Question

see attachment, please help me at problem 2 which links to problem 1. I don't know how to project vectors onto the region

Answer 1

Help please

Answer 2

Connections?

Answer 3

what do you mean? I know I have to plug the <u,v> = 2u1v1 + 3 u2v2 onto somewhere but really don't know where

Answer 4

I don't do this math sorry try Mathway.com

Answer 5

ok, thanks for reply.

Answer 6

Notice that \(u=(u_1,u_2)\) and \(v=(v_1,v_2)\)
so in the problema 2 there is defind a "weighted" product as \(2u_1v_1+3u_2v_2\)
So just do the operations taking this as definition of the dot product

Answer 7

I don't get. please do a) as sample.

Answer 8

no, sorry, I need b)

Answer 9

let v=(3,2).
As you know |v|=v.v, so:
|v|=2(3x3)+3(2x2)=18+12=30

Answer 10

sry, I did d)

Answer 11

how |v| =v.v?

Answer 12

b)
v=(3,2), w=(0,-1), k=3
so:
3v=(9,6),
and 3v.w=2(9x0)+3(6x(-1))=-18

Answer 13

is it not \[\sqrt{v1^2 + v2^2}\] = \[\sqrt{9+4}\]

Answer 14

Oh, you did 1b, not 2b

Answer 15

I am ok with 1, just "weighted euclidean inner product" part

Answer 16

\( v.v = v_1^2+v_2^2 =|v|^2\)
so \(v=\sqrt {v_1^2+v_2^2}=\sqrt{v.v}\)

Answer 17

I did 2b

Answer 18

2b)
v=(3,2), w=(0,-1), k=3
so:
3v=(9,6),
and 3v.w=2(9x0)+3(6x(-1))=-18

Answer 19

so, it doesn't relate to <u,v> = 2u1v1+3u2v2?

Answer 20

que. 2 states, "use this new definition for inner product"
\[<kv,w>=2(kv)_1(w)_1+3(kv)_2(w)_2\\
<kv,w>=2(3\times3)(0)+3(3\times2)(-1)\]

Answer 21

GOT IT!!!!:) :)

Answer 22

yep

Answer 23

ok thanks kid. !!!

Answer 24

this projection is used in computer graphics and electronic 3D systems, etc

Answer 25

\[\sin \alpha =2\]