you're a bro right?
Ok.. so Next problem?
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
|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...
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
sorry,but let me smile a little...I understand arabic language more than maths..how can you communicate with this language?have fun!
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?
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?
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))
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.
what is that?*whistle*
yellow card!
what is that?