Mathematics: Post your doubts here!

[PhyZac]

lawl then how did u get the answer

Umm, it is just finding points on a line, and I go for two points always, when x is 0 and when y is 0.

 

[syed1995]

dude.. and guys at random... post krne se pehle dekkh liya karo ke kisi aur ne swal ka jawab tou nhi dde dya.. besides i was asking fr sumone else... thnkoo anyways buddy

didn't feel like reading 3 pages

<--- is too lazy

 

[Rutzaba]

didn't feel like reading 3 pages

<--- is too lazy

u need not go bak three pages... just the top of this page actually

 

[Rutzaba]

Umm, it is just finding points on a line, and I go for two points always, when x is 0 and when y is 0.

le mai kiya hare kerri rocket science ay! xD

 

[Sanis]

i always have trouble in trigonometry ... i reached into 3sinxcosx=1 .. i checked the mark scheme x=0.365 radians .. when verifying it turns to be correct but however i don't know how I am supposed to reach this value when sin and cos are multiplied .. PLZ HELP URGENTLY thanks further

 

[PhyZac]

i always have trouble in trigonometry ... i reached into 3sinxcosx=1 .. i checked the mark scheme x=0.365 radians .. when verifying it turns to be correct but however i don't know how I am supposed to reach this value when sin and cos are multiplied .. PLZ HELP URGENTLY thanks further

Can you give the very first step?

 

[Sanis]

Can you give the very first step?

3tanx.(cos^2)x=1

 

[PhyZac]

3tanx.(cos^2)x=1

Well, see, i got x=0.365 as one of the answers, but I am not sure my way is correct, very complicated as well as, not the only answer! Was there a range?

 

[Sanis]

Well, see, i got x=0.365 as one of the answers, but I am not sure my way is correct, very complicated as well as, not the only answer! Was there a range?

1.206 .. anyways nvm abt it im sure there is another solution for this .. thank you

 

[PhyZac]

1.206 .. anyways nvm abt it im sure there is another solution for this .. thank you

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)

 

[Sanis]

Well, see, i got x=0.365 as one of the answers, but I am not sure my way is correct, very complicated as well as, not the only answer! Was there a range?

well i found it 3sinxcosx=1 .. u gotta pretend that sinxcosx = half sin 2 theta .. since since sin 2 theta = 2sincosx .. so we actually get 1.5 sin 2x = 1 .. sin 2x = 2/3 .. then x= 20.90 degrees and it turns to be correct that if u took math a2 u will know this formula

 

[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
:/