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# P3 may 2005 Mark Scheme? watch

1. Hi there, I'm looking for the answers for p3 may 2005.
If no one has them could someone help me with:
http://img233.echo.cx/img233/603/picture72vt.png

Now there seems to be a typo, let's say that vector equation of line l is is:
r=5i+11k....

This is what I thought:
(2i + j + 5k).(xi + yk + zk)=0
2x + y + 5z = 0

Maybe I should find the cartesian equation?
2. if its edexcel u r lookin for, iv got it..pm me ur email addy n i'll mail it

iv got the jan'05 one!

3. oh , would you like to attempt it
4. (I guess you're talking about part b.)

Since P lies on l, then it has position vector (2t, 5+t, 11+5t) relative to O, and since OP is perpendicular to l:
OP . direction of l = 0
(2t, 5+t, 11+5t) . (2, 1, 5) = 0
4t + (5+t) + 5(11+5t) = 0
t = -2

So:
OP = (-4, 3, 11)
5. there u go, someones done it already

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