Mathematics: Post your doubts here!

[iKhaled]

7(ii) and (iii), of http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_31.pdf

Detail on 7 (ii) i can make out |z-u|<1 will be circle with radius 1 n centre 2,2 but cant make out the other inequality!!!
Please help!

Esme or any1?

use this rule for the other inequality

 

[iKhaled]

And then shade the area inside your circle and below the perpendicular bisector. i.e. the area closer to the point 1

y do we shade the region below the prepen and not above it?

 

[Esme]

y do we shade the region below the prepen and not above it?

The inequality is |z-1|=<|z-i|
So I always shade the area which is closer to the " =< " which in this case is |z-1|

Sorry this isn't exactly the proper explanation

 

[Anika Raisa]

Thank u iKhaled and Esme i got it now!
May Allah Bless u both!

 

[iKhaled]

Thank u iKhaled and Esme i got it now!
May Allah Bless u both!

anytime

 

[surinr]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w07_qp_3.pdf

Please help with entire Q10
thank you

 

[iKhaled]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w07_qp_3.pdf

Please help with entire Q10
thank you

alright, fortunately i only know the answer to the first part and third part but i am sorry i can't find the correct solution of the second part..

(i) u substitute the equation of line l in the equation of plane p since they both intersect so

(i+6j-3k+s(i-2j+2k)-3i)(2i-3j+6k) = 0
((-2+s)i+(6-2s)j+(-3+2s)k)(2i-3j+6k) = o

do the scalar product of these 2 and u will get s = 2
now x = 1+2 y = 6-2(2) z=-3+2(2)
so the position vector of A is 3i+2j+k

part iii do uk this common prependicular normal thingy..apply it to these 2 i-2j+2k and 2i-3j+6k to find the direction vector of the line which lies in P
the direction vector u will get is -6i-2j+k and since it passes through A it will have a position vector of 3i+2j+k

now the equation of the line will be 3i+2j+k+t(-6i-2j+k)

now i am off to bed peace!!

 

[mohamad malak]

hi i hope anyone would help me in solving number 5ii) pleasee http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s07_qp_3.pdf thanks

 

[Adorkableme]

someone please help me solve Qs 6i from http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_32.pdf ......... I get upto the part of finding dy/dx but I am getting stuck somewhere in the next part not sure!!

and Qs 5i from ... http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_31.pdf

 

[Alice123]

Anika Raisa and rosogolla993
Sorry for the late reply.. Net was too slow last night


Let Q be the closest point on the line to P... Then PQ is perpendicular to the direction of the line.....
as Q lies on the line, the position vector of Q
OQ=(1 3 -4) - t(2 1 3)= (1+2t 3+t -4+3t)

Now PQ=OQ-OP
=(1+2t 3+t -4+3t)- (-1 4 11)=(2+2t -1+t -15+3t)
Since PQ is perpendicular to the line
PQ* direction of l=0

(2+2t -1+t -15+3t)(2 1 3)=0
u get the value of t ie 3
sub value of t in PQ
(2+2t -1+t -15+3t)= (8 2 -6)
lPQl=sqroot(8^2+2^2+6^2)= 10.2

 

[surinr]

alright, fortunately i only know the answer to the first part and third part but i am sorry i can't find the correct solution of the second part..

(i) u substitute the equation of line l in the equation of plane p since they both intersect so

(i+6j-3k+s(i-2j+2k)-3i)(2i-3j+6k) = 0
((-2+s)i+(6-2s)j+(-3+2s)k)(2i-3j+6k) = o

do the scalar product of these 2 and u will get s = 2
now x = 1+2 y = 6-2(2) z=-3+2(2)
so the position vector of A is 3i+2j+k

part iii do uk this common prependicular normal thingy..apply it to these 2 i-2j+2k and 2i-3j+6k to find the direction vector of the line which lies in P
the direction vector u will get is -6i-2j+k and since it passes through A it will have a position vector of 3i+2j+k

now the equation of the line will be 3i+2j+k+t(-6i-2j+k)

now i am off to bed peace!!

how did you get s=2 exactly?

 

[Jiyad Ahsan]

Lamda will be denoted as t
You know that P is a point on L , so it has x,y,z coordinates as follows : (1 + 2t), (1 + t), (-1 + 2t) , and m = x + 2y -2z = 1 and n = 2x - 2y + z = 7
Using the formula they provided, you find the distance of P to m and of P to n keeping in mind that they should be equal, so distance of P to m
| (1 + 2t) + 2 (1 +t) -2(-1 + 2t) - 1 | / square root ( 1^2 + 2^2 + (-2)^2 ) to get |4|/3
distance of P to n using the same equation will get you |4t - 8|/3
These two answers are equal, their numerators are equal so you square both sides to get rid of the modulus, leaving you with an equation such as :
4^2 = (4t - 8)^2
16 = 16t^2 - 64t - 64, solve for t and obtain t = 3 or t =1
substitute those two value in the original coordinates of P to get the two position vectors that are (7i + 4j + 5k) and (3i +2j + k) and then find the distance between these two points using your skill from P1! Good luck!

thnx ! there was so much going on i sort of phased out

 

[Jiyad Ahsan]

Chem champ having prob with maths???
Here u go.... This was done by Physac
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
ask if u dont get

yeah.. i mean did you look at the stupid question.. there was so much going on ! i looked at it and phased out.. literally i've never seen such a .. dash.. question
awesome explanation btw .. Thnx !

 

[Mubariz Ahmad]

Can anyone share a document which contains all the important/key notes for Math P3. Thanks

 

[Jiyad Ahsan]

Can anyone share a document which contains all the important/key notes for Math P3. Thanks

umm i have some on complex numbers and vectors that some people shared here

 

[WayneRooney10]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_qp_31.pdf
Why is the answer to Q1 only one value? Why isn't it -1/2<x<5/4? And do question Q2 as well please! Thanks.

 

[layapaxx]

i need help in this vector questionss

 

[Mubariz Ahmad]

Any notes on differentiation and integration?

 

[Jiyad Ahsan]

Any notes on differentiation and integration?

http://www.s-cool.co.uk/a-level/maths/integration/revise-it/differential-equations
i dnt have notes but this link ought to help you out..

 

[LindaKim123]

I am not the ryt guy to do that..
I just visualise situations of draw them to understand what is going on in vectors question and then act accordingly..

haha it's okay then!! Don't worry!

btw, can you help me solve this one :
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w04_qp_3.pdf

question 9 ii) please!!