OCR C4 (not mei) 18th June 2013 revision

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Reply 80

DingDong!

Jan 13 paper and MS

4724-01Jan13 (1).pdf206.4KB

4724_MS_Jan13.pdf165.9KB

Reply 81

erniiee

Original post by Genesis2703

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Hey man, could you help me on part i of question 9 on the June 08? http://www.ocr.org.uk/Images/62967-question-paper-4724-core-mathematics-4.pdf

Reply 82

Genesis2703

Original post by erniiee

Hey man, could you help me on part i of question 9 on the June 08? http://www.ocr.org.uk/Images/62967-question-paper-4724-core-mathematics-4.pdf

B is at an intersection with the x-axis, meaning that y = 0, A it at the highest points possible (highest y-value possible). Does that help?

Reply 83

erniiee

Original post by Genesis2703

B is at an intersection with the x-axis, meaning that y = 0, A it at the highest points possible (highest y-value possible). Does that help?

Well I understand that, and I also appreciate what the mark scheme says in terms of the answer, but as we've been given the parametric equations, do we literally just have to assume the graph has the same "limits" in terms of range and domain as the parametric equations?

The way I found the values was assuming the graph followed the same pattern as the sine graph in terms of highest value and value at y=0, but only because that's the trig function that was in the parametric equations..so lets say those were cos instead of sine but everything else the same (except for the graph shifting to the right appropriately) we'd assume theta = 0 and 3pi over 2?

Reply 84

a10

Original post by Genesis2703

B is at an intersection with the x-axis, meaning that y = 0, A it at the highest points possible (highest y-value possible). Does that help?

I saw your post on the FP2 thread, congrats on slaying the beast :wink:

hows ur c4 revision going?

Reply 85

mrmccarl

hi guys...you know when a question which involves expanding a binomial series....usually the question right after is...for what x values is this valid for?...can someone please explain the approach to answering that sort of question. thanks.

Reply 86

stirkee

Original post by mrmccarl

http://examsolutions.net/maths-revision/core-maths/sequences-series/binomial/formula/validity/tutorial-1.php

Reply 87

bellaanne

can someone help me with this vectors question please? (june 2012 core 4 not mei Q10)

lines L1 and L2 have vector equations:

r= -i+2j+7k + t(2i+2j+k) and r=2i+9j-4k + s(i+3j-2k)
respectively. The point A has coordinates (-3,0,6) relative to the origin O

i)) show that A lies on L1 and that OA is perpendicular to L1
ii) show that the line through O and A intersects L2
iii) Given that the point of intersection in part (ii) is B find the ratio (magnitude of OA): (magnitude of BA)

(edited 10 years ago)

Reply 88

joostan

Original post by bellaanne

can someone help me with this vectors question please? (june 2012 core 4 not mei Q10)

lines L1 and L2 have vector equations:

r= -i+2j+7k + t(2i+2j+k) and r=2i+9j-4k + s(i+3j-2k)
respectively. the point A lies on L1 and has coordinates (-3,0,6) relative to the origin O

i)) show that A lies on L1 and that OA is perpendicular to L1
ii) show that the line through O and A intersects L2
iii) Given that the point of intersection in part (ii) is B find the ratio (magnitude of OA): (magnitude of BA)

Which part? :smile:

Reply 89

bellaanne

Original post by joostan

Which part? :smile:

Part ii)
Im unsure what to do. Do we have to find the vector equation of A?

I know it's position vector is the coordinates that are given: (-3,0,6)
but Im unsure what it's direction vector is. I was thinking of using the same direction vector of the line L2 for A but im not 100% sure

Thanks :smile:

Reply 90

joostan

Original post by bellaanne

The direction vector of OA is

\begin{pmatrix} -3 \\ 0 \\ 6 \end{pmatrix}

Does this help? :smile:

Reply 91

bellaanne

Original post by joostan

The direction vector of OA is

\begin{pmatrix} -3 \\ 0 \\ 6 \end{pmatrix}

Does this help? :smile:

what? i thought that was it's position vector :frown:

Reply 92

joostan

Original post by bellaanne

what? i thought that was it's position vector :frown:

Yes, but it is also the direction vector of A from the origin if that makes sense? :smile:

Reply 93

bellaanne

Ohhhh ok, i get you now. I just wanted to quickly ask that if a point (position vector) LIES on a line does that mean that it has the SAME direction vector as the line?

Reply 94

joostan

Original post by bellaanne

Ohhhh ok, i get you now. I just wanted to quickly ask that if a point (position vector) LIES on a line does that mean that it has the SAME direction vector as the line?

Not necessarily, but if the line passes through the origin then yes. :smile:

Reply 95

m00c0w

Could someone explain 4ii on June 2008 paper please? I hate vectors!

Reply 96

a10

Original post by m00c0w

Could someone explain 4ii on June 2008 paper please? I hate vectors!

the position vector P will have co-ordinates (x,y,z)

therefore OP = p - o ------> so P remains unchanged (x,y,z)

if P lies on the line then you can say:

(x,y,z) = (3, 2, 3) + t(-2,1,1)

therefore you know the values of x, y and z

you are then told that they are perpendicular so the scalar prod = 0

using that information you can deduce that -2x+y+z = 0

you already know what x,y and z are so just sub in and solve to find t; then you can work out the co-ordinates of position vector P.

Reply 97

m00c0w

Original post by a10

Surely you would need to work out t to be able to find x, y and z. How would you do that?
Thanks!

Reply 98

a10

Original post by m00c0w

Surely you would need to work out t to be able to find x, y and z. How would you do that?
Thanks!

well you know what x, y and z equal to, so sub those values in and you can work out t then use that value of t to work out the co-ordinates of the point P.

Reply 99

m00c0w

Original post by a10

well you know what x, y and z equal to, so sub those values in and you can work out t then use that value of t to work out the co-ordinates of the point P.