Question

I have another question for the same topic. that is: u = (2,k,6), v = (l, 5,3) w = (1,2,3) the question is do there exist scalars k, and l such that the vectors are mutually orthogonal .it means i have to find k, l and u,v,w orthogonal to each other?

Answer 1

You have to calculate ||u-v||, using the fact that:\[\|v\|=\sqrt{\langle v,v\rangle}\]where <*,*> is the inner product. You also need to use the bilinearity of the inner product. \[\|u-v\|=\sqrt{\langle u-v,u-v\rangle}=\sqrt{\langle u,u\rangle-2\langle u,v\rangle+\langle v,v\rangle} \]Since u and v are orthogonal, <u,v>=0. Since they are unit vectors, <u,u>=1 and <v,v>=1. So you get root 2 on the right hand side of the equation.

Answer 2

hello joe!!

Answer 3

just says it is an inner product space, doesn't say what the metric is

Answer 4

oh nvm i misread your answer sorry

Answer 5

no no i misread it that is all

Answer 6

It depends on whether or not we can assume the knowledge that when you have an inner product on a space, it induces a norm. I assumed that was the norm in question >.< my bad.

Answer 7

you are right, ignore me

Answer 8

of course it is the norm, i just made a reading error

Answer 9

They are unit vectors, so their magnitude is \(1\)

Answer 10

\[\|u+v\|^2=\langle u-v,u-v\rangle=\|u\|^2+\|v\|^2-\langle uv\rangle-\langle v,u\rangle\]

Answer 11

is it that hard, friends?

Answer 12

what if i make up a matrix as \[\left[\begin{matrix}2 & k & 6 \\ l & 5 & 3 \\1 & 2 & 3\end{matrix}\right]\] and solve the homogenous system to get the condition, if no solution, so the answer is no.
what do you think?

Answer 13

if they are mutually orthogoanl, does that mean they form a bases in R^3?

Answer 14

if so, then the determinant would need to NOT be zero

Answer 15

or, you can setup dots

2 k 6
l 5 3
------
2+5k+18=0


l 5 3
1 2 3
-----
L+10+9=0

2 k 6
1 2 3
-----
2+2k+18 = 0

Answer 16

i spose that first one id 2L ... l looks like a one to me :)

Answer 17

ok, change the letter from l to m to be clear. but you are solving the matrix?

Answer 18

so they are end up at L = -19 and K = 10. ?

Answer 19

im solving the orthogonal property of vectors

Answer 20

ok, got it

Answer 21

2 k 6
l 5 3
------
2L+5k+18=0


l 5 3
1 2 3
-----
L+10+9=0; L = -19

2 k 6
1 2 3
-----
2+2k+18 = 0 ; K = -10

2(-19) + 5(-10) + 18 = -70, so not othogonal

Answer 22

yeap.!! got it, they can othogonal to (1,2,3) but they are not orthogonal to each other. so , the solution is there not exists any k,l satisfy the problem? am I right?

Answer 23

thanks a lot

Answer 24

i believe that is correct :)

Answer 25

@99smartscore person: no need to get medal, right? sorry for that.

Answer 26

you can do a grahm schmidt process to find an orthogonal basis but i dont see a way that you can make it out of these 3 vectors :)

Answer 27

that part is on the previous part of Grahm . so, I am not allowed to use it

Answer 28

lol, medals are pointless. the thankyou is good enough ;)

Answer 29

thank you!