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Core 4 Soloman paper g

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Hi , for question 6, i dont understand the correct notation or method to use to prove that two lines are parallel and go through the same point.
https://5c59854d0ccd29d489c9e5e689a8bbadf49aa0f0.googledrive.com/host/0B1ZiqBksUHNYREhxMHhfam1IQm8/for-OCR/Solomon%20G%20QP%20-%20C4%20OCR.pdf

If anyone could explain this it would be great.
Thanks
Reply 1
Original post by SamuelN98
IMG_0257.jpg

Hi , for question 6, i dont understand the correct notation or method to use to prove that two lines are parallel and go through the same point.
https://5c59854d0ccd29d489c9e5e689a8bbadf49aa0f0.googledrive.com/host/0B1ZiqBksUHNYREhxMHhfam1IQm8/for-OCR/Solomon%20G%20QP%20-%20C4%20OCR.pdf

If anyone could explain this it would be great.
Thanks


All you have to do is show that AC\vec{AC} is parallel to AB\vec{AB} by calculating each of those two vectors and showing one is a multiple of the other. i.e: AC=λAB\vec{AC} =\lambda \vec{AB} for which λ\lambda?

Then, since you've shown these two lines are parallel and they both pass through A, then they must be a single straight line.
Reply 2
Original post by Zacken
All you have to do is show that AC\vec{AC} is parallel to AB\vec{AB} by calculating each of those two vectors and showing one is a multiple of the other. i.e: AC=λAB\vec{AC} =\lambda \vec{AB} for which λ\lambda?

Then, since you've shown these two lines are parallel and they both pass through A, then they must be a single straight line.


Thanks!
Its not in the question but how would i get the r=.......... equation of the line would they both have a position vector of OA?
Reply 3
Original post by SamuelN98
Thanks!
Its not in the question but how would i get the r=.......... equation of the line would they both have a position vector of OA?


Well you know that the point OA\vec{OA} is on the line and that the line is parallel to the vector ABAB so you can write it in a number of different ways: r=OA+λAB=OA+μAC=OB+κBC=\mathbf{r} = \vec{OA} + \lambda \vec{AB} = \vec{OA} + \mu \vec{AC} = \vec{OB} + \kappa \vec{BC} = \cdots.
Reply 4
Original post by Zacken
Well you know that the point OA\vec{OA} is on the line and that the line is parallel to the vector ABAB so you can write it in a number of different ways: r=OA+λAB=OA+μAC=OB+κBC=\mathbf{r} = \vec{OA} + \lambda \vec{AB} = \vec{OA} + \mu \vec{AC} = \vec{OB} + \kappa \vec{BC} = \cdots.


Cheers.
Reply 5
Original post by SamuelN98
Cheers.


:smile:

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