cdot \hat{\mathbf {x}}= a_1 x_1 + a_2 x_2 + a_3 x_3 and

\hat{\mathbf {b}} \cdot \hat{\mathbf {x}}= b_1 x_1 + b_2 x_2 + b_3 x_3.

So if a_1 x_1 + a_2 x_2 + a_3 x_3 = b_1 x_1 + b_2 x_2 + b_3 x_3 for any x_1, x_2, and x_3, does that mean that a_1=b_1, a_2=b_2, and a_3=b_3?

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?

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.