Ttdfjt

What initially awakened the Balrog?

As a dual citizen, my US passport will expire one day after traveling to the US. Will this work?

New Order #6: Easter Egg

In musical terms, what properties are varied by the human voice to produce different words / syllables?

"klopfte jemand" or "jemand klopfte"?

Would color changing eyes affect vision?

The test team as an enemy of development? And how can this be avoided?

My mentor says to set image to Fine instead of RAW — how is this different from JPG?

What does 丫 mean? 丫是什么意思？

What adaptations would allow standard fantasy dwarves to survive in the desert?

Why complex landing gears are used instead of simple,reliability and light weight muscle wire or shape memory alloys?

One-one communication

Why not send Voyager 3 and 4 following up the paths taken by Voyager 1 and 2 to re-transmit signals of later as they fly away from Earth?

retrieve food groups from food item list

Printing attributes of selection in ArcPy?

What is a more techy Technical Writer job title that isn't cutesy or confusing?

What does it mean that physics no longer uses mechanical models to describe phenomena?

Can two person see the same photon?

Test print coming out spongy

Relating to the President and obstruction, were Mueller's conclusions preordained?

Tannaka duality for semisimple groups

How does light 'choose' between wave and particle behaviour?

How can a team of shapeshifters communicate?

Show current row "win streak"

If $vec u$ is orthogonal to both $vec v $ and $vec w$, and $vec u$ not equal to $0$, prove $vec u$ is not in the span of $ vec v$ and $vec w$.

3

1

$begingroup$

QN: If u is orthogonal to both v and w, and u ≠ 0, argue that u is not in the span of v and w.

Where I am at:

I get stuck when it comes to solving my augmented matrix with Gauss Jordan Elimination.

I also tried formulating the following steps to solve the problem.

1. Create instances of u, v and w that pertain to the question. My visualisation in Geogebra can be viewed here: https://ggbm.at/b6xvwhpa




2. Set u = av + bw = u (where a and b are constants)



3. Disprove (2)

However, I could not get past step 1.

Any pointers would be greatly appreciated.

linear-algebra matrices

share|cite|improve this question

edited 2 days ago

YuiTo Cheng

2,58641037

asked 2 days ago

Dimen3ionalDimen3ional

232

New contributor

Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

3

1

$begingroup$

QN: If u is orthogonal to both v and w, and u ≠ 0, argue that u is not in the span of v and w.

Where I am at:

I get stuck when it comes to solving my augmented matrix with Gauss Jordan Elimination.

I also tried formulating the following steps to solve the problem.

1. Create instances of u, v and w that pertain to the question. My visualisation in Geogebra can be viewed here: https://ggbm.at/b6xvwhpa




2. Set u = av + bw = u (where a and b are constants)



3. Disprove (2)

However, I could not get past step 1.

Any pointers would be greatly appreciated.

linear-algebra matrices

share|cite|improve this question

edited 2 days ago

YuiTo Cheng

2,58641037

asked 2 days ago

Dimen3ionalDimen3ional

232

New contributor

Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

QN: If u is orthogonal to both v and w, and u ≠ 0, argue that u is not in the span of v and w.

Where I am at:

I get stuck when it comes to solving my augmented matrix with Gauss Jordan Elimination.

I also tried formulating the following steps to solve the problem.

1. Create instances of u, v and w that pertain to the question. My visualisation in Geogebra can be viewed here: https://ggbm.at/b6xvwhpa




2. Set u = av + bw = u (where a and b are constants)



3. Disprove (2)

However, I could not get past step 1.

Any pointers would be greatly appreciated.

linear-algebra matrices

share|cite|improve this question

edited 2 days ago

YuiTo Cheng

2,58641037

asked 2 days ago

Dimen3ionalDimen3ional

232

New contributor

Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 2 days ago

YuiTo Cheng

2,58641037

asked 2 days ago

Dimen3ionalDimen3ional

232

New contributor

Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 2 days ago

YuiTo Cheng

2,58641037

asked 2 days ago

Dimen3ionalDimen3ional

232

New contributor

Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 2 days ago

YuiTo Cheng

2,58641037

edited 2 days ago

YuiTo Cheng

2,58641037

asked 2 days ago

Dimen3ionalDimen3ional

232

New contributor

Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 days ago

Dimen3ionalDimen3ional

232

2 Answers
2

active

oldest

votes

7

$begingroup$

$defmyvec#1bf#1$
This is the same as saying

if $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, and $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=myvec 0$.

So, if $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=amyvec v+bmyvec w$ for some scalars $a,b$. Assuming also $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, this means $myvec ucdot myvec v=0$ and $myvec ucdotmyvec w=0$, so
$$myvec ucdot myvec u=myvec ucdot(amyvec v+bmyvec w)=a(myvec ucdot myvec v)+b(myvec ucdot myvec w)=a0+b0=0 .$$
Since $myvec ucdot myvec u=0$ we have $myvec u=myvec 0$.

share|cite|improve this answer

answered 2 days ago

DavidDavid

70.2k668131

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
  $endgroup$
  – Dimen3ional
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Have you studied proof by contradiction or by contrapositive? That's what this is.
  $endgroup$
  – David
  2 days ago
  
  

  
  
  
  

add a comment | 

2

$begingroup$

If $u$ belongs to the span of $v$ and $w$ the $u=av+bw$ for some scalars $a$ and $b$. Since $langle u, v rangle=0$ and $langle u, w rangle=0$ we get $langle u, (av+bw) rangle=0$ so $langle u, u rangle=0$. This means $u=0$ which is a contradiction.

share|cite|improve this answer

answered 2 days ago

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

$endgroup$

add a comment | 

Your Answer

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',
autoActivateHeartbeat: false,
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
);



);

Dimen3ional is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3192025%2fif-vec-u-is-orthogonal-to-both-vec-v-and-vec-w-and-vec-u-not-equal%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

7

$begingroup$

$defmyvec#1bf#1$
This is the same as saying

if $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, and $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=myvec 0$.

So, if $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=amyvec v+bmyvec w$ for some scalars $a,b$. Assuming also $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, this means $myvec ucdot myvec v=0$ and $myvec ucdotmyvec w=0$, so
$$myvec ucdot myvec u=myvec ucdot(amyvec v+bmyvec w)=a(myvec ucdot myvec v)+b(myvec ucdot myvec w)=a0+b0=0 .$$
Since $myvec ucdot myvec u=0$ we have $myvec u=myvec 0$.

share|cite|improve this answer

answered 2 days ago

DavidDavid

70.2k668131

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
  $endgroup$
  – Dimen3ional
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Have you studied proof by contradiction or by contrapositive? That's what this is.
  $endgroup$
  – David
  2 days ago
  
  

  
  
  
  

add a comment | 

7

$begingroup$

$defmyvec#1bf#1$
This is the same as saying

if $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, and $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=myvec 0$.

So, if $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=amyvec v+bmyvec w$ for some scalars $a,b$. Assuming also $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, this means $myvec ucdot myvec v=0$ and $myvec ucdotmyvec w=0$, so
$$myvec ucdot myvec u=myvec ucdot(amyvec v+bmyvec w)=a(myvec ucdot myvec v)+b(myvec ucdot myvec w)=a0+b0=0 .$$
Since $myvec ucdot myvec u=0$ we have $myvec u=myvec 0$.

share|cite|improve this answer

answered 2 days ago

DavidDavid

70.2k668131

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
  $endgroup$
  – Dimen3ional
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Have you studied proof by contradiction or by contrapositive? That's what this is.
  $endgroup$
  – David
  2 days ago
  
  

  
  
  
  

add a comment | 

$begingroup$

$defmyvec#1bf#1$
This is the same as saying

if $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, and $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=myvec 0$.

So, if $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=amyvec v+bmyvec w$ for some scalars $a,b$. Assuming also $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, this means $myvec ucdot myvec v=0$ and $myvec ucdotmyvec w=0$, so
$$myvec ucdot myvec u=myvec ucdot(amyvec v+bmyvec w)=a(myvec ucdot myvec v)+b(myvec ucdot myvec w)=a0+b0=0 .$$
Since $myvec ucdot myvec u=0$ we have $myvec u=myvec 0$.

share|cite|improve this answer

answered 2 days ago

DavidDavid

70.2k668131

$endgroup$

share|cite|improve this answer

answered 2 days ago

DavidDavid

70.2k668131

answered 2 days ago

DavidDavid

70.2k668131

answered 2 days ago

DavidDavid

70.2k668131

2

$begingroup$

If $u$ belongs to the span of $v$ and $w$ the $u=av+bw$ for some scalars $a$ and $b$. Since $langle u, v rangle=0$ and $langle u, w rangle=0$ we get $langle u, (av+bw) rangle=0$ so $langle u, u rangle=0$. This means $u=0$ which is a contradiction.

share|cite|improve this answer

answered 2 days ago

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

$endgroup$

add a comment | 

2

$begingroup$

If $u$ belongs to the span of $v$ and $w$ the $u=av+bw$ for some scalars $a$ and $b$. Since $langle u, v rangle=0$ and $langle u, w rangle=0$ we get $langle u, (av+bw) rangle=0$ so $langle u, u rangle=0$. This means $u=0$ which is a contradiction.

share|cite|improve this answer

answered 2 days ago

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

$endgroup$

add a comment | 

$begingroup$

If $u$ belongs to the span of $v$ and $w$ the $u=av+bw$ for some scalars $a$ and $b$. Since $langle u, v rangle=0$ and $langle u, w rangle=0$ we get $langle u, (av+bw) rangle=0$ so $langle u, u rangle=0$. This means $u=0$ which is a contradiction.

share|cite|improve this answer

answered 2 days ago

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

$endgroup$

share|cite|improve this answer

answered 2 days ago

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

answered 2 days ago

Kavi Rama MurthyKavi Rama Murthy

76.4k53370

answered 2 days ago

Kavi Rama MurthyKavi Rama Murthy

76.4k53370