$u, v, n$ are vectors in $mathbbR^2$, show that $(u-v)$ is $perp n$

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
-1
down vote

favorite












Let $l : Ax+By+C= 0$ be a straight line. Let $overrightarrowu=u_1overrightarrowi+u_2overrightarrowj$ and $overrightarrowv=v_1overrightarrowi+v_2overrightarrowj$ be two points on $l$ with $overrightarrowu-overrightarrowvneq 0$, $overrightarrown=Aoverrightarrowi+Boverrightarrowj$. Show that$(overrightarrowu-overrightarrowv) perp overrightarrown$.




vector-spaces




share|cite|improve this question





















  • What have you done so far? Tell us where you are stuck.
    – Kavi Rama Murthy
    Jul 25 at 8:28










  • n should be on the line l, shouldn't it? that means it's impossible for l to perpendicular to n?
    – Astrid L
    Jul 25 at 8:32










  • It is not on the line; it is perpendicular to the line.
    – Kavi Rama Murthy
    Jul 25 at 8:34










  • @KaviRamaMurthy but as you can see, n=Ai+Bj. It's at least parallel to l...?
    – Astrid L
    Jul 25 at 8:36














up vote
-1
down vote

favorite












Let $l : Ax+By+C= 0$ be a straight line. Let $overrightarrowu=u_1overrightarrowi+u_2overrightarrowj$ and $overrightarrowv=v_1overrightarrowi+v_2overrightarrowj$ be two points on $l$ with $overrightarrowu-overrightarrowvneq 0$, $overrightarrown=Aoverrightarrowi+Boverrightarrowj$. Show that$(overrightarrowu-overrightarrowv) perp overrightarrown$.




vector-spaces




share|cite|improve this question





















  • What have you done so far? Tell us where you are stuck.
    – Kavi Rama Murthy
    Jul 25 at 8:28










  • n should be on the line l, shouldn't it? that means it's impossible for l to perpendicular to n?
    – Astrid L
    Jul 25 at 8:32










  • It is not on the line; it is perpendicular to the line.
    – Kavi Rama Murthy
    Jul 25 at 8:34










  • @KaviRamaMurthy but as you can see, n=Ai+Bj. It's at least parallel to l...?
    – Astrid L
    Jul 25 at 8:36












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











Let $l : Ax+By+C= 0$ be a straight line. Let $overrightarrowu=u_1overrightarrowi+u_2overrightarrowj$ and $overrightarrowv=v_1overrightarrowi+v_2overrightarrowj$ be two points on $l$ with $overrightarrowu-overrightarrowvneq 0$, $overrightarrown=Aoverrightarrowi+Boverrightarrowj$. Show that$(overrightarrowu-overrightarrowv) perp overrightarrown$.




vector-spaces




share|cite|improve this question













Let $l : Ax+By+C= 0$ be a straight line. Let $overrightarrowu=u_1overrightarrowi+u_2overrightarrowj$ and $overrightarrowv=v_1overrightarrowi+v_2overrightarrowj$ be two points on $l$ with $overrightarrowu-overrightarrowvneq 0$, $overrightarrown=Aoverrightarrowi+Boverrightarrowj$. Show that$(overrightarrowu-overrightarrowv) perp overrightarrown$.





vector-spaces





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 25 at 8:30









Jack J.

3661317




3661317









asked Jul 25 at 8:24









Astrid L

23




23











  • What have you done so far? Tell us where you are stuck.
    – Kavi Rama Murthy
    Jul 25 at 8:28










  • n should be on the line l, shouldn't it? that means it's impossible for l to perpendicular to n?
    – Astrid L
    Jul 25 at 8:32










  • It is not on the line; it is perpendicular to the line.
    – Kavi Rama Murthy
    Jul 25 at 8:34










  • @KaviRamaMurthy but as you can see, n=Ai+Bj. It's at least parallel to l...?
    – Astrid L
    Jul 25 at 8:36
















  • What have you done so far? Tell us where you are stuck.
    – Kavi Rama Murthy
    Jul 25 at 8:28










  • n should be on the line l, shouldn't it? that means it's impossible for l to perpendicular to n?
    – Astrid L
    Jul 25 at 8:32










  • It is not on the line; it is perpendicular to the line.
    – Kavi Rama Murthy
    Jul 25 at 8:34










  • @KaviRamaMurthy but as you can see, n=Ai+Bj. It's at least parallel to l...?
    – Astrid L
    Jul 25 at 8:36















What have you done so far? Tell us where you are stuck.
– Kavi Rama Murthy
Jul 25 at 8:28




What have you done so far? Tell us where you are stuck.
– Kavi Rama Murthy
Jul 25 at 8:28












n should be on the line l, shouldn't it? that means it's impossible for l to perpendicular to n?
– Astrid L
Jul 25 at 8:32




n should be on the line l, shouldn't it? that means it's impossible for l to perpendicular to n?
– Astrid L
Jul 25 at 8:32












It is not on the line; it is perpendicular to the line.
– Kavi Rama Murthy
Jul 25 at 8:34




It is not on the line; it is perpendicular to the line.
– Kavi Rama Murthy
Jul 25 at 8:34












@KaviRamaMurthy but as you can see, n=Ai+Bj. It's at least parallel to l...?
– Astrid L
Jul 25 at 8:36




@KaviRamaMurthy but as you can see, n=Ai+Bj. It's at least parallel to l...?
– Astrid L
Jul 25 at 8:36










1 Answer
1






active

oldest

votes

















up vote
0
down vote













Show that the scalar product $(overrightarrowu-overrightarrowv) cdot overrightarrown=0$.



Use that $overrightarrowu cdot overrightarrown= -C$ and $overrightarrowv cdot overrightarrown= -C$.






share|cite|improve this answer





















  • What is the second line doing?
    – Astrid L
    Jul 25 at 12:02










  • $overrightarrowu$ is on $l$, hence $Au_1+Bu_2+C=0$.
    – Fred
    Jul 26 at 5:57










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: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
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%2f2862171%2fu-v-n-are-vectors-in-mathbbr2-show-that-u-v-is-perp-n%23new-answer', 'question_page');

);

Post as a guest

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote













Show that the scalar product $(overrightarrowu-overrightarrowv) cdot overrightarrown=0$.



Use that $overrightarrowu cdot overrightarrown= -C$ and $overrightarrowv cdot overrightarrown= -C$.






share|cite|improve this answer





















  • What is the second line doing?
    – Astrid L
    Jul 25 at 12:02










  • $overrightarrowu$ is on $l$, hence $Au_1+Bu_2+C=0$.
    – Fred
    Jul 26 at 5:57














up vote
0
down vote













Show that the scalar product $(overrightarrowu-overrightarrowv) cdot overrightarrown=0$.



Use that $overrightarrowu cdot overrightarrown= -C$ and $overrightarrowv cdot overrightarrown= -C$.






share|cite|improve this answer





















  • What is the second line doing?
    – Astrid L
    Jul 25 at 12:02










  • $overrightarrowu$ is on $l$, hence $Au_1+Bu_2+C=0$.
    – Fred
    Jul 26 at 5:57












up vote
0
down vote










up vote
0
down vote









Show that the scalar product $(overrightarrowu-overrightarrowv) cdot overrightarrown=0$.



Use that $overrightarrowu cdot overrightarrown= -C$ and $overrightarrowv cdot overrightarrown= -C$.






share|cite|improve this answer













Show that the scalar product $(overrightarrowu-overrightarrowv) cdot overrightarrown=0$.



Use that $overrightarrowu cdot overrightarrown= -C$ and $overrightarrowv cdot overrightarrown= -C$.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 25 at 8:29









Fred

37.2k1237




37.2k1237











  • What is the second line doing?
    – Astrid L
    Jul 25 at 12:02










  • $overrightarrowu$ is on $l$, hence $Au_1+Bu_2+C=0$.
    – Fred
    Jul 26 at 5:57
















  • What is the second line doing?
    – Astrid L
    Jul 25 at 12:02










  • $overrightarrowu$ is on $l$, hence $Au_1+Bu_2+C=0$.
    – Fred
    Jul 26 at 5:57















What is the second line doing?
– Astrid L
Jul 25 at 12:02




What is the second line doing?
– Astrid L
Jul 25 at 12:02












$overrightarrowu$ is on $l$, hence $Au_1+Bu_2+C=0$.
– Fred
Jul 26 at 5:57




$overrightarrowu$ is on $l$, hence $Au_1+Bu_2+C=0$.
– Fred
Jul 26 at 5:57












 

draft saved


draft discarded


























 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2862171%2fu-v-n-are-vectors-in-mathbbr2-show-that-u-v-is-perp-n%23new-answer', 'question_page');

);

Post as a guest
































































Comments

Post a Comment

Popular posts from this blog

What is the equation of a 3D cone with generalised tilt?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
Read more

Color the edges and diagonals of a regular polygon

Image
Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
Read more

Relationship between determinant of matrix and determinant of adjoint?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
Read more