Geometric result proof using Vector Algebra

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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

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asked Jul 28 at 14:27

Quasar

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
  

  
  
  
  

add a comment | 

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

asked Jul 28 at 14:27

Quasar

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
  

  
  
  
  

add a comment | 

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

asked Jul 28 at 14:27

Quasar

697412

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

asked Jul 28 at 14:27

Quasar

697412

share|cite|improve this question

share|cite|improve this question

asked Jul 28 at 14:27

Quasar

697412

asked Jul 28 at 14:27

Quasar

697412

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
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  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

add a comment | 

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