How to start this question?

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Kulla_9289

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I know it's a guideline to show my work, but I don't even know how to start this problem, so please make an exception. I do know how to do the questions before but I have no clue how to start the last question. How would you solve without knowing the ratios? A guide to the right path will be highly appreciated.
 

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Kulla_9289 said:
I know it's a guideline to show my work, but I don't even know how to start this problem, so please make an exception. I do know how to the questions before but I have no clue how to start the last question. How would you solve without knowing the ratios? A guide to the right path will be highly appreciated.
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Please show us - how you have solved (i) and (ii)
 
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Now show us your sketch of OXYF, and state the definition of a trapezium. (It varies from place to place.) What has to be true of point Y?

(That's how you start!)
 
A trapezium has two parallel sides.
 

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Kulla_9289 said:
it's a ratio. BC + CD = 1. We can assign BC to be lambda and CD to be mu.
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I don't think that makes sense. Don't you already know that the vector BC is q? Why give it another name? And how can two vectors add to a number?

The condition I was asking about is that PY has to be parallel to OX, just as you have drawn it (since clearly OF and XY can't be parallel, which would be the other option -- this is why a sketch was important).

Now, what have you learned about dealing with parallelism in vectors?
 
Kulla_9289 said:
scalar multiple?
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Yes (and collinear is essentially the same thing, since vectors have no specific location).

So write a vector equation expressing the fact that these vectors are parallel, in terms of p and q (and a scalar).

(Don't just ask questions; show some work, so we can move forward.)
 
Kulla_9289 said:
OX = aFY
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Dr.Peterson said:
So write a vector equation expressing the fact that these vectors are parallel, in terms of p and q (and a scalar).
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Keep going. Express OX in terms of p and q.

And then express FY in terms of p and q and another scalar, using the fact that Y is on BD.

Or if you can't do this, show an example you were given similar to this one, so we can see what you are being taught about this sort of problem.
 
Kulla_9289 said:
OX = (3p/2) + q

FY = 2p + 2q + zBD

BD = 2q

So, (3p/2) + q = a(2p + 2q + z(2q))
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Check the details; I don't think everything is exactly right, but I don't have time to examine it in detail. (I'm preparing for a trip, and will be offline for a while.)

But you will eventually have an equation that says a linear combination of p and q equals zero, which can only be true if the coefficients are both zero. Then you'll have two equations in two unknown scalars.
 
Kulla_9289 said:
This is my actual work. I substituted OX and FY into this: OX = aFY
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I thought you were saying that was the next thing, since I told you what to do next. I assumed you were following my advice, and either changing some detail because you'd found an error, or describing what you did next, without showing any details. In any case, it didn't help us see what help you need now.

Your work does look valid, so you should be able to continue; but it will be easier if you change a couple things.

The first thing that made me question your initial work was that you used BD where I would have used BC = q. What you did isn't wrong, though.

The second thing was where you put a, which makes things a little awkward. I used FY = a OX. But again, you can still get to an answer your way,

What you need to do it to continue as I suggested:
Dr.Peterson said:
But you will eventually have an equation that says a linear combination of p and q equals zero, which can only be true if the coefficients are both zero. Then you'll have two equations in two unknown scalars.
Click to expand...
Write that equation!
 
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