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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
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3 Answers

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
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@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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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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Thank you! I got the correct answer. If anyone else can help me further: There is not an example problem in my textbook that I can reference (I have said this on two other questions so far, but it's the truth. The example problems and solutions manual contain physics / engineering examples rather than abstract examples).

For this one, I did not know how to begin. Normally I would give the work I had done, even if it is incorrect.

For what values of t and so does the equality <4-2t, 2s-4t >= < s+4t, 3+2t>. Find t and s. At these values of t and s, find the resulting vector.

Can anyone point me in the correct direction or give me an idea on how to solve this?

Thank you in advance.

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Since this is an entirely different problem, you should post it as a separate question, not as an answer to your original question. Try deleting the answer (if you cannot do so, let me know and I'll flag it for moderator attention) and post the new question elsewhere. I'll reply there once you do. Don't forget to formally "accept" the answer you found most helpful in this one. –  Arturo Magidin Jan 26 '11 at 20:13
    
I honestly don't know how to formally "accept" a reply. I will figure out how and do so. Thank you. –  Math Student Jan 26 '11 at 23:50
    
you haven't voted up either... Please read the FAQ. In any case: on the left of each reply/question, you will see a number, an up arrow, and a down arrow. You have enough reputation now to vote up any answer you think is a good/informative answer, and any question (other than your own) that you think is a good question. In addition, under the number and arrows in the replies you questions you ask, there should be a check-mark. If you are satisfied that your question has been answered, then click on the check-mark next to the reply that you find most useful/informative. (cont...) –  Arturo Magidin Jan 27 '11 at 2:21
    
Clicking on the check-mark will "accept" that answer, and will mark your question on the site as "answered". (It will also give you some reputation). If you want to delete this "answer", there should be a bunch of links at the bottom left of your text, directly under Thank you in advance, which ought to include a delete one; if it doesn't please let me know in a comment, I'll flag the post for a moderator to delete it. –  Arturo Magidin Jan 27 '11 at 2:24
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