Question

What vector is the projection of u=(1;3) on v=(4;4)? The answer is supposed to be in the coordinates.
I got no clue how to go along. :/

Answer 1

Except for the theorem u_v=(u*v)/(v^2)*v
But I do not know how to properly use it.

Answer 2

v times |u| cos(a)

Answer 3

divided by that is .. not times :)

Answer 4

cos(a) = u.v/|u||v|

Answer 5

u=(1;3) |u|=sqrt(10)
v=(4;4) |v| = sqrt(32)

u.v = 4+12 = 16

cos(a) = 16/sqrt(320)

Answer 6

but then:

|u|*u.v
----- = u.v/|v|
|u||v|

v divided by 16/sqrt(32)

Answer 7

\[\frac{\overrightarrow{v}\cdot \overrightarrow{u}}{|\overrightarrow{u}|^2}\times \overrightarrow{u}\]
damn that took a long time to write !!

Answer 8

Would that be a matrix? ;o

Answer 9

<4,4>
---------- and simplify
16/sqrt(32)

Answer 10

I feel so embarassingly stupid regarding vector geometry. :C

Answer 11

<4/(16/sqrt32);4/(16/sqrt32)> or is this not how you're supposed to do it?

Answer 12

i think its good, just simplify the results

Answer 13

i think what i wrote was the projection of v on to u
lets try it.
\[\overrightarrow{u}\cdot\overrightarrow{v}=1\times 4+3\times 4=16\]
\[|\overrightarrow{u}|^2=32\] hmm i am getting a different answer. maybe i am wrong

Answer 14

But that fetches me <sqrt2;sqrt2> right? The answer is supposed to be <2;2>

Answer 15

looks like it must have magnitude of U.V and parallel to V

Answer 16

that is what i get yes.

Answer 17

\[\frac{\overrightarrow{u}\cdot\overrightarrow{v}}{|u^2|}=\frac{1\times 4+3\times 4}{4^2+4^2}=\frac{16}{32}=\frac{1}{2}\]

Answer 18

i dint unit length the v; thats prolly an error in mine

Answer 19

\[\frac{1}{2}<4,4>=<2,2>\]

Answer 20

\[\sqrt{x^2 + y ^2} = 16 \] and \[ y/x = 1\]

Answer 21

this be the formula
\[\frac{\overrightarrow{v}\cdot \overrightarrow{u}}{|\overrightarrow{u}|^2}\times \overrightarrow{u}\] as i have seen it written

Answer 22

u on v tho would be v/|v| not u/|u| right?

Answer 23

The |v^2| really explains alot. ^^ I do not know why they're skipping the abs operator in the demoninator - in the theorem.

Answer 24

\[|u|\frac{u.v}{|u||v|}*\frac{v}{|v|}\]

Answer 25

Superb. A thousand thanks! :)

Answer 26

I've never seen that formula ... what exactly is it??