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- Thread starter arzafar
- Start date

you're a bro right?

ya...lol.. i m bro...

Ok.. so Next problem?

yeah sure! go ahead and post one!

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

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..

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.

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:

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.

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?

^^^

any takers?

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)

Oh my Allah! Just about when i though i kicked complicated maths out of my life 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

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))

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!

what is that?*whistle*

yellow card!

what is that?

yellow card

A penalty card shown in football to player who breaks rules.