IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: Priya on June 14, 2009, 10:23:38 am
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(http://i42.tinypic.com/lcg3l.jpg)
For the second part,
I've worked it out like this:
Cross product with Normal vector of plane 1 & 2
I got (6, -10, -7)
It's right as per the mark scheme...
But then, i'm not sure whether the following steps are good because the point on the line is not the same..
Let the point have an x-cordinate=0
2y-2z=2
-3y+6z=3
Solve simultaneously to get y=7 and z=5
The point of the line is (0,7,5)?
I'm trying this method the first time. Pls help me out, thanks
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From the simultaneous equations i get z=2 and y=3
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I've checked it again... got it now.
Thank you!!
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Put z= t say then x+2y-2t=2 and 2x-3y+6t=3
so x+2y+2t+2 (1)
and 2x-3y=-6t+3 (2)
2*(1)-(2) gives 7y=10t+1 so y=1/7+10/7t
3*(1)+2*(3) 7x=-6t+12 so x=12/7+6/7t
Then r(t)=(12/7,1/7,0)+(6/7,1/7,1)t
on my last post I was on a mobile phone and couldnt see the attachm,ent