Question

Consider the vector v = (0;3;4) in 3D. Why are there infinite vectors that are perpendicular to v in 3D?

Answer 1

Because the vector is a line segment containing an infinite number of points - any of which may have a line (vector) drawn perpendicular to it.

Answer 2

Should I answer this question by saying what you just said?

Answer 3

I would answer the question the way I did.

You should understand what my answer means and decide whether I'm correct or not rather than trust me to tell you whether you should quote me or not.

I believe my concept is correct, but I don't know if I stated it in the terminology your teacher wants to hear.

Answer 4

hi friends, can I say something?

Answer 5

Sure. I don't rent the place out. :-)

Answer 6

yes sure

Answer 7

you have the formula to consider whether 2 vectors are perpendicular or not. if they are, <u,v> =0

Answer 8

sorry...just found that the actual question is why there are infinite UNIT vectors that are perpendicular to v in 3D

Answer 9

your u = (0,3,4)
let v =(a,b,c)
<u,v> = 0a +3b+4c =0
you have 1 equation, 3 variables the solution for that is infinite set

Answer 10

that is terminology of proving how to get and why it is infinite solution.
does it make sense?

Answer 11

but they are not necessarily unit vectors, right?

Answer 12

i think so, it's in general, of course including unit vector.

Answer 13

unit vector is just a vector with the length 1. that's it

Answer 14

moreover, your vector u=(0,3,4) is not unit vector

Answer 15

it's arbitrary vector, and no need to convert it to unit vector to consider

Answer 16

yes i know...what i want to know is why there are infinite unit vectors perpendicular to v

Answer 17

so, does mine make sense to you? 1 question, 3 variables, you have at least 2 free variables. base on that you can conclude the infinite solution for the stuff. I am sure.

Answer 18

sounds quite right but if we have 0a +3b+4c =0, the equation can be reduced to 3b+4c =0, and so only 2 variables left

Answer 19

and unit vector is of the length of 1, does it mean that we can set up another equation?

Answer 20

hey friend, I think you misunderstanding something in that. cannot reduce because if you do that, you miss a. I mean with any a, 0a =0 . for example if a =1, 0a=0, a =2 ,0a =0, too, a =3; 0a =0 and so on,... that mean infinite set of a, not a =0 is only one solution.
you know what I mean?

Answer 21

got it

Answer 22

hey, how about the asker, where is he?

Answer 23

that's me

Answer 24

ok, sorry friend.

Answer 25

as the unit vectors are of the length of 1, can we set up another equation ?

Answer 26

particular problem, please

Answer 27

as i said above, what i want to find out is UNIT vectors that are perpendicular to v=(0, 3, 4) in 3D

can i let u=(a,b,c) and then set up the following equation?

1/(a^2+b^2+c^2)(sqrt)=1

Answer 28

(a,b,c)/(a^2+b^2+c^2)(sqrt)=1

Answer 29

in the original question, you just ask the proof of infinite set of vectors which perpendicular to (0,3,4). I give you that proof, think of this, all unit vectors =1, generalize the vector perpendicular (a,b,c) in the set you have \[\frac{ <a,b,c> }{ \sqrt{a^2+b^2+c^2} }\] is unit vector where <(0,3,4),(a,b,c)>=0
I think so

Answer 30

sorry for my mistake...

Answer 31

even if we have 2 equations now, there are still infinite unit vectors perpendicular to v, right?

Answer 32

I don't understand.

Answer 33

@richard1024 exactly..
any combination of "b" and "c" that satisfies the above rule alongwith any real "a" will be the perpendicular to <u>

Answer 34

@qweqwe123123123123111 said that in plain English,
@hoa said the same exact thing in mathematical terms
:)
and I just restating them and connected them with a thread.

Answer 35

we have these 2 equations
(a,b,c)/(a^2+b^2+c^2)(sqrt)=1
and 0a+3b+4c=0

Answer 36

still there can be infinite solutions?

Answer 37

they both give the same equation. you wrote the first one incorrectly.

Answer 38

the numerator should be <(0,3,4),(a,b,c)>

Answer 39

which is the same exact thing as the other equation.
so, you've got only one equation. kapeesh?

Answer 40

why they both give the same equation?

Answer 41

eq# 1
<(0,3,4),(a,b,c)>=0

eq# 2
<(0,3,4),(a,b,c)>
--------------- = 0
sqrt(a^2+b^2+c^2)

Answer 42

now, aint they the same thing?

Answer 43

eq. 2 is just the unit vectors..

Answer 44

of the equation 1

Answer 45

i have just attached a file
are you able to open it?

Answer 46

what i don't know is question 1b(iii)

Answer 47

yes. that is infinite.

Answer 48

i don't understand why we have this equation
<(0,3,4),(a,b,c)>
--------------- = 0
sqrt(a^2+b^2+c^2)

Answer 49

then just forget it.. its misleading..
wipe it off...
:)

Answer 50

its not needed

Answer 51

just remember <(0,3,4),(a,b,c)>

Answer 52

as it is a unit vector, should we have an equation like this instead?

(a,b,c)
--------------- = 1
sqrt(a^2+b^2+c^2)

Answer 53

why it's (0,3,4),(a,b,c)?

Answer 54

well, thats the last step..
you are stuck in an infinite loop there... you first have to snap out of it

Answer 55

now i understand why
(a,b,c)
--------------- . (0,3,4)= 0
sqrt(a^2+b^2+c^2)

Answer 56

and unit vector has a length of 1, so is the following equation correct?
(a,b,c)
--------------- = 1
sqrt(a^2+b^2+c^2)

Answer 57

:)

Answer 58

\[(a,b,c)\div \sqrt{a ^{2}+b ^{2}+c ^{2}}=1\]

Answer 59

nice work

Answer 60

it's correct?

Answer 61

yep

Answer 62

that's why i said we have 2 equations

Answer 63

ok. but they are "parallel" and hence, same
but you got the idea.

Answer 64

but still there are infinite solutions because we have 2 equations but 3 unknowns, right?

Answer 65

yes.

Answer 66

now i understand everything
thank you very much for your patience, time, and all you told me

Answer 67

you are welcome.
in math, when you get stuck somewhere, snap out, clear your mind and look at it with fresh eyes.

Answer 68

okay :)