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Mathematics: Post your doubts here!

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Yes part a.....thanks! Jazaki Allah khairan.
ok so here goes :D
(r-3i) . (2i - 3j + 6k)=0
and r=((1+s) , (6-2s) , (-3+2s))
now you have to subtract r from 3i + oj + 0k
and then you get (-2+s , 6-2s , -3+2s)
now the dot product with (2i - 3j + 6k)
sooo (-2+s , 6-2s , -3+2s). (2 , - 3 , 6)=o
you get 2(-2+s) + -3(6-2s) + 6(-3+2s) = 0
-4 + 2s + 6s -18 -18 +12s=0
-40=-20s
s=2
replace s=2 in r=((1+s) , (6-2s) , (-3+2s))
and you get 3i + 2j + k :D
hope you got it (y)
 
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ok so here goes :D
(r-3i) . (2i - 3j + 6k)=0
and r=((1+s) , (6-2s) , (-3+2s))
now you have to subtract r from 3i + oj + 0k
and then you get (-2+s , 6-2s , -3+2s)
now the dot product with (2i - 3j + 6k)
sooo (-2+s , 6-2s , -3+2s). (2 , - 3 , 6)=o
you get 2(-2+s) + -3(6-2s) + 6(-3+2s) = 0
-4 + 2s + 6s -18 -18 +12s=0
-40=-20s
s=2
replace s=2 in r=((1+s) , (6-2s) , (-3+2s))
and you get 3i + 2j + k :D
hope you got it (y)
Thank You sooooooo much !!!!!!! And yes i get it very well Alhamdulilah...!! Jazaki Allah khairan...May Allah S.W.T reward you and your family with highest grades in the world and hereafter. May Allah S.W.T bless you, Aameen.!!! Thank ALOT
 
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I solved this before, very recently.!
We have to find the point p where the perpendicular distance to the two planes is same.


first we will for an equation the way the paper gave.
therefore
for plane m
|x+2y-2z-1|/sqrt((1^2) + (2^2) + (-2^2) = |x+2y-2z-1|/3 [P.S, sqrt is square root]
for plane n
|2x-2y+z-7|/sqrt((2^2)+(-2^2)+(1^2) = |2x-2y+z-7|/3


since you are finding a point where both distance is same , therefore.
|x+2y-2z-1|/3 = |2x-2y+z-7|/3 (3 can be cancelled both sides)
=
|x+2y-2z-1| = |2x-2y+z-7|

now we will sub the line values in the equation.
the x component (1+2t) [P.S, t is lamda ]
the y comp. (1+t)
the z comp. (-1+2t)

therefore.
| (1+2t) +2 ( 1+t) -2(-1+2t) -1 | = |2(1+2t) - 2(1+t) + (-1+2t) - 7 |
simplifying it you get
|4| = |-8+4t|
now to solve modulus, we do squaring method
16 = 64-64t+16t^2
simplify
2 = 8 - 8t +2t^2
2t^2 - 8t + 6 = 0
solve it and get
t = 3 or t= 1
when t=3 the position of point is [P.S, u do this by sub t value in line equation)
(7, 4, 5) => OA
when t = 1 the position of point is
(3, 2, 1) => OB

BA (or AB, same thing) = OA - OB
= (4 , 2, 4)

now find the mod
sqrt(4^2 + 2^2 + 4^2)
= 6

I dint do every step in detail, (very long) if u dont get any step, just ask. [Pray for me please.]
 
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Some question are already solved in thread.
Try searching before posting.
It works for me sometime!

Here is how I do it.

For example my doubt is in this link
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_32.pdf

You copy the last bit in this case "9709_s12_qp_32"

Now paste this in google and add the word doubt.

so you get

9707_s12_qp_32 doubt.

You might find many pages from xtremepapers, open all of them in new tabs (right click, newtab) and then look in each if it has same question, if yes, then follow up to see if anyone replied.
 
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I solved this before, very recently.!
We have to find the point p where the perpendicular distance to the two planes is same.
first we will for an equation the way the paper gave.
therefore
for plane m
|x+2y-2z-1|/sqrt((1^2) + (2^2) + (-2^2) = |x+2y-2z-1|/3 [P.S, sqrt is square root]
for plane n
|2x-2y+z-7|/sqrt((2^2)+(-2^2)+(1^2) = |2x-2y+z-7|/3
since you are finding a point where both distance is same , therefore.
|x+2y-2z-1|/3 = |2x-2y+z-7|/3 (3 can be cancelled both sides)
=
|x+2y-2z-1| = |2x-2y+z-7|
now we will sub the line values in the equation.
the x component (1+2t) [P.S, t is lamda ]
the y comp. (1+t)
the z comp. (-1+2t)
therefore.
| (1+2t) +2 ( 1+t) -2(-1+2t) -1 | = |2(1+2t) - 2(1+t) + (-1+2t) - 7 |
simplifying it you get
|4| = |-8+4t|
now to solve modulus, we do squaring method
16 = 64-64t+16t^2
simplify
2 = 8 - 8t +2t^2
2t^2 - 8t + 6 = 0
solve it and get
t = 3 or t= 1
when t=3 the position of point is [P.S, u do this by sub t value in line equation)
(7, 4, 5) => OA
when t = 1 the position of point is
(3, 2, 1) => OB
BA (or AB, same thing) = OA - OB
= (4 , 2, 4)
now find the mod
sqrt(4^2 + 2^2 + 4^2)
= 6
I dint do every step in detail, (very long) if u dont get any step, just ask. [Pray for me please.]
Dude you're a life saver! Thanks!
 
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Thank You sooooooo much !!!!!!! And yes i get it very well Alhamdulilah...!! Jazaki Allah khairan...May Allah S.W.T reward you and your family with highest grades in the world and hereafter. May Allah S.W.T bless you, Aameen.!!! Thank ALOT
Jazak Allah to you too :D
for helping me :D
InshaAllah i will remember you in my prayers :D
 
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okay first, move value to make two integration
(let ! be the integration sign)
dr/dt = 0.08r^2
1/r^2 dr = 0.08 dt
! 1/r^2 dr = !0.08 dt
! r^-2 dr = 0.08t +c
-1/r = 0.08t + c
sub values given in question
-1/5 = 0.08(0) + c
c = -1/5

-1/r = 0.08t - 1/5
r = -1 / (0.08 - 1/5)
r = -5 /0.4t - 1


Now for part iii
i am not sure
but what i think is the answer must be positive.
to do this the value shudnt be smaller than 0, (try calculating)
and even not more than 2.5
i got 2.5 by
0.4t-1 =0
0.4t =1
t= 2.5
 
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Okay i have the answering, but i dont know Q9 (i) if u can explain plzz.

so Q9 (iii)

If u got the a b values, 3 and 2

u sub them in the components.

3-λ = 4 + aµ
-2+2λ= 4 + bµ
1+λ = 2 - µ

so u get
3-λ = 4 + 3µ
-2+2λ= 4 + 2µ
1+λ = 2 - µ

choose any two and solve simultaneously

and u get µ = -1

no need to find λ or vice versa

take the value u got, i my case µ = -1
into its line equation and you willget
the i j k values.

about Q10 sorry i just saw it, will look into it, EDIT: i looked into it, i need to solve full question in order to answer tomorrow In Sha Allah
 
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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s09_qp_1.pdf

Q)4) help appreciated.


Anyone have trigonometry notes? Clearly my worst topic from P1.
You could try the trigonometry notes linked on the first page of this thread.
Here's what my guide says about the ques:

a: indicates the amplitude of the curve which is the max or min range of the curve from the mean position.
so, a=9-3=6

b: indicates the periodicity of the curve. we see that in the given figure there are two main cycles in a normal period of sine.
therefore, b=2

c: indicates the vertical shift in the graph from the horizontal axis, which in this case is 3 units.
thence, c=3 units
 
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