math 1002: linear algebra (1 Viewer)

[Wyvern]

im so stuck

 

[Carlito]

mah this is so vague. i asked an old high school friend for help and this is all he said "remember the relation between the inner product, vector lengths, and the cosine of the angle between the two vectors"

 

[Winston]

LOL wtf.. tell him to do the ques man!

 

[JUB JUB]

Originally posted by Carlito
mah this is so vague. i asked an old high school friend for help and this is all he said "remember the relation between the inner product, vector lengths, and the cosine of the angle between the two vectors"

bastard

 

[Carlito]

im guessing he doesnt have a clue but trying to sound like he does

 

[JUB JUB]

what question is everyone up to?

 

[Carlito]

still cant do 1(ii) b or c, and 2(iii) or (iv), although i have many pages of attempts and scribbles

 

[Winston]

ive made more squiggles than u that im starting to make a pattern, its turning out to be a map, and its pointing right at my place! AHHHHHHHHHHHHHHHHHHHHHHH

 

[JUB JUB]

Originally posted by Carlito
still cant do 1(ii) b

square it!!!!

 

[Carlito]

Originally posted by JUB JUB
square it!!!!

dude square what its already squared??

 

[Wyvern]

im with Carlito

i found this which sort of resembled 1 ii) part c but not really so i dont get it lol

Take a triangle whose vertices A, B, C have the position vectors a, b, and
c. Then the centre of the triangle is given by

(a+b+c)/3 = (2/3) * (b+c)/2 + (1/3) * a

Now (b+c)/2 is the mid point of the side BC. Call this point D. Thus the
centre of the triangle lies 2/3 of the way along the line joining the vertix
A to the midpoint D. This is true for any triangle.

Now if ABC is equilateral, then ABD will be a right-angled triangle (draw a
picture and you should see this). But we know that AB (the hypotenuse) = k
and BD = k/2. Hence by Pythagoras' Theorem:

AD^2 = K^2 - (k/2)^2 = 3(k^2)/4

So AD = sqrt(3) k / 2

The distance from a vertix to the centre is 2/3 of this, which is k /
sqrt(3).

 

[Carlito]

ok that makes no sense to me at all lol

 

[JUB JUB]

what the fuck is all that?

 

[Wyvern]

Originally posted by Carlito
ok that makes no sense to me at all lol

me either but without that i dont have anything :/..........maybe i will get a sympathy mark

 

[JUB JUB]

lol i think ur triangle image is wrong, whats the deal with the centre of the triangle

 

[Wyvern]

Originally posted by JUB JUB
lol i think ur triangle image is wrong, whats the deal with the centre of the triangle

lol i didnt do that, i was just looking for solutions to a question similar to that one on the net and thats the best i found

 

[lm1122]

hmmm... ok then... assignment all of a sudden looking a bit nicer...

 

[Carlito]

ok well we know that all the vertices must equal 1/2(a+b)
because thats what the thing to the midpoint equals so then we have to prove that AP and AB = 1/2(a+b)

but thats all i can come up with, no answer

 

[Wyvern]

Originally posted by lm1122
hmmm... ok then... assignment all of a sudden looking a bit nicer...

which ones have you done?

 

[JUB JUB]

use ur answers from 2ii...think...units squared and triangles....hmmmm