Question

Projecting Vertex.

Answer 1

@acxbox22 :)

Answer 2

sorry i dont know this stuff :(

Answer 3

@Nnesha

Answer 4

\(proj_{\vec{v}}\vec{u} = \frac{\vec{v} \bullet \vec{u}}{|\vec v|^2}\vec{v}\) is the *vector* projection of \(\vec u \) on \(\vec v \).

it's just a formula so calculate the various parts, eg \(\vec{v} \bullet \vec{u} = <11,3> \bullet <-6,-4>\)

Answer 5

\(| \vec v |^2 = \vec v \bullet \vec v\)

Answer 6

so
u * v = <-66,-12>

Answer 7

yeah?

Answer 8

simplified u*v=-78
v*v=52
is it right so far?

Answer 9

@IrishBoy123

Answer 10

@Loser66 :)

Answer 11

can you help me plz

Answer 12

Post your question on a new feed and i'll try :)

Answer 13

"u • v = <-66,-12>"

nah

u • v = <11,3).<-6,-4> = (11)(-6) + (3)(-4) = -66 - 12 = - 78

Answer 14

yep yep that's what I got for that one :)

Answer 15

\(\vec u \bullet \vec v = <u_x,u_y> \bullet <v_x,v_y> = (u_x \times v_x) + (u_y \times v_y) \)

Answer 16

sorry, bandwith issues, i am clearly behind times. let me read the thread again

Answer 17

It's all good :)

Answer 18

v.v = <-6,-4><-6,-4> = 36 + 16 = 52
seems we agree

Answer 19

what next?

Answer 20

uh..not sure haha

Answer 21

plug it into the formula?

Answer 22

yes
<9,6>?

Answer 23

so we have
\[\frac{ -78 }{ 52}v\]

Answer 24

woah where did you get<9,6> ?

Answer 25

v = <-6,-4>

and, -78/52 = -3/2

and, -3/2 * <-6,-4> = <9,6>

agree?

Answer 26

ah I see, thanks :)

Answer 27

so, is that the final answer then? o.0

Answer 28

or ..is that just what \[u _{1} \] equals?

Answer 29

that answers (a).
for (b) you have to .....!!! do something v similar

Answer 30

let's do it x)

Answer 31

we would use
(-75/52)v again, yeah?

Answer 32

you've already done \(\vec u_1\)

that's what you just calculated

so you need to do the same on \(\vec u_2\)

so what is \(\vec u_2\) ? it is orthogonal to \(\vec v\) ...

Answer 33

its \[u - u _{1}\]

Answer 34

<2,-3>?

Answer 35

<-6,-4>•<x,y> = 0
-6x-4y = 0
try:
<-6,-4>•<4,-6>

or
<-6,-4>•<-4,6>

both work

we can use either

Answer 36

does what I did work? o.o

Answer 37

I'm pretty sure that's what the prof did in class but ah well haha

Answer 38

let me actually do it

Answer 39

k

Answer 40

<2,-3>

Roger

Answer 41

Yay! fantastic :)

Answer 42

thanks for taking the time to calculate it all :)

Answer 43

so we have \[u _{1}=<9,6> u _{2}=<2,-3>\]

Answer 44

what goes into \[proj _{v}u\]?

Answer 45

nevermind, they just wanted <9,-6> there too :) thanks!!