Ggthjy

How does the math work when buying airline miles?

What do you call the main part of a joke?

How to tell that you are a giant?

Using et al. for a last / senior author rather than for a first author

Is the Standard Deduction better than Itemized when both are the same amount?

When the Haste spell ends on a creature, do attackers have advantage against that creature?

Crossing US/Canada Border for less than 24 hours

How do I make this wiring inside cabinet safer? (Pic)

Withdrew £2800, but only £2000 shows as withdrawn on online banking; what are my obligations?

Amount of permutations on an NxNxN Rubik's Cube

How to react to hostile behavior from a senior developer?

An adverb for when you're not exaggerating

Should I use a zero-interest credit card for a large one-time purchase?

Has negative voting ever been officially implemented in elections, or seriously proposed, or even studied?

How would a mousetrap for use in space work?

また usage in a dictionary

8 Prisoners wearing hats

How to write this math term? with cases it isn't working

How to compare two different files line by line in unix?

Is safe to use va_start macro with this as parameter?

What would be the ideal power source for a cybernetic eye?

Can a party unilaterally change candidates in preparation for a General election?

What does "lightly crushed" mean for cardamon pods?

How can I use the Python library networkx from Mathematica?

If u is orthogonal to both v and w, and u not equal to 0, argue that u is not in the span of v and w. (

2

$begingroup$

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 52 mins ago

YuiTo Cheng

2,52341037

asked 1 hour ago

Dimen3ionalDimen3ional

132

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 | 

2

$begingroup$

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 52 mins ago

YuiTo Cheng

2,52341037

asked 1 hour ago

Dimen3ionalDimen3ional

132

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 52 mins ago

YuiTo Cheng

2,52341037

asked 1 hour ago

Dimen3ionalDimen3ional

132

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 52 mins ago

YuiTo Cheng

2,52341037

asked 1 hour ago

Dimen3ionalDimen3ional

132

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 52 mins ago

YuiTo Cheng

2,52341037

asked 1 hour ago

Dimen3ionalDimen3ional

132

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 52 mins ago

YuiTo Cheng

2,52341037

edited 52 mins ago

YuiTo Cheng

2,52341037

asked 1 hour ago

Dimen3ionalDimen3ional

132

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 1 hour ago

Dimen3ionalDimen3ional

132

2 Answers
2

active

oldest

votes

3

$begingroup$

$defmyvec#1{{bf#1}}$
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 56 mins ago

DavidDavid

70.1k668131

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  49 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you studied proof by contradiction or by contrapositive? That's what this is.
  $endgroup$
  – David
  48 mins 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 55 mins ago

Kavi Rama MurthyKavi Rama Murthy

75.6k53270

$endgroup$

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-u-is-orthogonal-to-both-v-and-w-and-u-not-equal-to-0-argue-that-u-is-not-in%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

3

$begingroup$

$defmyvec#1{{bf#1}}$
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 56 mins ago

DavidDavid

70.1k668131

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  49 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

3

$begingroup$

$defmyvec#1{{bf#1}}$
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 56 mins ago

DavidDavid

70.1k668131

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  49 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

$defmyvec#1{{bf#1}}$
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 56 mins ago

DavidDavid

70.1k668131

$endgroup$

share|cite|improve this answer

answered 56 mins ago

DavidDavid

70.1k668131

answered 56 mins ago

DavidDavid

70.1k668131

answered 56 mins ago

DavidDavid

70.1k668131

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 55 mins ago

Kavi Rama MurthyKavi Rama Murthy

75.6k53270

$endgroup$

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 55 mins ago

Kavi Rama MurthyKavi Rama Murthy

75.6k53270

$endgroup$

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 55 mins ago

Kavi Rama MurthyKavi Rama Murthy

75.6k53270

$endgroup$

share|cite|improve this answer

answered 55 mins ago

Kavi Rama MurthyKavi Rama Murthy

75.6k53270

answered 55 mins ago

Kavi Rama MurthyKavi Rama Murthy

75.6k53270

answered 55 mins ago

Kavi Rama MurthyKavi Rama Murthy

75.6k53270