Author Topic: Math P1  (Read 2339 times)

Offline thecandydoll

  • SF Citizen
  • ***
  • Posts: 146
  • Reputation: 780
  • Gender: Female
Math P1
« on: May 09, 2010, 03:34:32 am »
Can anyone help me with the vectors questions IN CIE MATH
QP NOV 2008

Offline Meticulous

  • SF Master
  • ******
  • Posts: 2486
  • Reputation: 19401
Re: Math P1
« Reply #1 on: May 09, 2010, 06:30:55 am »
Which question?

Offline astarmathsandphysics

  • SF Overlord
  • *********
  • Posts: 11271
  • Reputation: 65534
  • Gender: Male
  • Free the exam papers!
Re: Math P1
« Reply #2 on: May 09, 2010, 06:31:47 am »
And paper

nid404

  • Guest
Re: Math P1
« Reply #3 on: May 09, 2010, 07:12:00 am »
Can anyone help me with the vectors questions IN CIE MATH
QP NOV 2008

check all the attachments...

PA=PE+EO+OA
   =-8j-6k-6i

PA= -6i-8j-6k

------------------------

PN=PE+EO+OD+DN

PN=-8j-6k+6i+10j
PN=6i+2j-6k

-------------------------

Calculating the angle

The dot product of the vector is given by
a.b=|a|.|b| cos \theta

PA.PN=|PA|.|PN| cos \theta

cos \theta= PA.PN/ |PA|.|PN|

PA= -6i-8j-6k

PN=6i+2j-6k

|PA|= \sqrt [(-6)^2+(-8)^2+(-6)^2]
      =\sqrt 136

|PN|= \sqrt [(6)^2+(2)^2+(-6)^2]
|PN|=\sqrt76
cos\theta= [(-6i-8j-6k) X (6i+2j-6k)] /  \sqrt136.\sqrt76

cos\theta= (-36)+(-16)+(36)/ \sqrt136.\sqrt76
                            =-16/ \sqrt136.\sqrt76
  cos\theta=  -0.157377895
   \theta   = 99.05473251°             

Hope this helps :)   

                   
                             




Offline thecandydoll

  • SF Citizen
  • ***
  • Posts: 146
  • Reputation: 780
  • Gender: Female
Re: Math P1
« Reply #4 on: May 10, 2010, 03:38:58 pm »
I really dont know to how to get the values?!!?!
i understand solving 2d vectors
But when the image is given i just cant.
how should i proceed!

cie as p1,nov 2007 :)
please,explain
Thank you.

nid404

  • Guest
Re: Math P1
« Reply #5 on: May 10, 2010, 03:49:08 pm »
I have to brush up on maths now. Will do it when I come back

Offline ny

  • Newbie
  • *
  • Posts: 40
  • Reputation: 9
Re: Math P1
« Reply #6 on: May 11, 2010, 02:53:34 am »
hello.i have this vector question for P1.
actually it was my mock exam question.

the question is:
the points A and B are such that the unit vector in the direction AB(from A to B) is 1/3i + pj +2/3k , where p is a negative constant.

i) find the value of p

then, the position vectors of A and B, relative to an origin O, are 9i +2j +qk and 12i -4j -5k respectively.

ii) find the value of q

thats all.thank you soo much.please reply asap..

Monica

  • Guest
Re: Math P1
« Reply #7 on: May 11, 2010, 03:42:01 am »
I really dont know to how to get the values?!!?!
i understand solving 2d vectors
But when the image is given i just cant.
how should i proceed!

cie as p1,nov 2007 :)
please,explain
Thank you.


10)i) PR= 2i + 2j +2k
       PQ= -2i+2j+4k

ii) QPR is PQ and PR. When they say scalar you have to do this a.b(a multiplied by b) so PQ.PR = 2(-2) + 2(2) + 2(4) = -4 + 4 +8 =8

Now to get an angle you should know this formula : Cos (theta) = a.b/ |a|x|b|

we got a.b, so now |a| and |b|

|PR| = square root of (2^2 + 2^2 +2^2) = square root of 12
|PQ|= square root of (-2^2 + 2^2 + 4^2) = square root of 24

cos QPR = 8 / square root of 12 x square root of 24

iii) to get the perimeter you will need to get the position vector of PR, PQ and QR, make them under square root each and then add. Now you already got for PR and PQ in part i) so only QR is missing which is 4i +0j - 2k.

square root each of them and add them to get the perimeter.

Nothing is there, all you need to do is look at the image and match lines. Just look carefully with full concentration.

@Nid, how do you write the square roots, brackets, theta, etc?

nid404

  • Guest
Re: Math P1
« Reply #8 on: May 11, 2010, 04:33:12 am »
Thanks Mony.

You use latex symbols...you see that symbol of pi...you type text within that...http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Commands

Monica

  • Guest
Re: Math P1
« Reply #9 on: May 11, 2010, 04:55:06 am »
hello.i have this vector question for P1.
actually it was my mock exam question.

the question is:
the points A and B are such that the unit vector in the direction AB(from A to B) is 1/3i + pj +2/3k , where p is a negative constant.

i) find the value of p

then, the position vectors of A and B, relative to an origin O, are 9i +2j +qk and 12i -4j -5k respectively.

ii) find the value of q

thats all.thank you soo much.please reply asap..

I solved all the vectors questions so can you just post the paper and I'll post my answer?

@Nid..ahh thank you! Seems confusing but I'll just bookmark this page to use. =]

nid404

  • Guest
Re: Math P1
« Reply #10 on: May 11, 2010, 05:33:32 am »
Yes, in the beginning it is...i didn't know until a few days back myself

Offline ny

  • Newbie
  • *
  • Posts: 40
  • Reputation: 9
Re: Math P1
« Reply #11 on: May 11, 2010, 07:15:37 am »
I solved all the vectors questions so can you just post the paper and I'll post my answer?

post u the paper?but i duh have the soft copy of the paper.


Monica

  • Guest
Re: Math P1
« Reply #12 on: May 11, 2010, 07:17:28 am »
post u the paper?but i duh have the soft copy of the paper.



hahahahaha..I like the "duh" part like I am a stupid person.  :P

I meant say the year? Is it 2005? 2006? etc etc.


Offline ny

  • Newbie
  • *
  • Posts: 40
  • Reputation: 9
Re: Math P1
« Reply #13 on: May 11, 2010, 07:21:05 am »
haha.it was typo actually.. its 'don't'

its not from past year papers. it was for my mocks.

nid404

  • Guest
Re: Math P1
« Reply #14 on: May 11, 2010, 07:30:03 am »
information given is not enough to find the values.