Question

Hyperplanes? Find a basis for the hyperplane vector A perp? A=(1,2,-3)

Answer 1

The answer is {(-2,1,0),(3,0,1)} if it helps.. I really don't get how they got this

Answer 2

doesnt ring a bell, srry

Answer 3

A is the cross of the answers....

Answer 4

U would go (a-b)X(a-c) to get a normal.

Answer 5

that is an answer...there are infinity many answers to your question.

Answer 6

(2,-1,0),(0,3,2)

is a second answer

Answer 7

Let \(v_1 = A\) then select two other vectors \(v_1, v_2\) such that \(\{v_1, v_2, v_3\}\) is linearly independent.Then perform the Gram-Schmidt orthogonalization process.

Answer 8

total overkill

Answer 9

just pick any two vectors that are linear independent and orthogonal to the vector given

Answer 10

for the vector given (1,2,-3)

make a vector that has one coordinate zero

(x,y,0)
now switch the 1 and 2 and make one of them negative

(-2,1,0)

you just found an orthogonal vector

now create another vector with the zero in a different location

(a,0,b)

switch 1 and -3 and change the sign of on of them

(3,0,1)

done!

Answer 11

Can I make the first one 0 then switch 2 and -3?

Answer 12

are there other possiblilities is what I'm trying to say, and is that how you go about for a hyperplane

Answer 13

there are infinity many possibilities...I think this is the easiest way to do it in this case

Answer 14

I am more used to it the other way about.

U have abc in the plane (in R3) and u get

p = b-a and q = c-b then n =pXq

then if u dot that with a (in xyz) u get equation for a plane.

Answer 15

"Can I make the first one 0 then switch 2 and -3?"

yes...see my first answer

Answer 16

Oh okay thank you! Estudier I get what you're saying but what's with the X in pXq

Answer 17

the X is for the cross product

Answer 18

Oh okay so once I get the equation for a plane how would I find the basis?

Answer 19

Sorry for the questions just studying for my final and this is like the one concept I don't quite understand

Answer 20

pick any two linearly independent vectors in you plane...they will form a basis

Answer 21

I like perpendicular ones...

Answer 22

those are nice...but you don't need them unless they specifically ask for vectors that are orthogonal

Answer 23

Okay so if there's more like 3 values, like a=(2,-1,4,1), I can do the same thing but I'd have three vectors?

Answer 24

more than 3*

Answer 25

just make each part 0 and switch the values but would one still need to be negative

Answer 26

cross is a problem in 4d

Answer 27

okay I'm lost again haha so how would do it with four

Answer 28

{(1,2,0,0),(0,4,1,0),(0,0,-1,4)}

would work

Answer 29

they are linearly independent and they are all orthogonal to the vector you gave above

Answer 30

This sort of thing is why I switched to Geometric Algebra...

Answer 31

the answer says (0,1,0,1) (0,0,1,-4),(0,4,1,0)

Answer 32

how are you getting this zarkon?

Answer 33

that is one answer...there are infinity many...as I stated before

Answer 34

the same way your book probably did it...make to entries 0 and switch the other two and change the sign on one of them

Answer 35

make two entries...

Answer 36

Then why does some of them have two zeroes.. My book doesn't explain hyperplanes like at all

Answer 37

Oh snap I reread your thing

Answer 38

both your answer and mine have exactly two zeros in each vector.

Answer 39

Okay if it was in 5 d, I'd make 3 entries zeroes?

Answer 40

yes

Answer 41

Oh okay I will try this thanks for clearing it, and thanks to everyone else who answered!

Answer 42

Eventually you will see the normal equation description of a hyperplane which makes life a lot simpler...

Answer 43

the answer you gave does not work

Answer 44

(0,1,0,1) (0,0,1,-4),(0,4,1,0) is not a basis

Answer 45

4(0,1,0,1)+(0,0,1,-4)=(0,4,1,0)

Answer 46

? That's the answer in my book {(0,1,0,1) (0,0,1,-4),(0,4,1,0)}

Answer 47

it's wrong

Answer 48

those vectors are dependent and therefore do not form a basis

Answer 49

Okay I understand that a basis needs to be independent and span all of R^n. So is that what a hyperplane is? Are we just taking the vector and find other vectors that can form a basis?

Answer 50

nvm I understand it now. Thanks for your help Zarkon