How to start this question?

I've been trying to get there but not sure how. I used XY = OY - OX.

OY = OF + FE + ED + DY
OY = 2p + 3q + (-2/3)*2q
OY = 2p + 5q/3

OX = (3p/2) + q

XY = 2p +5q/3 - (3p/2) - q
XY = (2q/3) + 1p/2
 
Kulla_9289 said:
I've been trying to get there but not sure how. I used XY = OY - OX.

OY = OF + FE + ED + DY
OY = 2p + 3q + (-2/3)*2q
OY = 2p + 5q/3

OX = (3p/2) + q

XY = 2p +5q/3 - (3p/2) - q
XY = (2q/3) + 1p/2
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That's almost what I got. I might be wrong, but you can either convince me, or discover an error in your work, by showing more details. Write out what each of OF, FE, ED, and DY is, and take small steps.

I used XY = OF + FY - OX, each of which had already been expressed in terms of p and q.
 
Kulla_9289 said:
This is my work. XY = OY - OX.
Click to expand...
However,

What is your answer?

OriginalProblem asks you to express XY in terms of p and q (vector ) as simply as possible (and no-more). You have shown it - but did not "flag" it as your answer.
 
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Kulla_9289 said:
I did get that.
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Yes, but then you said,
Kulla_9289 said:
If I change -2/3 to -1/3, then I get p/2 + 4q/3
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and I asked you (a) to show why you would do that -- in other words, your work for getting that value of a (you don't just change an answer without a reason), and then (b) to show the new work to get that value for XY (because that work could be mistaken). You need to convince yourself, as well as us, that you have the right answer.
Kulla_9289 said:
This is my work. XY = OY - OX.
Click to expand...
No, that is not work; that is just a first step, showing the general idea of what you did. It doesn't show anyone that you have made no mistakes.

Incidentally, did you try sketching each answer on the picture, to see which, if either, makes sense?
 
Kulla_9289 said:
1/3 makes sense by sketching but I am not sure mathematically. Could you explain to me why -2/3 can't be used?
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Apparently, because it's wrong.

Can you please show all your work from the start, so we can be sure which -2/3 you are talking about, and exactly where it came from? It's too much work to try and go back through all of this (across two pages) and be sure which things are right and which are wrong.
 
Kulla_9289 said:
Oh yes! I got it now. Could you summarize the logic used here and how you knew when to use a scalar?
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Because that's how parallel or collinear vectors are defined. (One scalar was needed to say that Y is on BD, and another to say that FY is parallel to OX.)

This is the usual logic for solving a problem: Start with whatever definitions or theorems apply.

(Now, technically, parallel vectors are defined as having a zero angle between them; so another approach could be to use the dot or cross product, if that fit the nature of the problem; it is a theorem that each will be a scalar multiple of the other. Or you could say these are two equivalent definitions.)

In any case, I knew because I've solved similar problems this way. Have you never seen an example like this?
 
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