Mixing
Euclidean distance of set of orthogonal vectors
$begingroup$
Let's define $x$ as a vector in $mathbb R^n$
Let's define $V$ as the set of all vectors orthogonal to $x$, i.e $V$={$y$ in $mathbb R^n$|$x·y=0$}
Let's define $z$ as another vector in $mathbb R^n$
Calculate the distance between $z$ and the nearest point to $z$ in $V$, i.e min||z-y|| for a vector $y$ in $V$.
After thinking about this, would the answer be $0$? For example, let's say x is the z-axis (0,0,1). So the vectors in V would be the ones around it of any length in the x-y axis. If z is any other vector in $mathbb R^n$, wouldn't the euclidean distance between z and a vector in V be $0$? Because you could find any vector in V that would intersect z or be infinitesimally close.
If my thinking process is wrong, any help would be great! I'm looking for a way to formalize my thoughts better :)
linear-algebra vector-spaces orthogonality
edited Jan 25 at 19:16
J. W. Tanner
3,4601320
asked Jan 25 at 17:51
AnthonyAnthony
35519
$endgroup$
add a comment |
$begingroup$
Let's define $x$ as a vector in $mathbb R^n$
Let's define $V$ as the set of all vectors orthogonal to $x$, i.e $V$={$y$ in $mathbb R^n$|$x·y=0$}
Let's define $z$ as another vector in $mathbb R^n$
Calculate the distance between $z$ and the nearest point to $z$ in $V$, i.e min||z-y|| for a vector $y$ in $V$.
After thinking about this, would the answer be $0$? For example, let's say x is the z-axis (0,0,1). So the vectors in V would be the ones around it of any length in the x-y axis. If z is any other vector in $mathbb R^n$, wouldn't the euclidean distance between z and a vector in V be $0$? Because you could find any vector in V that would intersect z or be infinitesimally close.
If my thinking process is wrong, any help would be great! I'm looking for a way to formalize my thoughts better :)
linear-algebra vector-spaces orthogonality
edited Jan 25 at 19:16
J. W. Tanner
3,4601320
asked Jan 25 at 17:51
AnthonyAnthony
35519
$endgroup$
add a comment |
$begingroup$
Let's define $x$ as a vector in $mathbb R^n$
Let's define $V$ as the set of all vectors orthogonal to $x$, i.e $V$={$y$ in $mathbb R^n$|$x·y=0$}
Let's define $z$ as another vector in $mathbb R^n$
Calculate the distance between $z$ and the nearest point to $z$ in $V$, i.e min||z-y|| for a vector $y$ in $V$.
After thinking about this, would the answer be $0$? For example, let's say x is the z-axis (0,0,1). So the vectors in V would be the ones around it of any length in the x-y axis. If z is any other vector in $mathbb R^n$, wouldn't the euclidean distance between z and a vector in V be $0$? Because you could find any vector in V that would intersect z or be infinitesimally close.
If my thinking process is wrong, any help would be great! I'm looking for a way to formalize my thoughts better :)
linear-algebra vector-spaces orthogonality
edited Jan 25 at 19:16
J. W. Tanner
3,4601320
asked Jan 25 at 17:51
AnthonyAnthony
35519
$endgroup$
Let's define $x$ as a vector in $mathbb R^n$
Let's define $V$ as the set of all vectors orthogonal to $x$, i.e $V$={$y$ in $mathbb R^n$|$x·y=0$}
Let's define $z$ as another vector in $mathbb R^n$
Calculate the distance between $z$ and the nearest point to $z$ in $V$, i.e min||z-y|| for a vector $y$ in $V$.
After thinking about this, would the answer be $0$? For example, let's say x is the z-axis (0,0,1). So the vectors in V would be the ones around it of any length in the x-y axis. If z is any other vector in $mathbb R^n$, wouldn't the euclidean distance between z and a vector in V be $0$? Because you could find any vector in V that would intersect z or be infinitesimally close.
If my thinking process is wrong, any help would be great! I'm looking for a way to formalize my thoughts better :)
linear-algebra vector-spaces orthogonality
linear-algebra vector-spaces orthogonality
edited Jan 25 at 19:16
J. W. Tanner
3,4601320
asked Jan 25 at 17:51
AnthonyAnthony
35519
edited Jan 25 at 19:16
J. W. Tanner
3,4601320
asked Jan 25 at 17:51
AnthonyAnthony
35519
edited Jan 25 at 19:16
J. W. Tanner
3,4601320
edited Jan 25 at 19:16
J. W. Tanner
3,4601320
edited Jan 25 at 19:16
J. W. Tanner
3,4601320
3,4601320
asked Jan 25 at 17:51
AnthonyAnthony
35519
asked Jan 25 at 17:51
AnthonyAnthony
35519
asked Jan 25 at 17:51
AnthonyAnthony
35519
35519
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Are you familiar with orthogonal decomposition?
If we take a vector x as you have done, then we could normalize it and obtain an orthonormal basis with $frac{x}{||x||}$ as a first vector.
Your V would then be the space created by all the other vectors in the orthonormal basis.
Also, what do u really mean by "intersection"? If what you defined as $z$ has a non-zero component in the $frac{x}{||x||}$ axis, then your minimum will be bigger than zero.
If you have any questions, feel free to ask
Edit: The correct generalization of the distance from a given vector to the given subspace is the projection of the given vector onto the subspace. The projection can be found by, for example, switching your vector into a basis where the some vectors are the ones that create the subspace and thus looking at the components of the vector in the remaining subspaces.
answered Jan 25 at 18:31
Marat AlievMarat Aliev
1312
$endgroup$
$begingroup$
Hi, thanks for the answer! I think I remember orthogonal decomposition (it's been a while) and your explanation makes sense. To make sure I'm understanding this correctly, the other vectors in the orthonormal basis would be (1,0,0) and (0,1,0) and (x,y,0) where root(x^2+y^2)=1, right? And can you explain a bit more about the minimum being greater than zero? Thanks!
$endgroup$
– Anthony
Jan 25 at 18:42
$begingroup$
Yes, for your example basis you can complete it with (1,0,0) and (0,1,0).For the minimum, the key thing to realize is that the distance between a vector and some subspace can only be zero when the vector itself is part of the subspace. In your example, if you have take a vector where the z component is non zero, it cannot be part of that subspace because it wouldn't necessarily be orthogonal.
$endgroup$
– Marat Aliev
Jan 25 at 18:45
$begingroup$
Ah, that totally makes sense to me. If I try to explain this more, my thoughts were that since the set V contains all vectors orthogonal to the z-axis, V contains all vectors of the form (x,y,0), where x,y∈ℝ. So if z was (0,0,1) and a vector in V was (x,0,0) where x was infinitesimally small, wouldn't the distance between the vector and z be almost 0?
$endgroup$
– Anthony
Jan 25 at 18:56
$begingroup$
Ah, well i understand your confusion. This isn't what people commonly mean by distance when vectors are involved. You seem to be thinking that the distance is 0 when vectors intersect when taken as lines from the origin. But in that case, since every vector is given relative to the origin, wouldnt all vectors intersect(at the origin)?
$endgroup$
– Marat Aliev
Jan 25 at 18:57
$begingroup$
You're right. I understand that now (my bad!). But am I still correct that V contains all vectors of the form (x,y,0), where x,y∈ℝ? Because if that's the case, I still find it difficult to generalize this answer without using this specific example :(
$endgroup$
– Anthony
Jan 25 at 19:07
|
show 2 more comments
$begingroup$
In your example you chose that $x=(0,0,1)Rightarrow y=(x,y,0)$ where $x,yinmathbb{R}$. If you choose that $z=(1,1,1)$ in your example, you will have that $min|z-y|=min|(1-x,1-y,1)|=1$.
answered Jan 25 at 18:41
kvangkvang
1
$endgroup$
$begingroup$
That makes sense. Do you have any idea on how to generalize that example? or would the minimum always be 1
$endgroup$
– Anthony
Jan 25 at 18:45
$begingroup$
The general idea is that the minimum distance between z and y is how much z "points" in the x-direction. If we let $z=(a,b,c)$ we have $min|(a-x,b-y,c)|=|c|$, since $x$ and $y$ are just arbitrary numbers we choose $x=a$ and $y=b$.
$endgroup$
– kvang
Jan 25 at 18:54
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',
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
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3087406%2feuclidean-distance-of-set-of-orthogonal-vectors%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Are you familiar with orthogonal decomposition?
If we take a vector x as you have done, then we could normalize it and obtain an orthonormal basis with $frac{x}{||x||}$ as a first vector.
Your V would then be the space created by all the other vectors in the orthonormal basis.
Also, what do u really mean by "intersection"? If what you defined as $z$ has a non-zero component in the $frac{x}{||x||}$ axis, then your minimum will be bigger than zero.
If you have any questions, feel free to ask
Edit: The correct generalization of the distance from a given vector to the given subspace is the projection of the given vector onto the subspace. The projection can be found by, for example, switching your vector into a basis where the some vectors are the ones that create the subspace and thus looking at the components of the vector in the remaining subspaces.
answered Jan 25 at 18:31
Marat AlievMarat Aliev
1312
$endgroup$
$begingroup$
Hi, thanks for the answer! I think I remember orthogonal decomposition (it's been a while) and your explanation makes sense. To make sure I'm understanding this correctly, the other vectors in the orthonormal basis would be (1,0,0) and (0,1,0) and (x,y,0) where root(x^2+y^2)=1, right? And can you explain a bit more about the minimum being greater than zero? Thanks!
$endgroup$
– Anthony
Jan 25 at 18:42
$begingroup$
Yes, for your example basis you can complete it with (1,0,0) and (0,1,0).For the minimum, the key thing to realize is that the distance between a vector and some subspace can only be zero when the vector itself is part of the subspace. In your example, if you have take a vector where the z component is non zero, it cannot be part of that subspace because it wouldn't necessarily be orthogonal.
$endgroup$
– Marat Aliev
Jan 25 at 18:45
$begingroup$
Ah, that totally makes sense to me. If I try to explain this more, my thoughts were that since the set V contains all vectors orthogonal to the z-axis, V contains all vectors of the form (x,y,0), where x,y∈ℝ. So if z was (0,0,1) and a vector in V was (x,0,0) where x was infinitesimally small, wouldn't the distance between the vector and z be almost 0?
$endgroup$
– Anthony
Jan 25 at 18:56
$begingroup$
Ah, well i understand your confusion. This isn't what people commonly mean by distance when vectors are involved. You seem to be thinking that the distance is 0 when vectors intersect when taken as lines from the origin. But in that case, since every vector is given relative to the origin, wouldnt all vectors intersect(at the origin)?
$endgroup$
– Marat Aliev
Jan 25 at 18:57
$begingroup$
You're right. I understand that now (my bad!). But am I still correct that V contains all vectors of the form (x,y,0), where x,y∈ℝ? Because if that's the case, I still find it difficult to generalize this answer without using this specific example :(
$endgroup$
– Anthony
Jan 25 at 19:07
|
show 2 more comments
$begingroup$
Are you familiar with orthogonal decomposition?
If we take a vector x as you have done, then we could normalize it and obtain an orthonormal basis with $frac{x}{||x||}$ as a first vector.
Your V would then be the space created by all the other vectors in the orthonormal basis.
Also, what do u really mean by "intersection"? If what you defined as $z$ has a non-zero component in the $frac{x}{||x||}$ axis, then your minimum will be bigger than zero.
If you have any questions, feel free to ask
Edit: The correct generalization of the distance from a given vector to the given subspace is the projection of the given vector onto the subspace. The projection can be found by, for example, switching your vector into a basis where the some vectors are the ones that create the subspace and thus looking at the components of the vector in the remaining subspaces.
answered Jan 25 at 18:31
Marat AlievMarat Aliev
1312
$endgroup$
$begingroup$
Hi, thanks for the answer! I think I remember orthogonal decomposition (it's been a while) and your explanation makes sense. To make sure I'm understanding this correctly, the other vectors in the orthonormal basis would be (1,0,0) and (0,1,0) and (x,y,0) where root(x^2+y^2)=1, right? And can you explain a bit more about the minimum being greater than zero? Thanks!
$endgroup$
– Anthony
Jan 25 at 18:42
$begingroup$
Yes, for your example basis you can complete it with (1,0,0) and (0,1,0).For the minimum, the key thing to realize is that the distance between a vector and some subspace can only be zero when the vector itself is part of the subspace. In your example, if you have take a vector where the z component is non zero, it cannot be part of that subspace because it wouldn't necessarily be orthogonal.
$endgroup$
– Marat Aliev
Jan 25 at 18:45
$begingroup$
Ah, that totally makes sense to me. If I try to explain this more, my thoughts were that since the set V contains all vectors orthogonal to the z-axis, V contains all vectors of the form (x,y,0), where x,y∈ℝ. So if z was (0,0,1) and a vector in V was (x,0,0) where x was infinitesimally small, wouldn't the distance between the vector and z be almost 0?
$endgroup$
– Anthony
Jan 25 at 18:56
$begingroup$
Ah, well i understand your confusion. This isn't what people commonly mean by distance when vectors are involved. You seem to be thinking that the distance is 0 when vectors intersect when taken as lines from the origin. But in that case, since every vector is given relative to the origin, wouldnt all vectors intersect(at the origin)?
$endgroup$
– Marat Aliev
Jan 25 at 18:57
$begingroup$
You're right. I understand that now (my bad!). But am I still correct that V contains all vectors of the form (x,y,0), where x,y∈ℝ? Because if that's the case, I still find it difficult to generalize this answer without using this specific example :(
$endgroup$
– Anthony
Jan 25 at 19:07
|
show 2 more comments
$begingroup$
Are you familiar with orthogonal decomposition?
If we take a vector x as you have done, then we could normalize it and obtain an orthonormal basis with $frac{x}{||x||}$ as a first vector.
Your V would then be the space created by all the other vectors in the orthonormal basis.
Also, what do u really mean by "intersection"? If what you defined as $z$ has a non-zero component in the $frac{x}{||x||}$ axis, then your minimum will be bigger than zero.
If you have any questions, feel free to ask
Edit: The correct generalization of the distance from a given vector to the given subspace is the projection of the given vector onto the subspace. The projection can be found by, for example, switching your vector into a basis where the some vectors are the ones that create the subspace and thus looking at the components of the vector in the remaining subspaces.
answered Jan 25 at 18:31
Marat AlievMarat Aliev
1312
$endgroup$
Are you familiar with orthogonal decomposition?
If we take a vector x as you have done, then we could normalize it and obtain an orthonormal basis with $frac{x}{||x||}$ as a first vector.
Your V would then be the space created by all the other vectors in the orthonormal basis.
Also, what do u really mean by "intersection"? If what you defined as $z$ has a non-zero component in the $frac{x}{||x||}$ axis, then your minimum will be bigger than zero.
If you have any questions, feel free to ask
Edit: The correct generalization of the distance from a given vector to the given subspace is the projection of the given vector onto the subspace. The projection can be found by, for example, switching your vector into a basis where the some vectors are the ones that create the subspace and thus looking at the components of the vector in the remaining subspaces.
answered Jan 25 at 18:31
Marat AlievMarat Aliev
1312
edited Jan 25 at 19:19
answered Jan 25 at 18:31
Marat AlievMarat Aliev
1312
answered Jan 25 at 18:31
Marat AlievMarat Aliev
1312
answered Jan 25 at 18:31
Marat AlievMarat Aliev
1312
1312
$begingroup$
Hi, thanks for the answer! I think I remember orthogonal decomposition (it's been a while) and your explanation makes sense. To make sure I'm understanding this correctly, the other vectors in the orthonormal basis would be (1,0,0) and (0,1,0) and (x,y,0) where root(x^2+y^2)=1, right? And can you explain a bit more about the minimum being greater than zero? Thanks!
$endgroup$
– Anthony
Jan 25 at 18:42
$begingroup$
Yes, for your example basis you can complete it with (1,0,0) and (0,1,0).For the minimum, the key thing to realize is that the distance between a vector and some subspace can only be zero when the vector itself is part of the subspace. In your example, if you have take a vector where the z component is non zero, it cannot be part of that subspace because it wouldn't necessarily be orthogonal.
$endgroup$
– Marat Aliev
Jan 25 at 18:45
$begingroup$
Ah, that totally makes sense to me. If I try to explain this more, my thoughts were that since the set V contains all vectors orthogonal to the z-axis, V contains all vectors of the form (x,y,0), where x,y∈ℝ. So if z was (0,0,1) and a vector in V was (x,0,0) where x was infinitesimally small, wouldn't the distance between the vector and z be almost 0?
$endgroup$
– Anthony
Jan 25 at 18:56
$begingroup$
Ah, well i understand your confusion. This isn't what people commonly mean by distance when vectors are involved. You seem to be thinking that the distance is 0 when vectors intersect when taken as lines from the origin. But in that case, since every vector is given relative to the origin, wouldnt all vectors intersect(at the origin)?
$endgroup$
– Marat Aliev
Jan 25 at 18:57
$begingroup$
You're right. I understand that now (my bad!). But am I still correct that V contains all vectors of the form (x,y,0), where x,y∈ℝ? Because if that's the case, I still find it difficult to generalize this answer without using this specific example :(
$endgroup$
– Anthony
Jan 25 at 19:07
|
show 2 more comments
$begingroup$
Hi, thanks for the answer! I think I remember orthogonal decomposition (it's been a while) and your explanation makes sense. To make sure I'm understanding this correctly, the other vectors in the orthonormal basis would be (1,0,0) and (0,1,0) and (x,y,0) where root(x^2+y^2)=1, right? And can you explain a bit more about the minimum being greater than zero? Thanks!
$endgroup$
– Anthony
Jan 25 at 18:42
$begingroup$
Yes, for your example basis you can complete it with (1,0,0) and (0,1,0).For the minimum, the key thing to realize is that the distance between a vector and some subspace can only be zero when the vector itself is part of the subspace. In your example, if you have take a vector where the z component is non zero, it cannot be part of that subspace because it wouldn't necessarily be orthogonal.
$endgroup$
– Marat Aliev
Jan 25 at 18:45
$begingroup$
Ah, that totally makes sense to me. If I try to explain this more, my thoughts were that since the set V contains all vectors orthogonal to the z-axis, V contains all vectors of the form (x,y,0), where x,y∈ℝ. So if z was (0,0,1) and a vector in V was (x,0,0) where x was infinitesimally small, wouldn't the distance between the vector and z be almost 0?
$endgroup$
– Anthony
Jan 25 at 18:56
$begingroup$
Ah, well i understand your confusion. This isn't what people commonly mean by distance when vectors are involved. You seem to be thinking that the distance is 0 when vectors intersect when taken as lines from the origin. But in that case, since every vector is given relative to the origin, wouldnt all vectors intersect(at the origin)?
$endgroup$
– Marat Aliev
Jan 25 at 18:57
$begingroup$
You're right. I understand that now (my bad!). But am I still correct that V contains all vectors of the form (x,y,0), where x,y∈ℝ? Because if that's the case, I still find it difficult to generalize this answer without using this specific example :(
$endgroup$
– Anthony
Jan 25 at 19:07
$begingroup$
Hi, thanks for the answer! I think I remember orthogonal decomposition (it's been a while) and your explanation makes sense. To make sure I'm understanding this correctly, the other vectors in the orthonormal basis would be (1,0,0) and (0,1,0) and (x,y,0) where root(x^2+y^2)=1, right? And can you explain a bit more about the minimum being greater than zero? Thanks!
$endgroup$
– Anthony
Jan 25 at 18:42
$begingroup$
Hi, thanks for the answer! I think I remember orthogonal decomposition (it's been a while) and your explanation makes sense. To make sure I'm understanding this correctly, the other vectors in the orthonormal basis would be (1,0,0) and (0,1,0) and (x,y,0) where root(x^2+y^2)=1, right? And can you explain a bit more about the minimum being greater than zero? Thanks!
$endgroup$
– Anthony
Jan 25 at 18:42
$begingroup$
Yes, for your example basis you can complete it with (1,0,0) and (0,1,0).For the minimum, the key thing to realize is that the distance between a vector and some subspace can only be zero when the vector itself is part of the subspace. In your example, if you have take a vector where the z component is non zero, it cannot be part of that subspace because it wouldn't necessarily be orthogonal.
$endgroup$
– Marat Aliev
Jan 25 at 18:45
$begingroup$
Yes, for your example basis you can complete it with (1,0,0) and (0,1,0).For the minimum, the key thing to realize is that the distance between a vector and some subspace can only be zero when the vector itself is part of the subspace. In your example, if you have take a vector where the z component is non zero, it cannot be part of that subspace because it wouldn't necessarily be orthogonal.
$endgroup$
– Marat Aliev
Jan 25 at 18:45
$begingroup$
Ah, that totally makes sense to me. If I try to explain this more, my thoughts were that since the set V contains all vectors orthogonal to the z-axis, V contains all vectors of the form (x,y,0), where x,y∈ℝ. So if z was (0,0,1) and a vector in V was (x,0,0) where x was infinitesimally small, wouldn't the distance between the vector and z be almost 0?
$endgroup$
– Anthony
Jan 25 at 18:56
$begingroup$
Ah, that totally makes sense to me. If I try to explain this more, my thoughts were that since the set V contains all vectors orthogonal to the z-axis, V contains all vectors of the form (x,y,0), where x,y∈ℝ. So if z was (0,0,1) and a vector in V was (x,0,0) where x was infinitesimally small, wouldn't the distance between the vector and z be almost 0?
$endgroup$
– Anthony
Jan 25 at 18:56
$begingroup$
Ah, well i understand your confusion. This isn't what people commonly mean by distance when vectors are involved. You seem to be thinking that the distance is 0 when vectors intersect when taken as lines from the origin. But in that case, since every vector is given relative to the origin, wouldnt all vectors intersect(at the origin)?
$endgroup$
– Marat Aliev
Jan 25 at 18:57
$begingroup$
Ah, well i understand your confusion. This isn't what people commonly mean by distance when vectors are involved. You seem to be thinking that the distance is 0 when vectors intersect when taken as lines from the origin. But in that case, since every vector is given relative to the origin, wouldnt all vectors intersect(at the origin)?
$endgroup$
– Marat Aliev
Jan 25 at 18:57
$begingroup$
You're right. I understand that now (my bad!). But am I still correct that V contains all vectors of the form (x,y,0), where x,y∈ℝ? Because if that's the case, I still find it difficult to generalize this answer without using this specific example :(
$endgroup$
– Anthony
Jan 25 at 19:07
$begingroup$
You're right. I understand that now (my bad!). But am I still correct that V contains all vectors of the form (x,y,0), where x,y∈ℝ? Because if that's the case, I still find it difficult to generalize this answer without using this specific example :(
$endgroup$
– Anthony
Jan 25 at 19:07
|
show 2 more comments
$begingroup$
In your example you chose that $x=(0,0,1)Rightarrow y=(x,y,0)$ where $x,yinmathbb{R}$. If you choose that $z=(1,1,1)$ in your example, you will have that $min|z-y|=min|(1-x,1-y,1)|=1$.
answered Jan 25 at 18:41
kvangkvang
1
$endgroup$
$begingroup$
That makes sense. Do you have any idea on how to generalize that example? or would the minimum always be 1
$endgroup$
– Anthony
Jan 25 at 18:45
$begingroup$
The general idea is that the minimum distance between z and y is how much z "points" in the x-direction. If we let $z=(a,b,c)$ we have $min|(a-x,b-y,c)|=|c|$, since $x$ and $y$ are just arbitrary numbers we choose $x=a$ and $y=b$.
$endgroup$
– kvang
Jan 25 at 18:54
add a comment |
$begingroup$
In your example you chose that $x=(0,0,1)Rightarrow y=(x,y,0)$ where $x,yinmathbb{R}$. If you choose that $z=(1,1,1)$ in your example, you will have that $min|z-y|=min|(1-x,1-y,1)|=1$.
answered Jan 25 at 18:41
kvangkvang
1
$endgroup$
$begingroup$
That makes sense. Do you have any idea on how to generalize that example? or would the minimum always be 1
$endgroup$
– Anthony
Jan 25 at 18:45
$begingroup$
The general idea is that the minimum distance between z and y is how much z "points" in the x-direction. If we let $z=(a,b,c)$ we have $min|(a-x,b-y,c)|=|c|$, since $x$ and $y$ are just arbitrary numbers we choose $x=a$ and $y=b$.
$endgroup$
– kvang
Jan 25 at 18:54
add a comment |
$begingroup$
In your example you chose that $x=(0,0,1)Rightarrow y=(x,y,0)$ where $x,yinmathbb{R}$. If you choose that $z=(1,1,1)$ in your example, you will have that $min|z-y|=min|(1-x,1-y,1)|=1$.
answered Jan 25 at 18:41
kvangkvang
1
$endgroup$
In your example you chose that $x=(0,0,1)Rightarrow y=(x,y,0)$ where $x,yinmathbb{R}$. If you choose that $z=(1,1,1)$ in your example, you will have that $min|z-y|=min|(1-x,1-y,1)|=1$.
answered Jan 25 at 18:41
kvangkvang
1
answered Jan 25 at 18:41
kvangkvang
1
answered Jan 25 at 18:41
kvangkvang
1
answered Jan 25 at 18:41
kvangkvang
1
1
$begingroup$
That makes sense. Do you have any idea on how to generalize that example? or would the minimum always be 1
$endgroup$
– Anthony
Jan 25 at 18:45
$begingroup$
The general idea is that the minimum distance between z and y is how much z "points" in the x-direction. If we let $z=(a,b,c)$ we have $min|(a-x,b-y,c)|=|c|$, since $x$ and $y$ are just arbitrary numbers we choose $x=a$ and $y=b$.
$endgroup$
– kvang
Jan 25 at 18:54
add a comment |
$begingroup$
That makes sense. Do you have any idea on how to generalize that example? or would the minimum always be 1
$endgroup$
– Anthony
Jan 25 at 18:45
$begingroup$
The general idea is that the minimum distance between z and y is how much z "points" in the x-direction. If we let $z=(a,b,c)$ we have $min|(a-x,b-y,c)|=|c|$, since $x$ and $y$ are just arbitrary numbers we choose $x=a$ and $y=b$.
$endgroup$
– kvang
Jan 25 at 18:54
$begingroup$
That makes sense. Do you have any idea on how to generalize that example? or would the minimum always be 1
$endgroup$
– Anthony
Jan 25 at 18:45
$begingroup$
That makes sense. Do you have any idea on how to generalize that example? or would the minimum always be 1
$endgroup$
– Anthony
Jan 25 at 18:45
$begingroup$
The general idea is that the minimum distance between z and y is how much z "points" in the x-direction. If we let $z=(a,b,c)$ we have $min|(a-x,b-y,c)|=|c|$, since $x$ and $y$ are just arbitrary numbers we choose $x=a$ and $y=b$.
$endgroup$
– kvang
Jan 25 at 18:54
$begingroup$
The general idea is that the minimum distance between z and y is how much z "points" in the x-direction. If we let $z=(a,b,c)$ we have $min|(a-x,b-y,c)|=|c|$, since $x$ and $y$ are just arbitrary numbers we choose $x=a$ and $y=b$.
$endgroup$
– kvang
Jan 25 at 18:54
add a comment |
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.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3087406%2feuclidean-distance-of-set-of-orthogonal-vectors%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
