Determining a vector orthogonal to $q_1=(1,1,1)$ and $q_3=(1,1,-2)$, why I'm wrong with my calculations?

up vote
1
down vote

favorite

Consider the vectors $q_1=(1,1,1)$ and $q_3=(1,1,-2)$. I need to find a third vector $q_2$ such that $q_1,q_2,q_3$ is a arthogonal basis for $mathbbR^3$.

My problem is the following: I did take $v=(1,0,0)$ and I did verify that $q_1,q_3,v$ is a basis for $mathbbR^3$. Then I did take $$q_2=v-langle v|q_1rangle q_1-langle v|q_3rangle q_3=(-1,-2,1)$$

And, by Gram-Schmidt process, $q_2$ must be orthogonal to $q_1$ and $q_3$. But, as we can see, it does not happen. So, where is my mistake?

edited Nov 22 at 18:50

quid♦

36.7k95093

asked Nov 22 at 16:51

Gödel

1,351319

add a comment |

up vote
1
down vote

favorite

Consider the vectors $q_1=(1,1,1)$ and $q_3=(1,1,-2)$. I need to find a third vector $q_2$ such that $q_1,q_2,q_3$ is a arthogonal basis for $mathbbR^3$.

My problem is the following: I did take $v=(1,0,0)$ and I did verify that $q_1,q_3,v$ is a basis for $mathbbR^3$. Then I did take $$q_2=v-langle v|q_1rangle q_1-langle v|q_3rangle q_3=(-1,-2,1)$$

And, by Gram-Schmidt process, $q_2$ must be orthogonal to $q_1$ and $q_3$. But, as we can see, it does not happen. So, where is my mistake?

edited Nov 22 at 18:50

quid♦

36.7k95093

asked Nov 22 at 16:51

Gödel

1,351319

add a comment |

up vote
1
down vote

favorite

Consider the vectors $q_1=(1,1,1)$ and $q_3=(1,1,-2)$. I need to find a third vector $q_2$ such that $q_1,q_2,q_3$ is a arthogonal basis for $mathbbR^3$.

My problem is the following: I did take $v=(1,0,0)$ and I did verify that $q_1,q_3,v$ is a basis for $mathbbR^3$. Then I did take $$q_2=v-langle v|q_1rangle q_1-langle v|q_3rangle q_3=(-1,-2,1)$$

And, by Gram-Schmidt process, $q_2$ must be orthogonal to $q_1$ and $q_3$. But, as we can see, it does not happen. So, where is my mistake?

edited Nov 22 at 18:50

quid♦

36.7k95093

asked Nov 22 at 16:51

Gödel

1,351319

Consider the vectors $q_1=(1,1,1)$ and $q_3=(1,1,-2)$. I need to find a third vector $q_2$ such that $q_1,q_2,q_3$ is a arthogonal basis for $mathbbR^3$.

My problem is the following: I did take $v=(1,0,0)$ and I did verify that $q_1,q_3,v$ is a basis for $mathbbR^3$. Then I did take $$q_2=v-langle v|q_1rangle q_1-langle v|q_3rangle q_3=(-1,-2,1)$$

And, by Gram-Schmidt process, $q_2$ must be orthogonal to $q_1$ and $q_3$. But, as we can see, it does not happen. So, where is my mistake?

linear-algebra vectors orthogonality

edited Nov 22 at 18:50

quid♦

36.7k95093

asked Nov 22 at 16:51

Gödel

1,351319

edited Nov 22 at 18:50

quid♦

36.7k95093

asked Nov 22 at 16:51

Gödel

1,351319

edited Nov 22 at 18:50

quid♦

36.7k95093

edited Nov 22 at 18:50

quid♦

36.7k95093

edited Nov 22 at 18:50

quid♦

36.7k95093

asked Nov 22 at 16:51

Gödel

1,351319

asked Nov 22 at 16:51

Gödel

1,351319

asked Nov 22 at 16:51

Gödel

1,351319

add a comment |

3 Answers
3

active

oldest

votes

up vote
2
down vote

accepted

HINT

Since $q_1$ and $q_3$ are orthogonal it suffices to find $q_3$ by

$$q_2=q_1times q_3$$

As an alternative by GS we have

$$q_2=v-langle v|hat q_1rangle hat q_1-langle v| hat q_3rangle hat q_3=(1,0,0)-frac13(1,1,1)-frac16(1,1,-2)=left(frac12,-frac12,0right)$$

edited Nov 22 at 17:06

answered Nov 22 at 16:54

gimusi

88.2k74394

I don't know why I don't use cross product before, It is a simple way to solve the problem.
– Gödel
Nov 22 at 17:10

@Gödel Yes indeed it is the faster way in that case if we can't see that by inspection.
– gimusi
Nov 22 at 17:18

add a comment |

up vote
4
down vote

None of $q_1$ and $q_2$ are normalized. Hence, the formula would be
$$ q_2=v-q_1rangle over langle q_1 q_1-q_3rangle over langle q_3 q_3 = (1 over 2,-1 over 2,0)$$.

edited Nov 22 at 17:03

answered Nov 22 at 16:56

SinTan1729

2,387622

add a comment |

up vote
2
down vote

$$q_2=v-fraclangle vlangle q_1 q_1-fraclangle vlangle q_3 q_3$$

answered Nov 22 at 16:54

Siong Thye Goh

94.6k1462114

add a comment |

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\$","\$"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);

);

draft saved

draft discarded

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009351%2fdetermining-a-vector-orthogonal-to-q-1-1-1-1-and-q-3-1-1-2-why-im-wro%23new-answer', 'question_page');

);

Post as a guest

Name

Required, but never shown

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
2
down vote

accepted

HINT

Since $q_1$ and $q_3$ are orthogonal it suffices to find $q_3$ by

$$q_2=q_1times q_3$$

As an alternative by GS we have

$$q_2=v-langle v|hat q_1rangle hat q_1-langle v| hat q_3rangle hat q_3=(1,0,0)-frac13(1,1,1)-frac16(1,1,-2)=left(frac12,-frac12,0right)$$

edited Nov 22 at 17:06

answered Nov 22 at 16:54

gimusi

88.2k74394

I don't know why I don't use cross product before, It is a simple way to solve the problem.
– Gödel
Nov 22 at 17:10

@Gödel Yes indeed it is the faster way in that case if we can't see that by inspection.
– gimusi
Nov 22 at 17:18

add a comment |

up vote
2
down vote

accepted

HINT

Since $q_1$ and $q_3$ are orthogonal it suffices to find $q_3$ by

$$q_2=q_1times q_3$$

As an alternative by GS we have

$$q_2=v-langle v|hat q_1rangle hat q_1-langle v| hat q_3rangle hat q_3=(1,0,0)-frac13(1,1,1)-frac16(1,1,-2)=left(frac12,-frac12,0right)$$

edited Nov 22 at 17:06

answered Nov 22 at 16:54

gimusi

88.2k74394

I don't know why I don't use cross product before, It is a simple way to solve the problem.
– Gödel
Nov 22 at 17:10

@Gödel Yes indeed it is the faster way in that case if we can't see that by inspection.
– gimusi
Nov 22 at 17:18

add a comment |

up vote
2
down vote

accepted

HINT

Since $q_1$ and $q_3$ are orthogonal it suffices to find $q_3$ by

$$q_2=q_1times q_3$$

As an alternative by GS we have

$$q_2=v-langle v|hat q_1rangle hat q_1-langle v| hat q_3rangle hat q_3=(1,0,0)-frac13(1,1,1)-frac16(1,1,-2)=left(frac12,-frac12,0right)$$

edited Nov 22 at 17:06

answered Nov 22 at 16:54

gimusi

88.2k74394

HINT

Since $q_1$ and $q_3$ are orthogonal it suffices to find $q_3$ by

$$q_2=q_1times q_3$$

As an alternative by GS we have

$$q_2=v-langle v|hat q_1rangle hat q_1-langle v| hat q_3rangle hat q_3=(1,0,0)-frac13(1,1,1)-frac16(1,1,-2)=left(frac12,-frac12,0right)$$

edited Nov 22 at 17:06

answered Nov 22 at 16:54

gimusi

88.2k74394

edited Nov 22 at 17:06

answered Nov 22 at 16:54

gimusi

88.2k74394

answered Nov 22 at 16:54

gimusi

88.2k74394

answered Nov 22 at 16:54

gimusi

88.2k74394

I don't know why I don't use cross product before, It is a simple way to solve the problem.
– Gödel
Nov 22 at 17:10

@Gödel Yes indeed it is the faster way in that case if we can't see that by inspection.
– gimusi
Nov 22 at 17:18

add a comment |

I don't know why I don't use cross product before, It is a simple way to solve the problem.
– Gödel
Nov 22 at 17:10

@Gödel Yes indeed it is the faster way in that case if we can't see that by inspection.
– gimusi
Nov 22 at 17:18

I don't know why I don't use cross product before, It is a simple way to solve the problem.
– Gödel
Nov 22 at 17:10

@Gödel Yes indeed it is the faster way in that case if we can't see that by inspection.
– gimusi
Nov 22 at 17:18

add a comment |

up vote
4
down vote

None of $q_1$ and $q_2$ are normalized. Hence, the formula would be
$$ q_2=v-q_1rangle over langle q_1 q_1-q_3rangle over langle q_3 q_3 = (1 over 2,-1 over 2,0)$$.

edited Nov 22 at 17:03

answered Nov 22 at 16:56

SinTan1729

2,387622

add a comment |

up vote
4
down vote

None of $q_1$ and $q_2$ are normalized. Hence, the formula would be
$$ q_2=v-q_1rangle over langle q_1 q_1-q_3rangle over langle q_3 q_3 = (1 over 2,-1 over 2,0)$$.

edited Nov 22 at 17:03

answered Nov 22 at 16:56

SinTan1729

2,387622

add a comment |

up vote
4
down vote

None of $q_1$ and $q_2$ are normalized. Hence, the formula would be
$$ q_2=v-q_1rangle over langle q_1 q_1-q_3rangle over langle q_3 q_3 = (1 over 2,-1 over 2,0)$$.

edited Nov 22 at 17:03

answered Nov 22 at 16:56

SinTan1729

2,387622

None of $q_1$ and $q_2$ are normalized. Hence, the formula would be
$$ q_2=v-q_1rangle over langle q_1 q_1-q_3rangle over langle q_3 q_3 = (1 over 2,-1 over 2,0)$$.

edited Nov 22 at 17:03

answered Nov 22 at 16:56

SinTan1729

2,387622

edited Nov 22 at 17:03

answered Nov 22 at 16:56

SinTan1729

2,387622

answered Nov 22 at 16:56

SinTan1729

2,387622

answered Nov 22 at 16:56

SinTan1729

2,387622

add a comment |

up vote
2
down vote

$$q_2=v-fraclangle vlangle q_1 q_1-fraclangle vlangle q_3 q_3$$

answered Nov 22 at 16:54

Siong Thye Goh

94.6k1462114

add a comment |

up vote
2
down vote

$$q_2=v-fraclangle vlangle q_1 q_1-fraclangle vlangle q_3 q_3$$

answered Nov 22 at 16:54

Siong Thye Goh

94.6k1462114

add a comment |

up vote
2
down vote

$$q_2=v-fraclangle vlangle q_1 q_1-fraclangle vlangle q_3 q_3$$

answered Nov 22 at 16:54

Siong Thye Goh

94.6k1462114

$$q_2=v-fraclangle vlangle q_1 q_1-fraclangle vlangle q_3 q_3$$

answered Nov 22 at 16:54

Siong Thye Goh

94.6k1462114

answered Nov 22 at 16:54

Siong Thye Goh

94.6k1462114

answered Nov 22 at 16:54

Siong Thye Goh

94.6k1462114

answered Nov 22 at 16:54

Siong Thye Goh

94.6k1462114

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Mathematics Stack Exchange!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

Some of your past answers have not been well-received, and you're in danger of being blocked from answering.

Please pay close attention to the following guidance:

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

搜尋此網誌

mjhjmtu