vector q Watch

cooldudeman
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I need help with a). I can't do it because I don't know the relationship between the sizes of OA and OB

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ghostwalker
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(Original post by cooldudeman)
I need help with a). I can't do it because I don't know the relationship between the sizes of OA and OB

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You don't need to know the relationship between their "sizes". It would help to draw a good diagram though.

You're basically dealing with similar triangles.
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brianeverit
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(Original post by cooldudeman)
I need help with a). I can't do it because I don't know the relationship between the sizes of OA and OB

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PQ parallel to OB implies that  \overline{PQ}=\mu \overline{OB} \mathrm{\ for\ some\ } \mu and we know that  \overline{OP}=\frac{2}{3} \overline{OA}
Let  \overline {AQ}=\lambda \overline{AB} \mathrm{\ for\ some\ }\lambda
Now notice that the position vector of Q may be written as \overline{OP}+\overlinje{PQ} \mathrm{\ or\ }\overline{OA}+\overline{AQ}
Comparingt the two forms you should be able to deduce the value of \lambda and then answer the question.
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brianeverit
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(Original post by ghostwalker)
You don't need to know the relationship between their "sizes". It would help to draw a good diagram though.

You're basically dealing with similar triangles.
The question specified the method to be used which was not similar triangles
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ghostwalker
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(Original post by brianeverit)
The question specified the method to be used which was not similar triangles
I didn't say it was. It was a hint to the OP to get the relevant information on a diagram, rather than spelling out all the details :sigh:
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cooldudeman
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(Original post by ghostwalker)
I didn't say it was. It was a hint to the OP to get the relevant information on a diagram, rather than spelling out all the details :sigh:
nope I still can't do it. this is what I tried

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ghostwalker
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(Original post by cooldudeman)
nope I still can't do it. this is what I tried

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You want everything in terms of OA and OB.

So you know OP is 2/3 OA for example.

Rather than use OQ = OP+PA+AQ, use OQ = OP + PQ, because you know PQ is parallel to OB, and hence equal sOB for some unknown s.
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cooldudeman
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(Original post by ghostwalker)
You want everything in terms of OA and OB.

So you know OP is 2/3 OA for example.

Rather than use OQ = OP+PA+AQ, use OQ = OP + PQ, because you know PQ is parallel to OB, and hence equal sOB for some unknown s.
OK so the position vector can be written as 2/3OA + xOB, but what would you equate this to? I was thinking OA + AC but I can't get AC in terms of OA and OB

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ghostwalker
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(Original post by cooldudeman)
OK so the position vector can be written as 2/3OA + xOB, but what would you equate this to? I was thinking OA + AC but I can't get AC in terms of OA and OB

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AC is some, as yet unknown, multiple of AB, and you can get AB in terms of OA and OB.

This introduces a second unknown.

Since the two positon of Q are the same, you can equate these.

You now need to think how you can work out the unknowns - just be looking at, and possibly rearranging, the equation.
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cooldudeman
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(Original post by ghostwalker)
AC is some, as yet unknown, multiple of AB, and you can get AB in terms of OA and OB.

This introduces a second unknown.

Since the two positon of Q are the same, you can equate these.

You now need to think how you can work out the unknowns - just be looking at, and possibly rearranging, the equation.
I meant AQ instead of AC sorry.
anyway everything seem to be cancelling out

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Havaiana
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Hey, I think you can solve it using the intercept theorem.

http://en.wikipedia.org/wiki/Intercept_thm
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ghostwalker
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(Original post by cooldudeman)
I meant AQ instead of AC sorry.
anyway everything seem to be cancelling out

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You seem to have gone round the houses a bit.

Note that AB = OB - OA

So, AQ = p(OB-OA)

Edit: From scratch, I would have done:

OP+ PQ = OB + BQ

So, (2/3) OA + x OB = OB + y (BA)

Then, (2/3) OA + x OB = Ob + y (OB - OA)

etc.
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cooldudeman
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(Original post by ghostwalker)
You seem to have gone round the houses a bit.

Note that AB = OB - OA

So, AQ = p(OB-OA)

Edit: From scratch, I would have done:

OP+ PQ = OB + BQ

So, (2/3) OA + x OB = OB + y (BA)

Then, (2/3) OA + x OB = Ob + y (OB - OA)

etc.
i was just trying to look at your scratch one, and I still cant figure out how you work out the unknowns. please could you tell me, im no good at this.
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ghostwalker
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(Original post by cooldudeman)
i was just trying to look at your scratch one, and I still cant figure out how you work out the unknowns. please could you tell me, im no good at this.
I got that slightly wrong. Should have used BA = OA - OB,

Now, if you get all the OAs on one side and OBs on the other, then you end up with

[(2/3) - y] OA = (1-y-x) OB

Since OA and OB are not parallel, the only way this can be true is if the scalars are zero. So, (2/3) - y = 0 and 1-y-x = 0

So y = 2/3 and x = 1/3.

The x values is of no use to use.

From the y values we have BQ = 2/3 BA

From which we can deduce QA = 1/3 BA, since BQ + QA = BA

And you're virtually done.
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