Mathematics: Post your doubts here!

[Alice123]

Well i got this value too!

Okay here is the way.

P.S, i dint use more than Pure 1 knowledge.!

3tanx(cos^2)= 1

3 sinxcosx = 1

since , sinx = √1-cos^2 [ from sin^2 + cos^2 = 1]

3 √1-cos^2 cosx = 1

√1-cos^2 = 1/3cosx

1 - cos^2 = (1/3cosx)^2

1 - cos^2 = 1/9cosx^2

(9cosx^2)1 - cos^2 = 1

9cos^2 - 9cos^4 = 1

9cos^4 - 9cos^2 + 1 = 0

then using quaradictic formula u get.

cos^2(x) = 0.87267 or cos^2 = 0.12732

cos x = √0.87267 or cos x = √0.12732

x = cos^-1 (√0.87267) or x = cos^-1 (√0.12732)

SALUTE TO YOUR PATIENCE

 

[Soldier313]

Aoa wr wb
Can someone please help me with 7 ii (using the scalar product method) and 7 iii of this paper.....?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_31.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_ms_31.pdf

(PS: Please do provide a detailed explanation, vectors isn't exactly my cup of tea )
JazakAllah khair and thanx a ton!

 

[iKhaled]

ok for question 7(ii)

we got the equation of AB r = i +2j + 2k + t( 2i + 2j -2k) so any point lies on this line will have x= 1 +2t y = 2 + 2t z= 2-2t we can get the position vector of p from here. it says that OP (position vector) is perpendicular to the line AB which means that..

[(1+2t)i + (2+2t)j + (2-2t)k]X [2i + 2j -2k] = O from here u can find the value of t which will be t= -1/6 then substitute it to find x y and z and this will be the position vector of P

did u get it ?

 

[iKhaled]

Aoa wr wb
Can someone please help me with 7 ii (using the scalar product method) and 7 iii of this paper.....?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_31.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_ms_31.pdf

(PS: Please do provide a detailed explanation, vectors isn't exactly my cup of tea )
JazakAllah khair and thanx a ton!

do uk the common perpendicular method to find the normal of a plane from 2 other normals which is perpendicular to it ?

 

[Soldier313]

ok for question 7(ii)

we got the equation of AB r = i +2j + 2k + t( 2i + 2j -2k) so any point lies on this line will have x= 1 +2t y = 2 + 2t z= 2-2t we can get the position vector of p from here. it says that OP (position vector) is perpendicular to the line AB which means that..

[(1+2t)i + (2+2t)j + (2-2t)k]X [2i + 2j -2k] = O from here u can find the value of t which will be t= -1/6 then substitute it to find x y and z and this will be the position vector of P

did u get it ?

Thanx a lot, yes i did get it.....

do uk the common perpendicular method to find the normal of a plane from 2 other normals which is perpendicular to it ?

sorry, er what do you mean?

 

[iKhaled]

Thanx a lot, yes i did get it.....


sorry, er what do you mean?

which math book do u use?

 

[Soldier313]

which math book do u use?

Pure Mathematics 2 and 3 by Hugh Neil and Doughlas Quadling.

 

[PhyZac]

Soldier313

iKhaled means the the "cross product" do you know it?

 

[Soldier313]

Soldier313

iKhaled means the the "cross product" do you know it?

owww......yes i do
but i find it a lil' difficult to apply it
:/

 

[PhyZac]

owww......yes i do
but i find it a lil' difficult to apply it
:/

I will tell the easy which i think your book states it.

See, i did a sketch.

For i as I said cover the first row and do that multipliction then subtract.

For j do same thing but cover second row , and whatever value comes take its negative form (if +ve then -ve if -ve then +ve)

For k cover third row.

 

[iKhaled]

Pure Mathematics 2 and 3 by Hugh Neil and Doughlas Quadling.

open page 182 and check 13.4 the common perpendicular thingy because you need to know it so i can explain the third part..tell me if u had any doubts about that part ok ?

 

[iKhaled]

I will tell the easy which i think your book states it.

See, i did a sketch.

For i as I said cover the first row and do that multipliction then subtract.

For j do same thing but cover second row , and whatever value comes take its negative form (if +ve then -ve if -ve then +ve)

For k cover third row.

i think it would be better for him if he checked it from the book..isn't that better ?

 

[PhyZac]

i think it would be better for him if he checked it from the book..isn't that better ?

Yes ofcourse! I learned it from that book only.

Plus, when you learn it Soldier313, before solving that past paper question, I would recommend you to try some cross product question to make sure that chapter sticks to your head!

 

[Soldier313]

PhyZac iKhaled
JazakAllah khair bros!
Thanx a ton......
i got the concept alhamdulillah, if i do have more doubts will let you guys know......

 

[iKhaled]

PhyZac iKhaled
JazakAllah khair bros!
Thanx a ton......
i got the concept alhamdulillah, if i do have more doubts will let you guys know......

anytime bro

 

[abruzzi]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s06_qp_1.pdf
I need help with question 8 (i & ii).. I'm unable to find the unit vectors OD and OB. Someone please explain that.
Thanks

 

[PANDA-]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_1.pdf
I need help with question 8 (i & ii).. I'm unable to find the unit vectors OD and OB. Someone please explain that.
Thanks

OD = 4i+4j+5k

Now why? Well, start with 4i, this is because the length OA is 14m, and DE is 6, you need to subtract 6 from 14 and then divide by two, as this is how you'll get the length of i... Think about it, because I don't know how else to explain it . 4j because it goes halfway through the 8m, as the opposite side of the base rectangle are equal, so 8/2 = 4... so 4j. 5k is easy because they said the roof is at a height of 5cm, and k is the movement on the y axis, so 5k... 4i+4j+5k.

OB is much easier..
OB = 14i + 8j

14i because it moves 14m along from O to A, and then 8j because it moves 8m from A to B.

Now to find the angle...

a.b=|a||b|cosθ

4(14)+4(8) = √(4^2 + 4^2 + 5^2)√(14^2 + 8^2)cosθ
88 = √57 √260 cosθ
88=121.7cosθ
cosθ = 88/121.7
θ = cos^-1(88/121.7)
θ=43.7

... Yup correct

 

[peter brown]

ii) Probability of length less than 73 m and length greater than 77 m is the same since they are the same distance from mean. A normal distribution curve is symmetric about its mean.
P(X<73) = P(X>77) = 0.15

Let X represent the number of rolls having length greater than 77 m.
n = 8
p = 0.15
q = 0.85

P(X<3) = P(X = 0) + P(X = 1) + P(X = 2)
P(X<3) = (8C0) (0.85)^8 + (8C1) (0.15)(0.85)^7 + (8C2) (0.15)²(0.85)^6
P(X<3) = 0.895

thanks alot Dug. May God bless you. sorry for late reply

 

[syed1995]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_11.pdf

Question No. 11 part i..

Just do the differentiation for me.. and find dy/dx=0 I am getting gradient as -1/3 .. while the MS states the gradient is 1/3 :\

 

[iKhaled]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_11.pdf

Question No. 11 part i..

Just do the differentiation for me.. and find dy/dx=0 I am getting gradient as -1/3 .. while the MS states the gradient is 1/3 :\

dy/dx is 2(6x+2)^-2/3 is that what u got ?