Let $V=R^3.$ Let $W=operatorname{Span}({(1,-2,2)}).$ Find an orthonormal basis for $W^{bot}$ [closed]

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closed as off-topic by José Carlos Santos, Ali Caglayan, user302797, Jyrki Lahtonen, Brahadeesh Dec 3 at 4:19

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Google "Gram-Schmidt"
– Surb
Dec 2 at 19:04

Which inner product do you use? The usual one?
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math.stackexchange.com/questions/2307669/…
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I cant seem to figure this out, can anyone help me?

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Bernard

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closed as off-topic by José Carlos Santos, Ali Caglayan, user302797, Jyrki Lahtonen, Brahadeesh Dec 3 at 4:19

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Ali Caglayan, user302797, Jyrki Lahtonen, Brahadeesh

If this question can be reworded to fit the rules in the help center, please edit the question.

Google "Gram-Schmidt"
– Surb
Dec 2 at 19:04

Which inner product do you use? The usual one?
– Fakemistake
Dec 2 at 19:05

math.stackexchange.com/questions/2307669/…
– Boshu
Dec 2 at 19:08

add a comment |

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I cant seem to figure this out, can anyone help me?

edited Dec 2 at 19:27

Bernard

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Question
I cant seem to figure this out, can anyone help me?

linear-algebra vector-spaces orthogonality

edited Dec 2 at 19:27

Bernard

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asked Dec 2 at 19:00

Anas

165

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edited Dec 2 at 19:27

Bernard

117k637109

asked Dec 2 at 19:00

Anas

165

New contributor

edited Dec 2 at 19:27

Bernard

117k637109

edited Dec 2 at 19:27

Bernard

117k637109

edited Dec 2 at 19:27

Bernard

117k637109

asked Dec 2 at 19:00

Anas

165

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asked Dec 2 at 19:00

Anas

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asked Dec 2 at 19:00

Anas

165

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closed as off-topic by José Carlos Santos, Ali Caglayan, user302797, Jyrki Lahtonen, Brahadeesh Dec 3 at 4:19

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Ali Caglayan, user302797, Jyrki Lahtonen, Brahadeesh

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by José Carlos Santos, Ali Caglayan, user302797, Jyrki Lahtonen, Brahadeesh Dec 3 at 4:19

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Ali Caglayan, user302797, Jyrki Lahtonen, Brahadeesh

If this question can be reworded to fit the rules in the help center, please edit the question.

Google "Gram-Schmidt"
– Surb
Dec 2 at 19:04

Which inner product do you use? The usual one?
– Fakemistake
Dec 2 at 19:05

math.stackexchange.com/questions/2307669/…
– Boshu
Dec 2 at 19:08

add a comment |

Google "Gram-Schmidt"
– Surb
Dec 2 at 19:04

Which inner product do you use? The usual one?
– Fakemistake
Dec 2 at 19:05

math.stackexchange.com/questions/2307669/…
– Boshu
Dec 2 at 19:08

Google "Gram-Schmidt"
– Surb
Dec 2 at 19:04

Which inner product do you use? The usual one?
– Fakemistake
Dec 2 at 19:05

math.stackexchange.com/questions/2307669/…
– Boshu
Dec 2 at 19:08

add a comment |

1 Answer
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Let $ a= (1,-2,2)$ and let $b=(x,y,z)$ orthogonal on $a$, so $$acdot b= 0implies x-2y+2z=0$$

so we can take $b= (0,1,1)$. Now take $c= atimes b=...$ and then $b,c$ is orthogonal basis for $W^{bot}$.

answered Dec 2 at 19:05

greedoid

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1 Answer
1

active

oldest

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1 Answer
1

active

oldest

votes

up vote
-1
down vote

Let $ a= (1,-2,2)$ and let $b=(x,y,z)$ orthogonal on $a$, so $$acdot b= 0implies x-2y+2z=0$$

so we can take $b= (0,1,1)$. Now take $c= atimes b=...$ and then $b,c$ is orthogonal basis for $W^{bot}$.

answered Dec 2 at 19:05

greedoid

36.2k114591

add a comment |

up vote
-1
down vote

Let $ a= (1,-2,2)$ and let $b=(x,y,z)$ orthogonal on $a$, so $$acdot b= 0implies x-2y+2z=0$$

so we can take $b= (0,1,1)$. Now take $c= atimes b=...$ and then $b,c$ is orthogonal basis for $W^{bot}$.

answered Dec 2 at 19:05

greedoid

36.2k114591

add a comment |

up vote
-1
down vote

Let $ a= (1,-2,2)$ and let $b=(x,y,z)$ orthogonal on $a$, so $$acdot b= 0implies x-2y+2z=0$$

so we can take $b= (0,1,1)$. Now take $c= atimes b=...$ and then $b,c$ is orthogonal basis for $W^{bot}$.

answered Dec 2 at 19:05

greedoid

36.2k114591

Let $ a= (1,-2,2)$ and let $b=(x,y,z)$ orthogonal on $a$, so $$acdot b= 0implies x-2y+2z=0$$

so we can take $b= (0,1,1)$. Now take $c= atimes b=...$ and then $b,c$ is orthogonal basis for $W^{bot}$.

answered Dec 2 at 19:05

greedoid

36.2k114591

answered Dec 2 at 19:05

greedoid

36.2k114591

answered Dec 2 at 19:05

greedoid

36.2k114591

answered Dec 2 at 19:05

greedoid

36.2k114591

add a comment |

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