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expected grade boundries for As math p12

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No, OD was (12, 9,-2). For CD, you had to consider the vector in the direction of BA, not AB. Hence, OC was (8 4 -8) and you add up OC and CD to get the position vector of OD.
First of all that is axiomatically wrong. I'll agree that you had to consider BA but OC was ( 4 5 6) NOT ( 8 4 -8)
 
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yes Q2 (I)(a) was sin p= -sqrt(1-k^2)
and tan p was -(sqrr(1-k^2))/p
and sin2theta will be negative becoz sin theta will be negative and cos will be positve so sin times cos will always give a negative answer. This question could have been done very easily by those who have studied p3
I just realized that I forgot to put the negative sign. How many marks was q2 for?
 
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yes Q2 (I)(a) was sin p= -sqrt(1-k^2)
and tan p was -(sqrr(1-k^2))/p
and sin2theta will be negative becoz sin theta will be negative and cos will be positve so sin times cos will always give a negative answer. This question could have been done very easily by those who have studied p3
I had a very strong temptation to solve the sin 2theta question that way, but then chose the "P1 style", with full elaboration though. P3 methods reserved for 20th May. :D
 
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that vector question part 2 took like 20mins of my time but got the right answer in the end.
but still managed to finish before time. was good alhamdulillah
 
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