We love math

jahid89

Junior Member
next problem...

If ~x, ~y are non-zero vectors, show that |~x|~y + |~y|~x bisects the angle between ~x and ~y.
 

arzafar

Junior Member
next problem...

If ~x, ~y are non-zero vectors, show that |~x|~y + |~y|~x bisects the angle between ~x and ~y.

let X = ai + bj + ck
Y = pi + qj + rk
Z = |X|Y + |Y|X = (a|Y| + p|X|)i + (b|Y| + q|X|)j + (c|Y| + r|X|)k
A be the angle between X and Z
B be the angle between Y and Z
C be the angle between X and Z

|X| = (a^2 + b^2 + c^2)^(1/2)
|Y| = (p^2 + q^2 + r^2)^(1/2)
|Z| = [(a|Y| + p|X|)^2 + (b|Y| + q|X|)^2 + (c|Y| + r|X|)^2]^(1/2)

X.Z = |X||Z|cosA
cosA = [a(a|Y| + p|X|) + b(b|Y| + q|X|) + c(c|Y| + r|X|)]/|X||Z|
= [|X|(ap + bq + cr) + |Y|(a^2 + b^2 + c^2)]/|X||Z|
= [|X|(X.Y) + |Y||X||X|]/|X||Z|
= (X.Y + |X||Y|)/|Z|

Y.Z = |Y||Z|cosB
cosB = [p(a|Y| + p|X|) + q(b|Y| + q|X|) + r(c|Y| + r|X|)]/|Y||Z|
= [|Y|(ap + bq + cr) + |X|(p^2 + q^2 + r^2)]/|Y||Z|
= [|Y|(X.Y) + |X||Y||Y|]/|Y||Z|
= (X.Y + |X||Y|)/|Z|

thus cosA = cosB
=> A = B
also A + B = C
so A = B = C/2
hence proved
 

jahid89

Junior Member
let X = ai + bj + ck
Y = pi + qj + rk
Z = |X|Y + |Y|X = (a|Y| + p|X|)i + (b|Y| + q|X|)j + (c|Y| + r|X|)k
A be the angle between X and Z
B be the angle between Y and Z

|X| = (a^2 + b^2 + c^2)^(1/2)
|Y| = (p^2 + q^2 + r^2)^(1/2)
|Z| = [(a|Y| + p|X|)^2 + (b|Y| + q|X|)^2 + (c|Y| + r|X|)^2]^(1/2)

X.Z = |X||Z|cosA
cosA = [a(a|Y| + p|X|) + b(b|Y| + q|X|) + c(c|Y| + r|X|)]/|X||Z|
= [|X|(ap + bq + cr) + |Y|(a^2 + b^2 + c^2)]/|X||Z|
= [|X|(X.Y) + |Y||X||X|]/|X||Z|
= (|X||Y|cosB + |X||Y|)/|Z|

Y.Z = |Y||Z|cosB

i think ive missed something. Ill check later

to be continued...

waoo..good try..:)
:laughing-dancing:
 

arzafar

Junior Member
ok here is a nice problem

It takes x hours to fill a tank. After making a hole in the bottom of the tank, it now takes y hrs to fill the tank. how many hrs will this hole take to empty the tank?
 

jahid89

Junior Member
next problem...

If ~x, ~y are non-zero vectors, show that |~x|~y + |~y|~x bisects the angle between ~x and ~y.

consider the orthogonal projection of ~x on ~y and the orthogonal projection of ~y on ~x.
Let X be the angle between ~x and ~y so that ~x.~y =|~x||~y|cosX

Let ~u = |~x|~y + |~y|~x and Z be the angle between ~x and ~u

~x.~u = |~x||~u|cosZ where the magnitude of ~u is found to be

~x.~u = √2.|~x|^2.|~y|[√(1 + cosX)]cosZ

Also ~x.~u + ~x.[|~x|~y + |~y|~x] = |~x|^2.|~y|(1 + cosX)

This gives
√2.|~x|^2.|~y|[√(1 + cosX)]cosZ = |~x|^2.|~y|(1 + cosX)

or √2.cosZ =√(1 + cosX)

Squaring 2cos^2(Z) = 1 + cosX and from the double angle formula

2cos^2(X) = 1 + cos2X we conclude Z = X/2.

You can then perform a similar argument by considering the inner product of ~y and ~u and showing the angle is again X/2.
 

hayat84

I'm not what you believe
sorry,but let me smile a little...I understand arabic language more than maths..how can you communicate with this language?have fun!
 

jahid89

Junior Member
let X = ai + bj + ck
Y = pi + qj + rk
Z = |X|Y + |Y|X = (a|Y| + p|X|)i + (b|Y| + q|X|)j + (c|Y| + r|X|)k
A be the angle between X and Z
B be the angle between Y and Z
C be the angle between X and Z

|X| = (a^2 + b^2 + c^2)^(1/2)
|Y| = (p^2 + q^2 + r^2)^(1/2)
|Z| = [(a|Y| + p|X|)^2 + (b|Y| + q|X|)^2 + (c|Y| + r|X|)^2]^(1/2)

X.Z = |X||Z|cosA
cosA = [a(a|Y| + p|X|) + b(b|Y| + q|X|) + c(c|Y| + r|X|)]/|X||Z|
= [|X|(ap + bq + cr) + |Y|(a^2 + b^2 + c^2)]/|X||Z|
= [|X|(X.Y) + |Y||X||X|]/|X||Z|
= (X.Y + |X||Y|)/|Z|

Y.Z = |Y||Z|cosB
cosB = [p(a|Y| + p|X|) + q(b|Y| + q|X|) + r(c|Y| + r|X|)]/|Y||Z|
= [|Y|(ap + bq + cr) + |X|(p^2 + q^2 + r^2)]/|Y||Z|
= [|Y|(X.Y) + |X||Y||Y|]/|Y||Z|
= (X.Y + |X||Y|)/|Z|

thus cosA = cosB
=> A = B
also A + B = C
so A = B = C/2
hence proved

nice method....

:jumpclap::wavyarms:
 

arzafar

Junior Member
sorry,but let me smile a little...I understand arabic language more than maths..how can you communicate with this language?have fun!

math is really easy. you just have to study it for 12-16 grades.

fun part comes from the stimulation of mind.
 

arzafar

Junior Member
Let Vf be the constant speed of pouring liquid into the tank and
Ve be the constant leakage speed through the hole.
Because the tank does eventually fill up despite the leakage, we conclude that Vf > Ve

Let C be the capacity of the tank, i.e., its volume.

X = the time it takes to fill the tank without a hole = C / Vf
Y = the time is takes to fill the tank with the hole = C / (Vf - Ve)

Therefore,
Ve = Vf - C/Y
Thus, the time it takes to empty the tank after it's full
= C / Ve = C / (Vf - C/Y)
= C / (C/X - C/Y)
= 1 / (1/X - 1/Y)
= 1 / [(Y -X) / XY)]
= XY / (Y-X)

To illustrate,
If the tank capacity is C=100 m3 and the liquid is poured in at Vf=10 m3/hr then it normally takes X = 10 hours to fill.
If the leak is Ve = 1 m3/hr and it now takes Y=50 hours to fill then
The time it will take to empty the tank is 10*50/(50-10)=500/40=12.5 hours.

Did I get that right?

yeah that's right!

Alternatively,

rate1 = 1/x per hour
rate2 = 1/y per hour
rate of loss = rate1 - rate2 = (y-x)/xy per hour
so no of hours to empty the tank = 1/(y-x)/xy = xy(y-x)
 

Muslimah16

ServantOfAllah*
As-salaamu'alaykum

Oh my Allah! Just about when i though i kicked complicated maths out of my life :p i come accorss this thread. Lol.

MashaALlah we have so many smart brothers and sisters at TTI! *hoot hoot* :)

Keep it up guys and all the best lol

*I'm out of here*

Wassalaam
 

arzafar

Junior Member
You didn't say if it was OK to post a trigonometry question? But if it is, here is one:

Prove that 1/(1+tan(θ))=cotan(θ)/(1+cotan(θ))

let θ = x

1/(1+tan(θ))
= 1/(1+tanx)
= [1/(1+tanx)][tanx/tanx]
= (1/tanx)/[(1/tanx) + 1]
= cotanx/(1 + cotanx)
= cotan(θ)/(1+cotan(θ))
 

hayat84

I'm not what you believe
I'm sorry if I disturb you in this "brain breaking"game,but do I need to know how much does (a*b):x,to buy a kilo of bread!?I'm really sorry,but I can't stop laughing.forgive me please,I only joked.
 

arzafar

Junior Member
I'm sorry if I disturb you in this "brain breaking"game,but do I need to know how much does (a*b):x,to buy a kilo of bread!?I'm really sorry,but I can't stop laughing.forgive me please,I only joked.

*whistle*
yellow card!
 
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