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I've attempted to solve the problem, but I got $\langle \frac{1}{\sqrt{\frac{29}{4}}}, \frac{5}{\sqrt{\frac{29}{4}}}\rangle$, which is incorrect. There is not a similar problem in my textbook that I can reference.

I know that to find a unit vector, we first find the length/magnitude of the given vector, and multiply $$1/\sqrt{magnitude}$$ by the original vector.

$$L = \sqrt{x^2 + y^2}.$$

Can anyone give me any ideas on how to solve this problem?

Find the unit vector that has the same direction as the vector from the point A = (-1,2) to the point B = (3,3).

Thank you in advance.

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Please start formally accepting the answers you receive; see the last paragraph in the section How do I ask questions here? of the faq. – Arturo Magidin Jan 26 '11 at 20:25

2 Answers 2

The vector that goes from $A$ to $B$ is the vector $B-A$: to see this, notice that if you add vectors using the parallelogram rule, then adding the vector $V$ you are looking for to $A$ should give you $B$, so $A+V = B$, giving you $V=B-A$.

So the vector you are looking for is $V = B-A = (3,3) - (-1,2) = (4,1)$.

Now that you know the vector, finding the unit vector in the same direction is done as you indicate: find the magnitude of $V$, divide by the magnitude.

(Looks like you took $A+B$ instead of $B-A$)

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When you took the vector from A to B it looks like you added instead of subtracting. It should be (4,1).

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What does the answer (4,1) tell me? I understand how you calculate it but I don't know what to make of it. – zundarz Aug 9 '12 at 15:10
@zundarz: it means that to go from point A to point B you must move 4 units in the positive $x$ direction and 1 unit in the positive $y$ direction. If you plot the points on graph paper, you can confirm this by eye. – Ross Millikan Aug 9 '12 at 23:13

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