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i only passed once during school exam and thats lower 6 1st term exam ) , A- for physics ( disappointment, i was aiming for A ) , and A for every paper of maths and f maths. If only i could transfer some marks of maths to physics, it would be 3As already D:Anyway, i find this year's paper harder than last year. Last year's paper was the easiest of all i think. Quote:
Try being really good with some chapters. Make sure u score full marks if questions from those chapters came out. In my case, i always read a few books just for 1 chapter. |
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STPM 2007 Paper 2
Two straight lines l1 and l2 have equations -2x+4=2y-4=z-4 and 2x=y+1=-z+3 respectively. Determine whether l1 and l2 intersect. [7 marks] The points P & Q lie on l1 and l2 respectively, and the point R divides PQ in the ratio 2:3. Find the equation of the locus of R. [6 marks] Can anyone please tell me how to solve the 2nd part. Or rather, how does a locus equation in 3D looks like?
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A few links worth checking out: My Life Story Further Mathematics T Blog (2003-2011 Syllabus) |
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Less Junior Member
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use ratio theorem (3(l1)+2(l2))/5=..... then remove the variable inside the answer i get is z=6-2x but i not sure it is correct or not. havent checked yet. have to wait eurytos or francisfu come n check. it is a straight line. |
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A few links worth checking out: My Life Story Further Mathematics T Blog (2003-2011 Syllabus) |
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Less Junior Member
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for example, if coordinate A divides coordinate B and coordinate C in m:n ratio, then A=(nB+mC)/(m+n). it is same as in vector, the difference is just the word "coordinate" change to "vector" only. you will touch it in math T paper 2 later. |
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| The Following User Says Thank You to uoykcuf For This Useful Post: |
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Hey, i think i got the same answer as you! Thanks, I hope Eurytos will come and check it as well...
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A few links worth checking out: My Life Story Further Mathematics T Blog (2003-2011 Syllabus) |
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Super Junior Member
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Uoykcuf just showed me the workings to the question above. Below are the workings of his solution :
From ratio theorem, x = ... + ... t + ... s y = ... + ... t + ... s z = ... + ... t + ... s From equation 1, t = X - ... - ... s s = Y - ... - ... t Solving the equations above, you get s = Y - ... -----> Notice there no longer are t terms t = X - ... ------> Notice there no longer are s terms substitute both equations in Z= ... + ... t + ... s then you will get the answer. I forgot how to do it until i asked him. Anyway his solution is probably correct.
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If There Was A Race For 2nd Place, I Would Still Finish 2nd. |
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Hey brothers, another question:
Given 3 eigenvectors: v1 = (1 -1 0) v2 = (1 0 -1) v3 = (1 1 1) c) show that v1, v2 and v3 are linearly independent. I don't know what is it talking about... d) Express the vector (a b c) as a linear combination of v1, v2 and v3 with coefficients in terms of the constant a, b and c. The answer given was 1/3(a-2b+c)v1 + 1/3(a+b-2c)v2 + 1/3(a+b+c)v3 I've no idea how to show it...
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A few links worth checking out: My Life Story Further Mathematics T Blog (2003-2011 Syllabus) |
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