Dealing with vectors

[imperfectissimi]

Hello there, I'm new to this forum and can't find the 'FAQ' section or any introductory thread.
So let me introduce myself here.
I've already graduated from high school with a not-so-bad experience on math,
Neither am I good at it.

I want to improve my experience with math, so I'm still actively studying math and basic sciences
Because in my academic major, they are rudimentary.

Yesterday I came across some kind of problems about vectors, and after an hour of attempting to solve it,
I became confused.

Given a triangle ABC with AB = vector ā (a with a line over it) and AC = vector ƃ (b with a line over it).
The point D is on BC with BDC = 1:2
And the point E on AC with AE:EC = 2:1.

a. Express the vectors AE and AD in terms of ā and ƃ.

AE = (2/3)*b

To find AD we must discover CB first.

AC + CB = ā
ƃ + CB = ā

Thereby CB = ā - ƃ.

It is stated that CD:CB = 2:3, so
CD = (2/3)(ā - ƃ)

Then, we can conclude that
AD = (1/3)ƃ + (2/3)ā


b. If M is the intersection of the line AD and BE, express the vector AM in terms of ā and ƃ.

By using the vector addition and substraction technique, I've already managed to discover that (correct these if I'm wrong)

EB = a - (2/3)b

ED = (2/3)a - (1/3)b

AD = (1/3)b + (2/3)a

Then I tried to solve the problem by using AM = AM

AE + EM = ā + MB

But I ended up in (2/3)b + EM = (2/3)b + EM.

I tried the method using the different lines but the results are the same.

c. If the line CM is elongated so that it intersects the line AB on the point called F, how is the ratio of AF:FB?

d. If the lines DE and AB are elongated so that they intersect on the point called H, how is the ratio of AH:HB?


Recently I've tried using Wolfram Alpha to solve problems dealing with calculus, but I don't think they will solve this.

I would appreciate any solutions or clues given and try to be active in this forum, whether to help or to be helped.
Thanks

NB: How do I insert equations in this forum?

 

[mmm4444bot]

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can't find the 'FAQ' section or any introductory thread.

Please explore the links indicated below.

 

[Deleted member 4993]

Hello there, I'm new to this forum and can't find the 'FAQ' section or any introductory thread.
So let me introduce myself here.
I've already graduated from high school with a not-so-bad experience on math,
Neither am I good at it.

I want to improve my experience with math, so I'm still actively studying math and basic sciences
Because in my academic major, they are rudimentary.

Yesterday I came across some kind of problems about vectors, and after an hour of attempting to solve it,
I became confused.

Given a triangle ABC with AB = vector ā (a with a line over it) and AC = vector ƃ (b with a line over it).
The point D is on BC with BDC = 1:2
And the point E on AC with AE:EC = 2:1.

a. Express the vectors AE and AD in terms of ā and ƃ.

AE = (2/3)*b

To find AD we must discover CB first.

AC + CB = ā
ƃ + CB = ā

Thereby CB = ā - ƃ.

It is stated that CD:CB = 2:3, so
CD = (2/3)(ā - ƃ)

Then, we can conclude that
AD = (1/3)ƃ + (2/3)ā


b. If M is the intersection of the line AD and BE, express the vector AM in terms of ā and ƃ.

By using the vector addition and substraction technique, I've already managed to discover that (correct these if I'm wrong)

EB = a - (2/3)b

ED = (2/3)a - (1/3)b

AD = (1/3)b + (2/3)a

Then I tried to solve the problem by using AM = AM

AE + EM = ā + MB

But I ended up in (2/3)b + EM = (2/3)b + EM.

I tried the method using the different lines but the results are the same.

c. If the line CM is elongated so that it intersects the line AB on the point called F, how is the ratio of AF:FB?

d. If the lines DE and AB are elongated so that they intersect on the point called H, how is the ratio of AH:HB?


Recently I've tried using Wolfram Alpha to solve problems dealing with calculus, but I don't think they will solve this.

I would appreciate any solutions or clues given and try to be active in this forum, whether to help or to be helped.
Thanks

NB: How do I insert equations in this forum?

These types of problems are best solved by co-ordinate geometry.

Let us assign co-ordinates of ABC

A = [0,0]

B = [a, 0] .............................edited

C = [x_C , y_C]

Now find co-ordinates of all the points and define your vectors.

 

[imperfectissimi]

These types of problems are best solved by co-ordinate geometry.

Let us assign co-ordinates of ABC

A = [0,0]

B = [x_B, y_B]

B = [x_B, y_B]

C = [x_C , y_C]

Now find co-ordinates of all the points and define your vectors.

But... Let me see.
Not only they didn't give the specific values of a and b...
But also they don't state the angles.
Then, can it really done by geometric methods?
Would it make the problem more difficult to solve?

 

[Deleted member 4993]

But... Let me see.
Not only they didn't give the specific values of a and b...
But also they don't state the angles.
Then, can it really done by geometric methods?
Would it make the problem more difficult to solve?

That is the way I would tackle this problem. There could be some tricky cross-product and or dot product might make for an elegant solution - but I don't see it.

My next step would be

E = [2/3*x_C, 2/3*y_C]

and

|c| = c = √(x_C² + y_C²).... and so on....

 

[imperfectissimi]

That is the way I would tackle this problem. There could be some tricky cross-product and or dot product might make for an elegant solution - but I don't see it.

My next step would be

E = [2/3*x_C, 2/3*y_C]

and

|c| = c = √(x_C² + y_C²).... and so on....

So basically are you trying to say this?

Notes: xb and yb is one entity, not x*b nor y*b.

This means the first line has the points of
(a, 0) and ( (2/3)xb, (2/3)yb )

And the second line has the points of
(0, 0) and ( (a - ( (1/3) (a - xb) ), (1/3)yb )

Then they intersect in a point called M.

All we do is to create a system of two linear equations.

Equation of the first line:
(3x - 3a)/(2xb - 3a) = (3y - 2yb)/(-2yb)

Equation of the second line:
(3x)/(2a + 3xb) = (3y)/(yb)

By substituting x = x we get
ym = (4 xb yb)/( 3 (a + 8xb) )

By substituting y = y we get
xm = ( 4 xb (2a + 3xb) )/(3 (8xb + 3a) )

And I tried to put this into
SQRT((xm)^2 + (ym)^2)
But then I got stuck in

AM = (4xb/3)SQRT[ { ((yb)^2)((8xb + 3a)^2) + ((2a + 3xb)^2)((a + 8xb)^2) }/{ ((a + 8xb)^2)((8xb + 3a)^2) } ]

Which is practically can't be solved.

Are there any more suggestions?

 

[imperfectissimi]

Well, if you have values for a,b,x,y, then there's nothing to be "solved": it's only a calculation, right?

You can make the whole headache easier this way:
u = (8xb + 3a)^2
v = (8xb + a)^2
w = (2a + 3xb)^2

AM = (4xb / 3)SQRT{[u(yb)^2 + wv] / (uv)}

The question asks to express AM in terms of vector a and vector b.
Besides solving the stuff above, are there any easier ways to do it?

NB: The question that I mean is #b, refer to the first post.