#help-7|zen1thxyz
1 messages · Page 1 of 1 (latest)
nvm
Yes, basically
so the general case would be something that you're describing?
and for the coplanar thing they're talking about, is that just using 3 vectors that share the same plane and doing the math out for the distributives ?
Ordered Triples (because theyre conformed of 3 elements, with meaningful order) of Real numbers
v = (a,b,c) is in R^3
i'm assuming for 1.2 we can do the same thing and just use any 3 vectors to show that they're not associative
For 1.2 you can easily find a counter example
For 1.1 you cant
So you have to prove that the general case is true
and we do that without rigid vectors but by the ordered triples you're talking about?
rigid vectors and ordered triples i mentioned are two ways to represent the same object
and we're just using that to show the general case is true with the distributive equations ?
We cant use distributive properties because we are trying to prove they exist
okay so we're trying to prove that those equations are true in the general case
and the general case is the ordered tripled you mentioned
how does the ordered triples thingy prove that the distributive is true?
Because we define dot product and cross product using those too.
what is an ordered triple?
An object made by 3 elements, in order.
Say, (1,2,3) is an ordered triple
of 1 -> 2 -> 3
so, to start with, it isnt equal (7,4,2) because the elements arent the same
and isnt the same as (3,2,1) because its not the same order.
or like -1, -2, -3 would also be ordered?
Ordered has to do with the fact that the way they are presented is meaningful
not that they follow some order themselves
Bad example from me ig.
hm what does it mean to be presented meaningfully
i don't know if it's a bad example i'm just trying to understand haha
sorry if it sounds like i'm nitpicking
Points in the plane are also ordered pairs
if you read them as (x,y)
you have
Blue = (1,3)
Red = (2,2)
Grey = (2,1)
We usually say that "Points" are "position vectors"
but you could also draw them as arrows
Specially, when doing geometry with vectors, you assume that the vertices are like points, and there are vectors connecting them
This implies that theres some way to identify vertices to some arrow in space. (Coordinates)
i guess i don't understand how this proves the distributive thing
Im just giving you some introduction to the concept, has nothing to do with the problem itself.
oh i see
Just for a last thing
okay i'm following so far
If you say that the red point is at (1,2), and the vector connect the red point with the green is (1,3)
Then you can add (1,2) + (1,3)
Which works basically the same way as normal addition
but the arrow could be anywhere right?
(1,2) + (1,3) = (1+1 , 2+3) = (2,5)
tho, if you mean the one about the cross product, not at all
First i wanna see if you understand this, I just expect you to be able to see that these are simply 3 vectors, with (x,y,z) coordinates, without specifying.
r u saying that we plug in $\sqrt2 i$ for x?
Zen1thXYZ
yeah