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Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: thecandydoll on May 09, 2010, 03:34:32 am

Title: Math P1
Post by: thecandydoll on May 09, 2010, 03:34:32 am: Can anyone help me with the vectors questions IN CIE MATH
QP NOV 2008
Title: Re: Math P1
Post by: Meticulous on May 09, 2010, 06:30:55 am: Which question?
Title: Re: Math P1
Post by: astarmathsandphysics on May 09, 2010, 06:31:47 am: And paper
Title: Re: Math P1
Post by: nid404 on May 09, 2010, 07:12:00 am: Quote from: thecandydoll on May 09, 2010, 03:34:32 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 :)
Title: Re: Math P1
Post by: thecandydoll 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.
Title: Re: Math P1
Post by: nid404 on May 10, 2010, 03:49:08 pm: I have to brush up on maths now. Will do it when I come back
Title: Re: Math P1
Post by: ny 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..
Title: Re: Math P1
Post by: Monica on May 11, 2010, 03:42:01 am: Quote from: thecandydoll 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.

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?
Title: Re: Math P1
Post by: nid404 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
Title: Re: Math P1
Post by: Monica on May 11, 2010, 04:55:06 am: Quote from: ny 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..

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. =]
Title: Re: Math P1
Post by: nid404 on May 11, 2010, 05:33:32 am: Yes, in the beginning it is...i didn't know until a few days back myself
Title: Re: Math P1
Post by: ny on May 11, 2010, 07:15:37 am: Quote from: Shoshou..Mony on May 11, 2010, 04:55:06 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.
Title: Re: Math P1
Post by: Monica on May 11, 2010, 07:17:28 am: Quote from: ny on May 11, 2010, 07:15:37 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.
Title: Re: Math P1
Post by: ny 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.
Title: Re: Math P1
Post by: nid404 on May 11, 2010, 07:30:03 am: information given is not enough to find the values.
Title: Re: Math P1
Post by: ny on May 11, 2010, 07:39:55 am: but the questions are just like that.
most of us didnt answer well on that questions.
Title: Re: Math P1
Post by: nid404 on May 11, 2010, 07:48:53 am: Quote from: ny 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..

Then I would wish to make a change here 1/3i + pj +2/3k you mean 1/3 ( i+pj+ 2/3k)
by which you mean 3 is the modulus of the vector

then to your first part

i) |AB|= $\sqrt({1}^2+{-p}^2+{2/3}^2)$

|AB|=3
so $\sqrt({1}^2+{-p}^2+{2/3}^2)$ =3

You should get p when you solve this

but i should know what im assuming is right :-\
Title: Re: Math P1
Post by: ny on May 11, 2010, 08:19:41 am: okay2.
its a bit different from the one in the marking scheme.

i) (1/3)² + p² + (2/3)² = 1

p²= 4/9
p= -(2/3)
Title: Re: Math P1
Post by: immortal on May 11, 2010, 08:20:02 am: Quote from: ny 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..

I think u missed out something..
anyway da first part cn b done using nid404's method

ii>da ratio of i:k =1:2
   AB~(12-9)i+(-4-2)j+(-5-q)k
   3i-6j+(-5-q)k

Tere4 using 1:2
   -5-q=6
   q=-11
Now its also possible 2 find p
As da ratio is 3:6 =1:2
p=-2/3 (as p is a negative constant)
Title: Re: Math P1
Post by: immortal on May 11, 2010, 08:21:43 am: Quote from: ny on May 11, 2010, 08:19:41 am
okay2.
its a bit different from the one in the marking scheme.

i) (1/3)² + p² + (2/3)² = 1

p²= 4/9
p= -(2/3)
ya ,i got got da same ans using da method,but wasnt sure.so didint wont confuse u>
Title: Re: Math P1
Post by: immortal on May 11, 2010, 08:24:59 am: Bty wen u got ur p1 exam,mine is 2maro.
Title: Re: Math P1
Post by: ny on May 11, 2010, 08:26:56 am: Quote from: immortal on May 11, 2010, 08:20:02 am
I think u missed out something..
anyway da first part cn b done using nid404's method

ii>da ratio of i:k =1:2
   AB~(12-9)i+(-4-2)j+(-5-q)k
   3i-6j+(-5-q)k

Tere4 using 1:2
   -5-q=6
   q=-11
Now its also possible 2 find p
As da ratio is 3:6 =1:2
p=-2/3 (as p is a negative constant)

thanks..could u please tell me from where did u get the ratio and 6??
Title: Re: Math P1
Post by: immortal on May 11, 2010, 08:39:30 am: 1/3i + pj +2/3k
1/3:2/3
1:2(ignoring j)
3i-6j+(-5-q)k , as i=3 k=3*2(ratio)=6
Title: Re: Math P1
Post by: ny on May 11, 2010, 09:09:46 am: okay! thank you soo much,
yeah,my p1 is tmrow.
all the best!