Mixing
How do you find a point Q on the line L such that PQ is perpendicular to L
$begingroup$
P is the point (1,1,1) and the line L is given by the equation
x
¯
=
t
(
1
0
-
1
)
linear-algebra orthogonality
asked Jan 22 at 11:36
S KhanS Khan
1
$endgroup$
add a comment |
$begingroup$
P is the point (1,1,1) and the line L is given by the equation
x
¯
=
t
(
1
0
-
1
)
linear-algebra orthogonality
asked Jan 22 at 11:36
S KhanS Khan
1
$endgroup$
$begingroup$
Assume that $Q$ lies on the line, with parameter value $t_0$. Then you know the two lines. What is the angle between them? You can calculate their dot product.
$endgroup$
– Matti P.
Jan 22 at 11:39
add a comment |
$begingroup$
P is the point (1,1,1) and the line L is given by the equation
x
¯
=
t
(
1
0
-
1
)
linear-algebra orthogonality
asked Jan 22 at 11:36
S KhanS Khan
1
$endgroup$
P is the point (1,1,1) and the line L is given by the equation
x
¯
=
t
(
1
0
-
1
)
linear-algebra orthogonality
linear-algebra orthogonality
asked Jan 22 at 11:36
S KhanS Khan
1
asked Jan 22 at 11:36
S KhanS Khan
1
asked Jan 22 at 11:36
S KhanS Khan
1
asked Jan 22 at 11:36
S KhanS Khan
1
asked Jan 22 at 11:36
S KhanS Khan
1
1
$begingroup$
Assume that $Q$ lies on the line, with parameter value $t_0$. Then you know the two lines. What is the angle between them? You can calculate their dot product.
$endgroup$
– Matti P.
Jan 22 at 11:39
add a comment |
$begingroup$
Assume that $Q$ lies on the line, with parameter value $t_0$. Then you know the two lines. What is the angle between them? You can calculate their dot product.
$endgroup$
– Matti P.
Jan 22 at 11:39
$begingroup$
Assume that $Q$ lies on the line, with parameter value $t_0$. Then you know the two lines. What is the angle between them? You can calculate their dot product.
$endgroup$
– Matti P.
Jan 22 at 11:39
$begingroup$
Assume that $Q$ lies on the line, with parameter value $t_0$. Then you know the two lines. What is the angle between them? You can calculate their dot product.
$endgroup$
– Matti P.
Jan 22 at 11:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Point Q is a projection of point P onto line L.
You can find it calculating dot product of vector L and P:
L = (1 0 -1)
Lnorm = L / length(L)
P = (1 1 1)
Q = Lnorm * dot(P, Lnorm)
P.S. But this is not universal. I simplified formula because you line L starts at point (0 0 0). In general case formula will be Q = A + Lnorm * dot(P - A, Lnorm), where A is some point on line L.
answered Jan 22 at 11:49
Anton PetrovAnton Petrov
1262
$endgroup$
$begingroup$
I'm confused about why you need Lnorm and can't just use L, and why you multiply Lnorm with the dot product of Lnorm and P.
$endgroup$
– S Khan
Jan 22 at 23:28
add a comment |
$begingroup$
For $a,b,c,d,e,fin Bbb R$ and $tin Bbb R,$ let $v(t)=(at+b,ct+d,et+f).$ If $L={v(t):tin Bbb R}$ and $P=(x,y,z)$ then the square of the distance from $P$ to any $v(t)in L$ is $G(t)=(at+b-x)^2+(ct+d-y)^2+(et+f-z)^2,$ which is a quadratic function of $t$. When $G(t_0)=min {G(t):tin Bbb R}$ we have $v(t_0)=Q.$
answered Jan 22 at 14:44
DanielWainfleetDanielWainfleet
35.4k31648
$endgroup$
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%2f3083048%2fhow-do-you-find-a-point-q-on-the-line-l-such-that-pq-is-perpendicular-to-l%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
$begingroup$
Point Q is a projection of point P onto line L.
You can find it calculating dot product of vector L and P:
L = (1 0 -1)
Lnorm = L / length(L)
P = (1 1 1)
Q = Lnorm * dot(P, Lnorm)
P.S. But this is not universal. I simplified formula because you line L starts at point (0 0 0). In general case formula will be Q = A + Lnorm * dot(P - A, Lnorm), where A is some point on line L.
answered Jan 22 at 11:49
Anton PetrovAnton Petrov
1262
$endgroup$
$begingroup$
I'm confused about why you need Lnorm and can't just use L, and why you multiply Lnorm with the dot product of Lnorm and P.
$endgroup$
– S Khan
Jan 22 at 23:28
add a comment |
$begingroup$
Point Q is a projection of point P onto line L.
You can find it calculating dot product of vector L and P:
L = (1 0 -1)
Lnorm = L / length(L)
P = (1 1 1)
Q = Lnorm * dot(P, Lnorm)
P.S. But this is not universal. I simplified formula because you line L starts at point (0 0 0). In general case formula will be Q = A + Lnorm * dot(P - A, Lnorm), where A is some point on line L.
answered Jan 22 at 11:49
Anton PetrovAnton Petrov
1262
$endgroup$
$begingroup$
I'm confused about why you need Lnorm and can't just use L, and why you multiply Lnorm with the dot product of Lnorm and P.
$endgroup$
– S Khan
Jan 22 at 23:28
add a comment |
$begingroup$
Point Q is a projection of point P onto line L.
You can find it calculating dot product of vector L and P:
L = (1 0 -1)
Lnorm = L / length(L)
P = (1 1 1)
Q = Lnorm * dot(P, Lnorm)
P.S. But this is not universal. I simplified formula because you line L starts at point (0 0 0). In general case formula will be Q = A + Lnorm * dot(P - A, Lnorm), where A is some point on line L.
answered Jan 22 at 11:49
Anton PetrovAnton Petrov
1262
$endgroup$
Point Q is a projection of point P onto line L.
You can find it calculating dot product of vector L and P:
L = (1 0 -1)
Lnorm = L / length(L)
P = (1 1 1)
Q = Lnorm * dot(P, Lnorm)
P.S. But this is not universal. I simplified formula because you line L starts at point (0 0 0). In general case formula will be Q = A + Lnorm * dot(P - A, Lnorm), where A is some point on line L.
answered Jan 22 at 11:49
Anton PetrovAnton Petrov
1262
answered Jan 22 at 11:49
Anton PetrovAnton Petrov
1262
answered Jan 22 at 11:49
Anton PetrovAnton Petrov
1262
answered Jan 22 at 11:49
Anton PetrovAnton Petrov
1262
1262
$begingroup$
I'm confused about why you need Lnorm and can't just use L, and why you multiply Lnorm with the dot product of Lnorm and P.
$endgroup$
– S Khan
Jan 22 at 23:28
add a comment |
$begingroup$
I'm confused about why you need Lnorm and can't just use L, and why you multiply Lnorm with the dot product of Lnorm and P.
$endgroup$
– S Khan
Jan 22 at 23:28
$begingroup$
I'm confused about why you need Lnorm and can't just use L, and why you multiply Lnorm with the dot product of Lnorm and P.
$endgroup$
– S Khan
Jan 22 at 23:28
$begingroup$
I'm confused about why you need Lnorm and can't just use L, and why you multiply Lnorm with the dot product of Lnorm and P.
$endgroup$
– S Khan
Jan 22 at 23:28
add a comment |
$begingroup$
For $a,b,c,d,e,fin Bbb R$ and $tin Bbb R,$ let $v(t)=(at+b,ct+d,et+f).$ If $L={v(t):tin Bbb R}$ and $P=(x,y,z)$ then the square of the distance from $P$ to any $v(t)in L$ is $G(t)=(at+b-x)^2+(ct+d-y)^2+(et+f-z)^2,$ which is a quadratic function of $t$. When $G(t_0)=min {G(t):tin Bbb R}$ we have $v(t_0)=Q.$
answered Jan 22 at 14:44
DanielWainfleetDanielWainfleet
35.4k31648
$endgroup$
add a comment |
$begingroup$
For $a,b,c,d,e,fin Bbb R$ and $tin Bbb R,$ let $v(t)=(at+b,ct+d,et+f).$ If $L={v(t):tin Bbb R}$ and $P=(x,y,z)$ then the square of the distance from $P$ to any $v(t)in L$ is $G(t)=(at+b-x)^2+(ct+d-y)^2+(et+f-z)^2,$ which is a quadratic function of $t$. When $G(t_0)=min {G(t):tin Bbb R}$ we have $v(t_0)=Q.$
answered Jan 22 at 14:44
DanielWainfleetDanielWainfleet
35.4k31648
$endgroup$
add a comment |
$begingroup$
For $a,b,c,d,e,fin Bbb R$ and $tin Bbb R,$ let $v(t)=(at+b,ct+d,et+f).$ If $L={v(t):tin Bbb R}$ and $P=(x,y,z)$ then the square of the distance from $P$ to any $v(t)in L$ is $G(t)=(at+b-x)^2+(ct+d-y)^2+(et+f-z)^2,$ which is a quadratic function of $t$. When $G(t_0)=min {G(t):tin Bbb R}$ we have $v(t_0)=Q.$
answered Jan 22 at 14:44
DanielWainfleetDanielWainfleet
35.4k31648
$endgroup$
For $a,b,c,d,e,fin Bbb R$ and $tin Bbb R,$ let $v(t)=(at+b,ct+d,et+f).$ If $L={v(t):tin Bbb R}$ and $P=(x,y,z)$ then the square of the distance from $P$ to any $v(t)in L$ is $G(t)=(at+b-x)^2+(ct+d-y)^2+(et+f-z)^2,$ which is a quadratic function of $t$. When $G(t_0)=min {G(t):tin Bbb R}$ we have $v(t_0)=Q.$
answered Jan 22 at 14:44
DanielWainfleetDanielWainfleet
35.4k31648
answered Jan 22 at 14:44
DanielWainfleetDanielWainfleet
35.4k31648
answered Jan 22 at 14:44
DanielWainfleetDanielWainfleet
35.4k31648
answered Jan 22 at 14:44
DanielWainfleetDanielWainfleet
35.4k31648
35.4k31648
add a comment |
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%2f3083048%2fhow-do-you-find-a-point-q-on-the-line-l-such-that-pq-is-perpendicular-to-l%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
$begingroup$
Assume that $Q$ lies on the line, with parameter value $t_0$. Then you know the two lines. What is the angle between them? You can calculate their dot product.
$endgroup$
– Matti P.
Jan 22 at 11:39