Question

y=2 v1-4v2+v3

find ||y||

Answer 1

@amistre64

Answer 2

{v1,v2,v3} is orthonormal basis

Answer 3

@myininaya

Answer 4

@samjordon

Answer 5

isnt magnitude the sqrt of each compenent squared?

Answer 6

idk

Answer 7

sqrt(2^2+4^2+1^2)=sqrt(21)

Answer 8

hmmm if it was orthonormal then the magnitude wld be one if i remember correctly

Answer 9

what about v1 , v2, v3?

Answer 10

hmm let me think abt this one

Answer 11

@samjordon is right
You can have a look here @imranmeah91
http://tutorial.math.lamar.edu/Classes/LinAlg/OrthonormalBasis.aspx

Answer 12

neways it wldnt make a diff what v1 v2 or v3 is since they r orthonormal so when u wld find its magnitude it equals one so sqrt(21) is teh answer

Answer 13

haha still doubting my answer.
it seems correct but i am not sure i am gonna go get my linear algebra notes

Answer 14

y is a dot product so it has one value in the end

Answer 15

assume

y = a.v

what does |a.v| represent ?

Answer 16

magnitude of dot product

Answer 17

so absolute value of a dot product as well; and there are no values known for the vectors?

Answer 18

no, they just said the vector sets was orthonormall basis

Answer 19

the nulspace represents a plane equation

Answer 20

hmm, id prolly have to try out a few ortho normals and compare the results to see if the differ by a pattern

Answer 21

thanks to two of you

Answer 22

100
010
001 is the standard orthnormal

Answer 23

|-1| = 1 in that case

Answer 24

just looked up my notes i think i know what to do

Answer 25

when u have ||cv_1||
c||v_1|| like u pull out the c

Answer 26

does that make sense amistre?

Answer 27

1 3 -1
2 -1 4
1 -1 -7

is orthogonal; and if we normal it ....

1/sqrt(6)
2/sqrt(6)
1/sqrt(6)

3/sqrt(11)
-1/sqrt(11)
-1/sqrt(11)

-1\sqrt(66)
4\sqrt(66)
-7\sqrt(66)

if that dot product is 1 id say the likelihood of my random guess would match up by accident

Answer 28

the c thing? sure; pull out a scalar

Answer 29

sooo like wld u say that the number in front of the V_1, V_2 and V_3 are scalar?

Answer 30

yes

Answer 31

ohhhh so u wld just pull them out i guess the magnitude wld be 8

Answer 32

oh and c is absolute value when u pull it out it will always be positive that is what i saw

Answer 33

or just pull out the positive part and let the negative stay in

Answer 34

ya i guess

Answer 35

idk like i just took this course like 4 months ago and i am getting sooo confused

Answer 36

hahah the more i think abt this the more confused i am getting

Answer 37

as is the dotproduct represents a sphere or ball stuck at the origin on the end of a vector

Answer 38

cos(t) = a.v/|1||1| is the angle between any 2 vectors

so |a.v| </ 1

Answer 39

alright, I will read it over, thank

Answer 40

This will help http://www.wolframalpha.com/input/?i=y%3D2+v1-4v2%2Bv3++find+%7C%7Cy%7C%7C

Answer 41

lol, thanks lily