IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: The SMA on April 21, 2010, 09:38:15 am
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Hey guys, i need some help on P3 M/J 2003 Q9
and on P3 M/J 2002 Q8.
Both are Vectors questions and i found it hard + confusing to solve.
I hope someone can help me step by step :)
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mj 2003 q9
The angle between the planes is the angle between the normal
normal to x+2y-2z=2 is (1,2,-2) and normal to 2x-3y+6z=3 is (2,-3,6)
For b see http://astarmathsandphysics.com/a_level_maths_notes/C4/a_level_maths_notes_c4_finding_the_line_of_intersection_of_two_planes.html
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for mj 2002 p3 a) first you have to find the line passing through A and B
r=A +t(B-A) =(1,0,1) +t(3,-1,2)
Then follow this method
http://www.astarmathsandphysics.com/a_level_maths_notes/C4/a_level_maths_notes_c4_finding_the_point_of_intersection_of_a_line_with_a_plane.html
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I have done the mj 2003 q9 and the mj 2002 q8 (i) correctly.
But I still can't get the answer for q8 part (ii).
Since the new plane contains line l and it is perpendicular to p
[and by letting normal of the new plane to be (ax+by+cz)],
(3,-1,2).(a,b,c)=0 and (1,3,-2).(a,b,c)=0
I came up with two equations ; 3a-b+2c=0 --(1) and a+3b-2c=0 --(2)
and when I solve them simultaneously,all interms of b, i got (a,b,c) = (-1/2b,b,5/4b)
and then here is where i stucked. I need more help on here.
By the way, Thank you very much astar for your help . What an interesting site you have there!
It is very very helpful for me :)
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I dont understand your method. Home in few hours