MATH1002/1902 Discussion 2005 (1 Viewer)

[withoutaface]

I've done all but 12...

 

[acmilan]

Question 12 you just do external division from point Q to two of the quads sides. Then equate the values you get for vector q and rearrange to get an equation that shows how P divides the required lines internally.

 

[SmokedSalmon]

haha
beer before 3 hour aero lab = bad
luckily today it only went for like 20 minutes

being hungover and having a lab prac the same day is awful! I'm lucky my pracs this year are on tuesday and wednesdays.
I went to beachball last night and turned up to my 10am mblg2771 lecture feeling cruddy... the lights were soo bright. My head hurt

I'll be sure to ask you for advice if I decide I ever want to scrape a Pass at university.

I'll tell you now how to scrape a pass... do pretty average in your subjects and then cram for the exams during stuvac (1 week) and u'll pass all ur subjects. :uhhuh: The trick is to only do this in first year (preferably semester 1). You are nuts if you continue studying like this in 2nd and 3rd year. Those 2 years count far more (then first year) towards your degree.

 

[antarctic]

My proof for 12 was real dodgy, it was basically that if w=the position vector of P relative to O, and αu the position vector of Q relative to O, then βv is the position vector of P relative to Q.

If α is not unique then Q is variable. If Q is variable then the slope and hence direction of βv however, as it is scalar vector multiplications the slope of βv is constant (as it dictates direction) ∴ α is unique. Similarly for β.

Yeah, my proof for it is really dodgy too :-(. But yeah, thanks for sharing Xayma.
My idea is to show that you can make any vector w by multiplying vectors u and v by scalars.
First construct vectors u' and v' (where u'=u and v'=v), with the tail of u' at one end of w and the tail of v' at the other end of w. Then multiply u' and v' by scalars α and β so that the lengths of u' and v' are such that they coincide at a point, and the direction (changing direction if nessecary by multiplying by negative scalar) so that αu' + βv' =w by triangle method of vector addition. Because u=u' and v=v', then αu + βv must also equal w.
However, I asked my tutor about it, and he said that you should try to represent u,v and w on a cartesian plane (i.e. u = ai +bj, v=ci+dj, w=xi+yj) and then using the restricting conditions to get a set of equations which you solve to prove it, but I can't seem to prove it that way :-( Oh well, it did say in the intro that you should only try these q's if you've got time... and I've got alot of other things to do

Antarctic

 

[Jase]

i'm stuck on question one.. what the hell are the arrows point at.. i am confused.

 

[LoneShadow]

i'm stuck on question one.. what the hell are the arrows point at.. i am confused.

sit in front of each one...with ur backside towards the arrow head of the vector...then u shall be enlightened with the holy answer......................so stop making dumb jokes...

 

[acmilan]

Question 8(c) anyone? Its the only one i cant seem to do

 

[Xayma]

I don't think I have done that one, but what you want is the medians (a, b, c for simplicity) are of the form a+b=c with any terms being negative etc.

 

[antarctic]

This is my shot at 8(c):
Let ABC be any triangle. Let M,N and P be midpts of AB, BC and AC respectively, and a,b,c,m,n,p be the position vectors OA, OB, OC, OM, ON and OP respectively.
Therefore, we are required to prove that BP AN CM form a triangle
Now:
n=(1/2)b +(1/2)c
Similarly
p = (1/2)a + (1/2)c
m = (1/2)a +(1/2)b
Now, vector BP = p-b = (1/2)a + (1/2)c - b (by substitution)
Similarly
AN = n-a = (1/2)b + (1/2)c -a
CM = m-c = (1/2)b+(1/2)a -c
By inspection, BP + AN +CM = 0
Hence, BP, AN and CM must form a closed triangle, by triangle method of vector addition.

BTW. should someone start another thread for MATH1001/1901?

 

[zeropoint]

Question 8(c) anyone? Its the only one i cant seem to do

Let D, E and F be the midpoints of AB, BC and CA respectively. You need to prove that BF + AE + CD = 0.
BF + AE + CD = (BA + AF) + (AB + BE) + (CA + AD).
This simplifies to
BE + FA + AD = (1/2)(AB + BC + CA) = 0.

Does anyone have any clues for 8(e). I'd prefer if you didn't post the entire solution.

 

[Xayma]

Umm there is a way to do this that is slightly long.

But basically see 8) a) and 8) d)

More then likely a much quicker way that skips some formalities

 

[antarctic]

Alternatively, find the position vector of the midpoint of the line joining one pair of opposite midpoints of ABCD w.r.t a,b,c and d, and show that the midpoint of the line joining the other joining opposite midpoints of ABCD is the same vector.

 

[gordo]

just use simliar triangles

 

[zeropoint]

Alternatively, find the position vector of the midpoint of the line joining one pair of opposite midpoints of ABCD w.r.t a,b,c and d, and show that the midpoint of the line joining the other joining opposite midpoints of ABCD is the same vector.

Ahh position vectors, of course. I haven't even bothered to use them up until now.

 

[eventhorizon]

8(c)
u draw triangle ABC, where edges AB, BC and AC are bisected by D,E and F respectively
wat u need to prove is that:
(a-e)+(c-d) = (f-b)
since:
2e=b+c
2d=b+a
2f=a+c
RHS = 2a-2e+2c-2d
=2a-b-c+2c-b-a
=a-2b+c
LHS = 2f-2b
=a-2b+c
therefore RHS = LHS

how the hell r u meant to prove q12?

 

[acmilan]

how the hell r u meant to prove q12?

The question is how are you meant to prove it elegantly