MATHS P3 HELP!

[WellWIshER]

guys can someone tell me how to draw complicated complex numbers in an ARGAND DIAGRAM?

where r we suppose to learn it from????

 

[soumayya]

hmm...complicated like..??

 

[usman]

http://www.khanacademy.org/video/basic- ... recalculus

 

[WellWIshER]

boys n gurls,

how do u solve secx-tanx=0.5

range 0<x<0.5pie

 

[usman]

sec(x) - tan(x) = 0.5
1/cos(x) - sin(x)/cos(x) = 0.5
Multiplying throughout by cos(x) --->
1 - sin(x) = 0.5cos(x)
cos(x) = 2 - 2sin(x)
Squaring both sides --->
cos^2(x) = 4 - 8sin(x) 4sin^2(x)
1 - sin^2(x) = 4 - 8sin(x) 4sin^2(x)
Rearranging --->
5sin^2(x) - 8sin(x) + 3 = 0
[sin(x) - 1][5sin(x) - 3] = 0
(a) sin(x) = 1 ---> x = pi/2 (Extraneous root i.e. one wich doesn't satisfy the original equation bt is obtained due to squaring both sides in the process of solving it).
(b) 5sin(x) - 3 = 0 --> sin(x) = 3/5 ---> x = arcsin(3/5) or sin^-1(3/5) (This root seems to be the answer cos it satisfies the original equation)

 

[WellWIshER]

thanx man!

 

[WellWIshER]

1 more .......

do u have da p3 cie book...if so..then

page 187

q.10...on vectors...


plzz ans

 

[zahraahmed]

for argand diagrams use this:
http://store.aqa.org.uk/qual/pdf/AQA-MFP2-TEXTBOOK.PDF

 

[usman]

1 more .......

do u have da p3 cie book...if so..then

page 187

q.10...on vectors...


plzz ans

U shud try urself 1st den ask others. :x Anyways (wdout justifying my steps),
10.(a)
(i - j - 4k) x (2i + 5j + 6k) = 14i -14j + 7k wich is // to 2i - 2j + k
(i - j - 5k) + s(i - j - 4k) = (2i - 9j -14k) + t(2i + 5k + 6j)
[(1 + s)i + (-1 - s)j + (-5 - 4s)k] - [(2 + 2t)i + (-9 + 5t)j + (-14 + 6t)k] = u(2i -2j + k) ---> {s,t,u} = {1,1,-1}
----> P(2,-2,-9); Q(4,-4,-8) ---> PQ = 3 units

(b) (2i -2j + k) x (2i + 5j + 6k) = -17i -10j + 14k wich is // to 17i + 10j - 14k ----> 17x + 10y - 14z = 140 (plane equation)

(c) d(perp) = Abs[[17(1) + 10(-1) - 14(-5) - 140]/sqrt[17^2 + 10^2 + 14^2]] = 21/sqrt(65)

Now it's ur job to understand what i did :mrgreen:

 

[WellWIshER]

my dear friend i spent an hour on dis be4 asking u

 

[WellWIshER]

i cant figure out a thing!

thanx!!!

 

[MAVtKnmJ]

I thought the topic was about Complex Numbers then it changed into a Vectors thread?

Kindly post the problematic question or ask what you need help with, obviously we are not psychic

 

[WellWIshER]

fellas please tell me how to find da following using argand diagrams:

q.1 arg(z-2)=2/3pi and IzI=2 what is arg z?

q.2 if arg(1/3-z)=1/6pi what is arg(3z-1)?

q3 what is meaning of the term IxI<1 i binomial expansion?

1 vector q:
find the equation of the plane which passes through the point (1,2,3,) and contains the line of intersection of the planes
2x-y+z=4 and x+y+z=4?


thanx guys :mrgreen:

 

[MAVtKnmJ]

I've looked at the complex questions and I don't see how these are related to our syllabus. Moreover, I think you mean |z| which means modulus or distance in this case.

 

[usman]

10.(a)
(i - j - 4k) x (2i + 5j + 6k) = 14i -14j + 7k wich is // to 2i - 2j + k
(i - j - 5k) + s(i - j - 4k) = (2i - 9j -14k) + t(2i + 5k + 6j)
[(1 + s)i + (-1 - s)j + (-5 - 4s)k] - [(2 + 2t)i + (-9 + 5t)j + (-14 + 6t)k] = u(2i -2j + k) ---> {s,t,u} = {1,1,-1}
----> P(2,-2,-9); Q(4,-4,-8) ---> PQ = 3 units

(b) (2i -2j + k) x (2i + 5j + 6k) = -17i -10j + 14k wich is // to 17i + 10j - 14k ----> 17x + 10y - 14z = 140 (plane equation)

(c) d(perp) = Abs[[17(1) + 10(-1) - 14(-5) - 140]/sqrt[17^2 + 10^2 + 14^2]] = 21/sqrt(65)

In part (i) u can cross the two direction vectors to get a vector perp. to both the lines. Then, u can use the fact that PQ shud b parallel to this perp. vector, to get the vector PQ which turns out to be -2i + 2j - k

NB: This step is wrong. jux ignore it @ (i - j - 5k) + s(i - j - 4k) = (2i - 9j -14k) + t(2i + 5k + 6j)

 

[usman]

U may also project the vector AB (i - 8j - 9k) in the direction of the perp. vector (2i - 2j + k) i.e. [(i - 8j - 9k) x (2i - 2j + k)]/[|2i - 2j + k|] ----> = 3 units

 

[WellWIshER]

thanx a million.

do u know where to learn 3d vectors 4rm..

seriously i need some visualization to ans these questions.

the cie endorsed book is crap!!!

 

[MAVtKnmJ]

Have a look at the following link:

http://patrickjmt.com/

Press "control and f" together then type in vector to see the different types of topics.

 

[WellWIshER]

I DID THT long time ago but didnt find wat i was looking 4.

thanks anways

 

[MAVtKnmJ]

Well ask your questions here and I'll try to help.