Geometric result proof using Vector Algebra

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











up vote
2
down vote

favorite












I am a second year open university undergraduate math student and I am taking a first course on vector calculus. I am trying to prove the below result in geometry using vector algebra. I would like to have my proof verified/simplified if necessary, any inputs will be extremely valuable.



As an aside, what's a good physics book for a typical undergraduate student, that applies the ideas in Vector calculus, that I can pleasure read and solve along side the mathematics?




Lines drawn through a point $E$ and the vertices $A,B,C,D$ of a tetrahedron cut the planes of the opposite faces in $A',B',C',D'$. Show that the sum of the ratios in which these points divide $EA,EB,EC,ED$ is $-1$.




Proof.



From the theorem on co-planar points, we know that, we can always assign non-zero weights $alpha,beta,gamma,delta$ to four co-planar points $A,B,C,D$, such that they form a self-centroidal set.



$alpha A + beta B +gamma C + delta D = 0\
alpha + beta + gamma + delta = 0$



Consider the points $A',B',C',D'$ lying in the plane of the four triangular faces - $BCD$, $CDA$, $DAB$, $ABC$.



Since, $B,C,D,A'$ are co-planar, four non-zero weights can be assigned such that $B,C,D,A'$ form a self-centroidal set.



$beta b + gamma c + delta d + alpha' a' = 0\
beta + gamma + delta + alpha' = 0$



$impliesbeta b + gamma c + delta d = (beta + gamma + delta)a' $



On similar lines, we can write 4 equations for position vectors $a',b',c',d'$:



$beta b + gamma c + delta d = (beta + gamma + delta) a'$



$alpha a + gamma c + delta d = (alpha + gamma + delta) b'$



$alpha a + beta b + delta d = (alpha + beta + delta) c'$



$alpha a + beta b + gamma c = (alpha + beta + gamma) d'$



Subtracting expression (1) - (2),



$beta b - alpha a = (beta + gamma + delta) a' - (alpha + gamma + delta) b'\
beta b + (alpha + gamma + delta) b' = alpha a + (beta + gamma + delta) a'\
fracbeta b + (alpha + gamma + delta) b'alpha + beta + gamma + delta = fracalpha a + (beta + gamma + delta) a'alpha + beta + gamma + delta = e$



$E$ is the point of intersection of $AA,BB'$. Moreover,



$alpha a + (beta + gamma + delta) a' = (alpha + beta + gamma + delta)e\
a'=frac(alpha + beta + gamma + delta)e - alpha abeta + gamma + delta$



Thus, $A'$ divides $EA$ externally in the ratio $-alpha/(alpha + beta + gamma + delta)$. Likewise, $B',C',D'$ divide $EB,EC,ED$ externally in the ratios $-beta/(alpha + beta + gamma + delta), -gamma/(alpha + beta + gamma + delta), -delta/(alpha + beta + gamma + delta)$ respectively. The sum of these ratios is :



$$beginaligned&frac-alphaalpha + beta + gamma + delta+frac-betaalpha + beta + gamma + delta+frac-gammaalpha + beta + gamma + delta+frac-deltaalpha + beta + gamma + delta\&=-frac(alpha + beta + gamma + delta)alpha + beta + gamma + delta \&= -1 endaligned$$




multivariable-calculus proof-verification vectors




share|cite|improve this question















  • 2




    for the physics question, i thought about that today too, and i found Richard Feynman's Lectures on Physics. I started reading them today, and they are really good to read, and a lot of fun. I think there are exercises too, but you would have to look for them, i guess. I found this link b-ok.xyz/book/2139627/eed0be/?_ir=1
    – Pink Panther
    Jul 28 at 16:44






  • 1




    I agree, they build a gut, intuitive feel for physics and are absolute fun! Let me take a look at the exercises.
    – Quasar
    Jul 29 at 7:08






  • 1




    Spivak physics for mathematicians?
    – BCLC
    Aug 4 at 1:31














up vote
2
down vote

favorite












I am a second year open university undergraduate math student and I am taking a first course on vector calculus. I am trying to prove the below result in geometry using vector algebra. I would like to have my proof verified/simplified if necessary, any inputs will be extremely valuable.



As an aside, what's a good physics book for a typical undergraduate student, that applies the ideas in Vector calculus, that I can pleasure read and solve along side the mathematics?




Lines drawn through a point $E$ and the vertices $A,B,C,D$ of a tetrahedron cut the planes of the opposite faces in $A',B',C',D'$. Show that the sum of the ratios in which these points divide $EA,EB,EC,ED$ is $-1$.




Proof.



From the theorem on co-planar points, we know that, we can always assign non-zero weights $alpha,beta,gamma,delta$ to four co-planar points $A,B,C,D$, such that they form a self-centroidal set.



$alpha A + beta B +gamma C + delta D = 0\
alpha + beta + gamma + delta = 0$



Consider the points $A',B',C',D'$ lying in the plane of the four triangular faces - $BCD$, $CDA$, $DAB$, $ABC$.



Since, $B,C,D,A'$ are co-planar, four non-zero weights can be assigned such that $B,C,D,A'$ form a self-centroidal set.



$beta b + gamma c + delta d + alpha' a' = 0\
beta + gamma + delta + alpha' = 0$



$impliesbeta b + gamma c + delta d = (beta + gamma + delta)a' $



On similar lines, we can write 4 equations for position vectors $a',b',c',d'$:



$beta b + gamma c + delta d = (beta + gamma + delta) a'$



$alpha a + gamma c + delta d = (alpha + gamma + delta) b'$



$alpha a + beta b + delta d = (alpha + beta + delta) c'$



$alpha a + beta b + gamma c = (alpha + beta + gamma) d'$



Subtracting expression (1) - (2),



$beta b - alpha a = (beta + gamma + delta) a' - (alpha + gamma + delta) b'\
beta b + (alpha + gamma + delta) b' = alpha a + (beta + gamma + delta) a'\
fracbeta b + (alpha + gamma + delta) b'alpha + beta + gamma + delta = fracalpha a + (beta + gamma + delta) a'alpha + beta + gamma + delta = e$



$E$ is the point of intersection of $AA,BB'$. Moreover,



$alpha a + (beta + gamma + delta) a' = (alpha + beta + gamma + delta)e\
a'=frac(alpha + beta + gamma + delta)e - alpha abeta + gamma + delta$



Thus, $A'$ divides $EA$ externally in the ratio $-alpha/(alpha + beta + gamma + delta)$. Likewise, $B',C',D'$ divide $EB,EC,ED$ externally in the ratios $-beta/(alpha + beta + gamma + delta), -gamma/(alpha + beta + gamma + delta), -delta/(alpha + beta + gamma + delta)$ respectively. The sum of these ratios is :



$$beginaligned&frac-alphaalpha + beta + gamma + delta+frac-betaalpha + beta + gamma + delta+frac-gammaalpha + beta + gamma + delta+frac-deltaalpha + beta + gamma + delta\&=-frac(alpha + beta + gamma + delta)alpha + beta + gamma + delta \&= -1 endaligned$$




multivariable-calculus proof-verification vectors




share|cite|improve this question















  • 2




    for the physics question, i thought about that today too, and i found Richard Feynman's Lectures on Physics. I started reading them today, and they are really good to read, and a lot of fun. I think there are exercises too, but you would have to look for them, i guess. I found this link b-ok.xyz/book/2139627/eed0be/?_ir=1
    – Pink Panther
    Jul 28 at 16:44






  • 1




    I agree, they build a gut, intuitive feel for physics and are absolute fun! Let me take a look at the exercises.
    – Quasar
    Jul 29 at 7:08






  • 1




    Spivak physics for mathematicians?
    – BCLC
    Aug 4 at 1:31












up vote
2
down vote

favorite









up vote
2
down vote

favorite











I am a second year open university undergraduate math student and I am taking a first course on vector calculus. I am trying to prove the below result in geometry using vector algebra. I would like to have my proof verified/simplified if necessary, any inputs will be extremely valuable.



As an aside, what's a good physics book for a typical undergraduate student, that applies the ideas in Vector calculus, that I can pleasure read and solve along side the mathematics?




Lines drawn through a point $E$ and the vertices $A,B,C,D$ of a tetrahedron cut the planes of the opposite faces in $A',B',C',D'$. Show that the sum of the ratios in which these points divide $EA,EB,EC,ED$ is $-1$.




Proof.



From the theorem on co-planar points, we know that, we can always assign non-zero weights $alpha,beta,gamma,delta$ to four co-planar points $A,B,C,D$, such that they form a self-centroidal set.



$alpha A + beta B +gamma C + delta D = 0\
alpha + beta + gamma + delta = 0$



Consider the points $A',B',C',D'$ lying in the plane of the four triangular faces - $BCD$, $CDA$, $DAB$, $ABC$.



Since, $B,C,D,A'$ are co-planar, four non-zero weights can be assigned such that $B,C,D,A'$ form a self-centroidal set.



$beta b + gamma c + delta d + alpha' a' = 0\
beta + gamma + delta + alpha' = 0$



$impliesbeta b + gamma c + delta d = (beta + gamma + delta)a' $



On similar lines, we can write 4 equations for position vectors $a',b',c',d'$:



$beta b + gamma c + delta d = (beta + gamma + delta) a'$



$alpha a + gamma c + delta d = (alpha + gamma + delta) b'$



$alpha a + beta b + delta d = (alpha + beta + delta) c'$



$alpha a + beta b + gamma c = (alpha + beta + gamma) d'$



Subtracting expression (1) - (2),



$beta b - alpha a = (beta + gamma + delta) a' - (alpha + gamma + delta) b'\
beta b + (alpha + gamma + delta) b' = alpha a + (beta + gamma + delta) a'\
fracbeta b + (alpha + gamma + delta) b'alpha + beta + gamma + delta = fracalpha a + (beta + gamma + delta) a'alpha + beta + gamma + delta = e$



$E$ is the point of intersection of $AA,BB'$. Moreover,



$alpha a + (beta + gamma + delta) a' = (alpha + beta + gamma + delta)e\
a'=frac(alpha + beta + gamma + delta)e - alpha abeta + gamma + delta$



Thus, $A'$ divides $EA$ externally in the ratio $-alpha/(alpha + beta + gamma + delta)$. Likewise, $B',C',D'$ divide $EB,EC,ED$ externally in the ratios $-beta/(alpha + beta + gamma + delta), -gamma/(alpha + beta + gamma + delta), -delta/(alpha + beta + gamma + delta)$ respectively. The sum of these ratios is :



$$beginaligned&frac-alphaalpha + beta + gamma + delta+frac-betaalpha + beta + gamma + delta+frac-gammaalpha + beta + gamma + delta+frac-deltaalpha + beta + gamma + delta\&=-frac(alpha + beta + gamma + delta)alpha + beta + gamma + delta \&= -1 endaligned$$




multivariable-calculus proof-verification vectors




share|cite|improve this question











I am a second year open university undergraduate math student and I am taking a first course on vector calculus. I am trying to prove the below result in geometry using vector algebra. I would like to have my proof verified/simplified if necessary, any inputs will be extremely valuable.



As an aside, what's a good physics book for a typical undergraduate student, that applies the ideas in Vector calculus, that I can pleasure read and solve along side the mathematics?




Lines drawn through a point $E$ and the vertices $A,B,C,D$ of a tetrahedron cut the planes of the opposite faces in $A',B',C',D'$. Show that the sum of the ratios in which these points divide $EA,EB,EC,ED$ is $-1$.




Proof.



From the theorem on co-planar points, we know that, we can always assign non-zero weights $alpha,beta,gamma,delta$ to four co-planar points $A,B,C,D$, such that they form a self-centroidal set.



$alpha A + beta B +gamma C + delta D = 0\
alpha + beta + gamma + delta = 0$



Consider the points $A',B',C',D'$ lying in the plane of the four triangular faces - $BCD$, $CDA$, $DAB$, $ABC$.



Since, $B,C,D,A'$ are co-planar, four non-zero weights can be assigned such that $B,C,D,A'$ form a self-centroidal set.



$beta b + gamma c + delta d + alpha' a' = 0\
beta + gamma + delta + alpha' = 0$



$impliesbeta b + gamma c + delta d = (beta + gamma + delta)a' $



On similar lines, we can write 4 equations for position vectors $a',b',c',d'$:



$beta b + gamma c + delta d = (beta + gamma + delta) a'$



$alpha a + gamma c + delta d = (alpha + gamma + delta) b'$



$alpha a + beta b + delta d = (alpha + beta + delta) c'$



$alpha a + beta b + gamma c = (alpha + beta + gamma) d'$



Subtracting expression (1) - (2),



$beta b - alpha a = (beta + gamma + delta) a' - (alpha + gamma + delta) b'\
beta b + (alpha + gamma + delta) b' = alpha a + (beta + gamma + delta) a'\
fracbeta b + (alpha + gamma + delta) b'alpha + beta + gamma + delta = fracalpha a + (beta + gamma + delta) a'alpha + beta + gamma + delta = e$



$E$ is the point of intersection of $AA,BB'$. Moreover,



$alpha a + (beta + gamma + delta) a' = (alpha + beta + gamma + delta)e\
a'=frac(alpha + beta + gamma + delta)e - alpha abeta + gamma + delta$



Thus, $A'$ divides $EA$ externally in the ratio $-alpha/(alpha + beta + gamma + delta)$. Likewise, $B',C',D'$ divide $EB,EC,ED$ externally in the ratios $-beta/(alpha + beta + gamma + delta), -gamma/(alpha + beta + gamma + delta), -delta/(alpha + beta + gamma + delta)$ respectively. The sum of these ratios is :



$$beginaligned&frac-alphaalpha + beta + gamma + delta+frac-betaalpha + beta + gamma + delta+frac-gammaalpha + beta + gamma + delta+frac-deltaalpha + beta + gamma + delta\&=-frac(alpha + beta + gamma + delta)alpha + beta + gamma + delta \&= -1 endaligned$$





multivariable-calculus proof-verification vectors





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 28 at 14:27









Quasar

697412




697412







  • 2




    for the physics question, i thought about that today too, and i found Richard Feynman's Lectures on Physics. I started reading them today, and they are really good to read, and a lot of fun. I think there are exercises too, but you would have to look for them, i guess. I found this link b-ok.xyz/book/2139627/eed0be/?_ir=1
    – Pink Panther
    Jul 28 at 16:44






  • 1




    I agree, they build a gut, intuitive feel for physics and are absolute fun! Let me take a look at the exercises.
    – Quasar
    Jul 29 at 7:08






  • 1




    Spivak physics for mathematicians?
    – BCLC
    Aug 4 at 1:31












  • 2




    for the physics question, i thought about that today too, and i found Richard Feynman's Lectures on Physics. I started reading them today, and they are really good to read, and a lot of fun. I think there are exercises too, but you would have to look for them, i guess. I found this link b-ok.xyz/book/2139627/eed0be/?_ir=1
    – Pink Panther
    Jul 28 at 16:44






  • 1




    I agree, they build a gut, intuitive feel for physics and are absolute fun! Let me take a look at the exercises.
    – Quasar
    Jul 29 at 7:08






  • 1




    Spivak physics for mathematicians?
    – BCLC
    Aug 4 at 1:31







2




2




for the physics question, i thought about that today too, and i found Richard Feynman's Lectures on Physics. I started reading them today, and they are really good to read, and a lot of fun. I think there are exercises too, but you would have to look for them, i guess. I found this link b-ok.xyz/book/2139627/eed0be/?_ir=1
– Pink Panther
Jul 28 at 16:44




for the physics question, i thought about that today too, and i found Richard Feynman's Lectures on Physics. I started reading them today, and they are really good to read, and a lot of fun. I think there are exercises too, but you would have to look for them, i guess. I found this link b-ok.xyz/book/2139627/eed0be/?_ir=1
– Pink Panther
Jul 28 at 16:44




1




1




I agree, they build a gut, intuitive feel for physics and are absolute fun! Let me take a look at the exercises.
– Quasar
Jul 29 at 7:08




I agree, they build a gut, intuitive feel for physics and are absolute fun! Let me take a look at the exercises.
– Quasar
Jul 29 at 7:08




1




1




Spivak physics for mathematicians?
– BCLC
Aug 4 at 1:31




Spivak physics for mathematicians?
– BCLC
Aug 4 at 1:31















active

oldest

votes











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%2f2865289%2fgeometric-result-proof-using-vector-algebra%23new-answer', 'question_page');

);

Post as a guest






















active

oldest

votes













active

oldest

votes









active

oldest

votes






active

oldest

votes










 

draft saved


draft discarded


























 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2865289%2fgeometric-result-proof-using-vector-algebra%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

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

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