Author Topic: Math P1  (Read 2478 times)

Offline ny

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Re: Math P1
« Reply #15 on: May 11, 2010, 07:39:55 am »
but the questions are just like that.
most of us didnt answer well on that questions.

nid404

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Re: Math P1
« Reply #16 on: May 11, 2010, 07:48:53 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  :-\
               

Offline ny

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Re: Math P1
« Reply #17 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)

Offline immortal

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Re: Math P1
« Reply #18 on: May 11, 2010, 08:20:02 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)
Life is short...so live it to da fullest :)

Offline immortal

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Re: Math P1
« Reply #19 on: May 11, 2010, 08:21:43 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>
Life is short...so live it to da fullest :)

Offline immortal

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Re: Math P1
« Reply #20 on: May 11, 2010, 08:24:59 am »
Bty wen u got ur p1 exam,mine is 2maro.
Life is short...so live it to da fullest :)

Offline ny

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Re: Math P1
« Reply #21 on: May 11, 2010, 08:26:56 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??

Offline immortal

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Re: Math P1
« Reply #22 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
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Offline ny

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Re: Math P1
« Reply #23 on: May 11, 2010, 09:09:46 am »
okay! thank you soo much,
yeah,my p1 is tmrow.
all the best!