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Basic theorem from vector analysis - Condition that 4 points $A,B,C,D$ no three of which are collinear, will lie in a plane
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I am taking a first course in Vector calculus. I am attempting to write a proof for the following theorem.
The proof in the book splits this into two cases -
(i) either the $vec AB$ is parallel to $vec CD$, so $vec AB = k cdotvecCD$
(ii) $vec AB$ intersects $vec CD$ at some point $P$, so $p=fraca+lambda b1+lambda=fracc+lambda' d1+lambda'$
Although I know the section formula, I couldn't come up with the above formulation on my own. The below is my proof. Could someone comment, on whether the proof is elegant, and also acceptable?
Theorem. Four points $A,B,C,D,$ no three of which are collinear, will lie in a plane when and only when there exist four non-zero numbers $alpha,beta,gamma,delta,$ such that
$$alpha a+ beta b + gamma c + delta d = 0 \
alpha + beta + gamma + delta = 0 $$
Proof.
Idea: If the four points are coplanar, we should be able to construct a parallelogram of the diagonal on $vecCD$, with sides as scalar multiples of $vec AB$ and $vec CD$.
Let
$beginaligned
vec CD &= lambda vecAB + lambda' vecBC\
(d - c) &= lambda (b - a) + lambda'(c - b)\
d &= -lambda a + (lambda - lambda')b + (1 + lambda')c
endaligned$
We would like the coefficients of $a,b,c$ to be $alpha,beta,gamma$ respective.
Set : $-lambda = fracalphaalpha + beta + gamma,
1+lambda' = fracgammaalpha + beta + gamma$. Solving for $lambda,lambda'$ we get :
$$beginaligned
lambda' &= -frac(alpha + beta)alpha + beta + gamma\
lambda - lambda' &= fracbetaalpha + beta + gamma
endaligned$$
So, we have :
$$beginaligned
d &= fracalphaalpha + beta + gammaa + fracbetaalpha + beta + gammab + fracgammaalpha + beta + gammac \\
(alpha + beta + gamma)d &= alpha a + beta b + gamma c
endaligned$$
Let $delta = -(alpha + beta + gamma)$. Thus,
$$
alpha a + beta b + gamma c + delta d = 0
$$
Further, these constants are non-zero, since $lambda,lambda'$ cannot be $0,infty$.
multivariable-calculus proof-verification proof-writing vectors
asked Jul 15 at 3:19
Quasar
697412
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up vote
0
down vote
favorite
I am taking a first course in Vector calculus. I am attempting to write a proof for the following theorem.
The proof in the book splits this into two cases -
(i) either the $vec AB$ is parallel to $vec CD$, so $vec AB = k cdotvecCD$
(ii) $vec AB$ intersects $vec CD$ at some point $P$, so $p=fraca+lambda b1+lambda=fracc+lambda' d1+lambda'$
Although I know the section formula, I couldn't come up with the above formulation on my own. The below is my proof. Could someone comment, on whether the proof is elegant, and also acceptable?
Theorem. Four points $A,B,C,D,$ no three of which are collinear, will lie in a plane when and only when there exist four non-zero numbers $alpha,beta,gamma,delta,$ such that
$$alpha a+ beta b + gamma c + delta d = 0 \
alpha + beta + gamma + delta = 0 $$
Proof.
Idea: If the four points are coplanar, we should be able to construct a parallelogram of the diagonal on $vecCD$, with sides as scalar multiples of $vec AB$ and $vec CD$.
Let
$beginaligned
vec CD &= lambda vecAB + lambda' vecBC\
(d - c) &= lambda (b - a) + lambda'(c - b)\
d &= -lambda a + (lambda - lambda')b + (1 + lambda')c
endaligned$
We would like the coefficients of $a,b,c$ to be $alpha,beta,gamma$ respective.
Set : $-lambda = fracalphaalpha + beta + gamma,
1+lambda' = fracgammaalpha + beta + gamma$. Solving for $lambda,lambda'$ we get :
$$beginaligned
lambda' &= -frac(alpha + beta)alpha + beta + gamma\
lambda - lambda' &= fracbetaalpha + beta + gamma
endaligned$$
So, we have :
$$beginaligned
d &= fracalphaalpha + beta + gammaa + fracbetaalpha + beta + gammab + fracgammaalpha + beta + gammac \\
(alpha + beta + gamma)d &= alpha a + beta b + gamma c
endaligned$$
Let $delta = -(alpha + beta + gamma)$. Thus,
$$
alpha a + beta b + gamma c + delta d = 0
$$
Further, these constants are non-zero, since $lambda,lambda'$ cannot be $0,infty$.
multivariable-calculus proof-verification proof-writing vectors
asked Jul 15 at 3:19
Quasar
697412
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am taking a first course in Vector calculus. I am attempting to write a proof for the following theorem.
The proof in the book splits this into two cases -
(i) either the $vec AB$ is parallel to $vec CD$, so $vec AB = k cdotvecCD$
(ii) $vec AB$ intersects $vec CD$ at some point $P$, so $p=fraca+lambda b1+lambda=fracc+lambda' d1+lambda'$
Although I know the section formula, I couldn't come up with the above formulation on my own. The below is my proof. Could someone comment, on whether the proof is elegant, and also acceptable?
Theorem. Four points $A,B,C,D,$ no three of which are collinear, will lie in a plane when and only when there exist four non-zero numbers $alpha,beta,gamma,delta,$ such that
$$alpha a+ beta b + gamma c + delta d = 0 \
alpha + beta + gamma + delta = 0 $$
Proof.
Idea: If the four points are coplanar, we should be able to construct a parallelogram of the diagonal on $vecCD$, with sides as scalar multiples of $vec AB$ and $vec CD$.
Let
$beginaligned
vec CD &= lambda vecAB + lambda' vecBC\
(d - c) &= lambda (b - a) + lambda'(c - b)\
d &= -lambda a + (lambda - lambda')b + (1 + lambda')c
endaligned$
We would like the coefficients of $a,b,c$ to be $alpha,beta,gamma$ respective.
Set : $-lambda = fracalphaalpha + beta + gamma,
1+lambda' = fracgammaalpha + beta + gamma$. Solving for $lambda,lambda'$ we get :
$$beginaligned
lambda' &= -frac(alpha + beta)alpha + beta + gamma\
lambda - lambda' &= fracbetaalpha + beta + gamma
endaligned$$
So, we have :
$$beginaligned
d &= fracalphaalpha + beta + gammaa + fracbetaalpha + beta + gammab + fracgammaalpha + beta + gammac \\
(alpha + beta + gamma)d &= alpha a + beta b + gamma c
endaligned$$
Let $delta = -(alpha + beta + gamma)$. Thus,
$$
alpha a + beta b + gamma c + delta d = 0
$$
Further, these constants are non-zero, since $lambda,lambda'$ cannot be $0,infty$.
multivariable-calculus proof-verification proof-writing vectors
asked Jul 15 at 3:19
Quasar
697412
I am taking a first course in Vector calculus. I am attempting to write a proof for the following theorem.
The proof in the book splits this into two cases -
(i) either the $vec AB$ is parallel to $vec CD$, so $vec AB = k cdotvecCD$
(ii) $vec AB$ intersects $vec CD$ at some point $P$, so $p=fraca+lambda b1+lambda=fracc+lambda' d1+lambda'$
Although I know the section formula, I couldn't come up with the above formulation on my own. The below is my proof. Could someone comment, on whether the proof is elegant, and also acceptable?
Theorem. Four points $A,B,C,D,$ no three of which are collinear, will lie in a plane when and only when there exist four non-zero numbers $alpha,beta,gamma,delta,$ such that
$$alpha a+ beta b + gamma c + delta d = 0 \
alpha + beta + gamma + delta = 0 $$
Proof.
Idea: If the four points are coplanar, we should be able to construct a parallelogram of the diagonal on $vecCD$, with sides as scalar multiples of $vec AB$ and $vec CD$.
Let
$beginaligned
vec CD &= lambda vecAB + lambda' vecBC\
(d - c) &= lambda (b - a) + lambda'(c - b)\
d &= -lambda a + (lambda - lambda')b + (1 + lambda')c
endaligned$
We would like the coefficients of $a,b,c$ to be $alpha,beta,gamma$ respective.
Set : $-lambda = fracalphaalpha + beta + gamma,
1+lambda' = fracgammaalpha + beta + gamma$. Solving for $lambda,lambda'$ we get :
$$beginaligned
lambda' &= -frac(alpha + beta)alpha + beta + gamma\
lambda - lambda' &= fracbetaalpha + beta + gamma
endaligned$$
So, we have :
$$beginaligned
d &= fracalphaalpha + beta + gammaa + fracbetaalpha + beta + gammab + fracgammaalpha + beta + gammac \\
(alpha + beta + gamma)d &= alpha a + beta b + gamma c
endaligned$$
Let $delta = -(alpha + beta + gamma)$. Thus,
$$
alpha a + beta b + gamma c + delta d = 0
$$
Further, these constants are non-zero, since $lambda,lambda'$ cannot be $0,infty$.
multivariable-calculus proof-verification proof-writing vectors
asked Jul 15 at 3:19
Quasar
697412
edited Jul 15 at 5:07
asked Jul 15 at 3:19
Quasar
697412
asked Jul 15 at 3:19
Quasar
697412
asked Jul 15 at 3:19
Quasar
697412
697412
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
The proof in the book splits this into two cases
Don't quite follow this. There is just two cases only if the points are assumed to be coplanar already. Otherwise, it is entirely possible that $AB$ and $CD$ are skew i.e. neither parallel, nor intersecting.
The more direct way, in my opinion, would be to rewrite it in the equivalent form:
Theorem. Four points $A,B,C,D,$ no three of which are collinear, will lie in a plane when and only when there exist three non-zero numbers $alpha,beta,gamma,,$ with $alpha+beta+gamma ne 0,$ such that
$$alpha a+ beta b + gamma c - (alpha+beta+gamma) d = 0$$
This reduces to:
$$
d= fracalpha a+ beta b + gamma calpha+beta+gamma = a + fracbetaalpha+beta+gamma(b-a) + fracgammaalpha+beta+gamma(c-a) \[20px]
iffquadquad d-a = lambda (b-a) + mu(c-a)
$$
The latter means $vecAD$ is a linear combination of $vecAB$ and $,vecAC,$, which is the necessary and sufficient condition for the points to be coplanar.
answered Jul 15 at 4:09
dxiv
54.2k64797
The proof in the book starts with those two possibilities. I have listed my proof however - beginning with the basic idea, the it should be possible to construct a parallelogram with diagonal as $vecCD$ and sides $lambda vecAB$ and $mu vecBC$ - triangle law of addition.
– Quasar
Jul 15 at 4:15
1
@Quasar I didn't say it didn't ;-) Just that I don't understand their reasoning. Neither of those possibilities covers the case where $AB,CD$ are the opposite edges of a tetrahedron, for example.
– dxiv
Jul 15 at 4:19
1
Yes, I meant coplanar! Corrected the typo.
– Quasar
Jul 15 at 5:06
Sorry, about being outright careless, while representing my attempt.
– Quasar
Jul 15 at 5:13
add a comment |Â
up vote
1
down vote
Let $x_0, x_1, x_2, x_3$ be the points. Then all four points are on the same plane if array whos rows are the coordinates of $x_1-x_0, x_2-x_0, x_3-x_0$ has a rank of two.
answered Jul 15 at 3:31
steven gregory
16.5k22055
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The proof in the book splits this into two cases
Don't quite follow this. There is just two cases only if the points are assumed to be coplanar already. Otherwise, it is entirely possible that $AB$ and $CD$ are skew i.e. neither parallel, nor intersecting.
The more direct way, in my opinion, would be to rewrite it in the equivalent form:
Theorem. Four points $A,B,C,D,$ no three of which are collinear, will lie in a plane when and only when there exist three non-zero numbers $alpha,beta,gamma,,$ with $alpha+beta+gamma ne 0,$ such that
$$alpha a+ beta b + gamma c - (alpha+beta+gamma) d = 0$$
This reduces to:
$$
d= fracalpha a+ beta b + gamma calpha+beta+gamma = a + fracbetaalpha+beta+gamma(b-a) + fracgammaalpha+beta+gamma(c-a) \[20px]
iffquadquad d-a = lambda (b-a) + mu(c-a)
$$
The latter means $vecAD$ is a linear combination of $vecAB$ and $,vecAC,$, which is the necessary and sufficient condition for the points to be coplanar.
answered Jul 15 at 4:09
dxiv
54.2k64797
The proof in the book starts with those two possibilities. I have listed my proof however - beginning with the basic idea, the it should be possible to construct a parallelogram with diagonal as $vecCD$ and sides $lambda vecAB$ and $mu vecBC$ - triangle law of addition.
– Quasar
Jul 15 at 4:15
1
@Quasar I didn't say it didn't ;-) Just that I don't understand their reasoning. Neither of those possibilities covers the case where $AB,CD$ are the opposite edges of a tetrahedron, for example.
– dxiv
Jul 15 at 4:19
1
Yes, I meant coplanar! Corrected the typo.
– Quasar
Jul 15 at 5:06
Sorry, about being outright careless, while representing my attempt.
– Quasar
Jul 15 at 5:13
add a comment |Â
up vote
1
down vote
accepted
The proof in the book splits this into two cases
Don't quite follow this. There is just two cases only if the points are assumed to be coplanar already. Otherwise, it is entirely possible that $AB$ and $CD$ are skew i.e. neither parallel, nor intersecting.
The more direct way, in my opinion, would be to rewrite it in the equivalent form:
Theorem. Four points $A,B,C,D,$ no three of which are collinear, will lie in a plane when and only when there exist three non-zero numbers $alpha,beta,gamma,,$ with $alpha+beta+gamma ne 0,$ such that
$$alpha a+ beta b + gamma c - (alpha+beta+gamma) d = 0$$
This reduces to:
$$
d= fracalpha a+ beta b + gamma calpha+beta+gamma = a + fracbetaalpha+beta+gamma(b-a) + fracgammaalpha+beta+gamma(c-a) \[20px]
iffquadquad d-a = lambda (b-a) + mu(c-a)
$$
The latter means $vecAD$ is a linear combination of $vecAB$ and $,vecAC,$, which is the necessary and sufficient condition for the points to be coplanar.
answered Jul 15 at 4:09
dxiv
54.2k64797
The proof in the book starts with those two possibilities. I have listed my proof however - beginning with the basic idea, the it should be possible to construct a parallelogram with diagonal as $vecCD$ and sides $lambda vecAB$ and $mu vecBC$ - triangle law of addition.
– Quasar
Jul 15 at 4:15
1
@Quasar I didn't say it didn't ;-) Just that I don't understand their reasoning. Neither of those possibilities covers the case where $AB,CD$ are the opposite edges of a tetrahedron, for example.
– dxiv
Jul 15 at 4:19
1
Yes, I meant coplanar! Corrected the typo.
– Quasar
Jul 15 at 5:06
Sorry, about being outright careless, while representing my attempt.
– Quasar
Jul 15 at 5:13
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The proof in the book splits this into two cases
Don't quite follow this. There is just two cases only if the points are assumed to be coplanar already. Otherwise, it is entirely possible that $AB$ and $CD$ are skew i.e. neither parallel, nor intersecting.
The more direct way, in my opinion, would be to rewrite it in the equivalent form:
Theorem. Four points $A,B,C,D,$ no three of which are collinear, will lie in a plane when and only when there exist three non-zero numbers $alpha,beta,gamma,,$ with $alpha+beta+gamma ne 0,$ such that
$$alpha a+ beta b + gamma c - (alpha+beta+gamma) d = 0$$
This reduces to:
$$
d= fracalpha a+ beta b + gamma calpha+beta+gamma = a + fracbetaalpha+beta+gamma(b-a) + fracgammaalpha+beta+gamma(c-a) \[20px]
iffquadquad d-a = lambda (b-a) + mu(c-a)
$$
The latter means $vecAD$ is a linear combination of $vecAB$ and $,vecAC,$, which is the necessary and sufficient condition for the points to be coplanar.
answered Jul 15 at 4:09
dxiv
54.2k64797
The proof in the book splits this into two cases
Don't quite follow this. There is just two cases only if the points are assumed to be coplanar already. Otherwise, it is entirely possible that $AB$ and $CD$ are skew i.e. neither parallel, nor intersecting.
The more direct way, in my opinion, would be to rewrite it in the equivalent form:
Theorem. Four points $A,B,C,D,$ no three of which are collinear, will lie in a plane when and only when there exist three non-zero numbers $alpha,beta,gamma,,$ with $alpha+beta+gamma ne 0,$ such that
$$alpha a+ beta b + gamma c - (alpha+beta+gamma) d = 0$$
This reduces to:
$$
d= fracalpha a+ beta b + gamma calpha+beta+gamma = a + fracbetaalpha+beta+gamma(b-a) + fracgammaalpha+beta+gamma(c-a) \[20px]
iffquadquad d-a = lambda (b-a) + mu(c-a)
$$
The latter means $vecAD$ is a linear combination of $vecAB$ and $,vecAC,$, which is the necessary and sufficient condition for the points to be coplanar.
answered Jul 15 at 4:09
dxiv
54.2k64797
edited Jul 15 at 5:13
answered Jul 15 at 4:09
dxiv
54.2k64797
answered Jul 15 at 4:09
dxiv
54.2k64797
answered Jul 15 at 4:09
dxiv
54.2k64797
54.2k64797
The proof in the book starts with those two possibilities. I have listed my proof however - beginning with the basic idea, the it should be possible to construct a parallelogram with diagonal as $vecCD$ and sides $lambda vecAB$ and $mu vecBC$ - triangle law of addition.
– Quasar
Jul 15 at 4:15
1
@Quasar I didn't say it didn't ;-) Just that I don't understand their reasoning. Neither of those possibilities covers the case where $AB,CD$ are the opposite edges of a tetrahedron, for example.
– dxiv
Jul 15 at 4:19
1
Yes, I meant coplanar! Corrected the typo.
– Quasar
Jul 15 at 5:06
Sorry, about being outright careless, while representing my attempt.
– Quasar
Jul 15 at 5:13
add a comment |Â
The proof in the book starts with those two possibilities. I have listed my proof however - beginning with the basic idea, the it should be possible to construct a parallelogram with diagonal as $vecCD$ and sides $lambda vecAB$ and $mu vecBC$ - triangle law of addition.
– Quasar
Jul 15 at 4:15
1
@Quasar I didn't say it didn't ;-) Just that I don't understand their reasoning. Neither of those possibilities covers the case where $AB,CD$ are the opposite edges of a tetrahedron, for example.
– dxiv
Jul 15 at 4:19
1
Yes, I meant coplanar! Corrected the typo.
– Quasar
Jul 15 at 5:06
Sorry, about being outright careless, while representing my attempt.
– Quasar
Jul 15 at 5:13
The proof in the book starts with those two possibilities. I have listed my proof however - beginning with the basic idea, the it should be possible to construct a parallelogram with diagonal as $vecCD$ and sides $lambda vecAB$ and $mu vecBC$ - triangle law of addition.
– Quasar
Jul 15 at 4:15
The proof in the book starts with those two possibilities. I have listed my proof however - beginning with the basic idea, the it should be possible to construct a parallelogram with diagonal as $vecCD$ and sides $lambda vecAB$ and $mu vecBC$ - triangle law of addition.
– Quasar
Jul 15 at 4:15
1
1
@Quasar I didn't say it didn't ;-) Just that I don't understand their reasoning. Neither of those possibilities covers the case where $AB,CD$ are the opposite edges of a tetrahedron, for example.
– dxiv
Jul 15 at 4:19
@Quasar I didn't say it didn't ;-) Just that I don't understand their reasoning. Neither of those possibilities covers the case where $AB,CD$ are the opposite edges of a tetrahedron, for example.
– dxiv
Jul 15 at 4:19
1
1
Yes, I meant coplanar! Corrected the typo.
– Quasar
Jul 15 at 5:06
Yes, I meant coplanar! Corrected the typo.
– Quasar
Jul 15 at 5:06
Sorry, about being outright careless, while representing my attempt.
– Quasar
Jul 15 at 5:13
Sorry, about being outright careless, while representing my attempt.
– Quasar
Jul 15 at 5:13
add a comment |Â
up vote
1
down vote
Let $x_0, x_1, x_2, x_3$ be the points. Then all four points are on the same plane if array whos rows are the coordinates of $x_1-x_0, x_2-x_0, x_3-x_0$ has a rank of two.
answered Jul 15 at 3:31
steven gregory
16.5k22055
add a comment |Â
up vote
1
down vote
Let $x_0, x_1, x_2, x_3$ be the points. Then all four points are on the same plane if array whos rows are the coordinates of $x_1-x_0, x_2-x_0, x_3-x_0$ has a rank of two.
answered Jul 15 at 3:31
steven gregory
16.5k22055
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $x_0, x_1, x_2, x_3$ be the points. Then all four points are on the same plane if array whos rows are the coordinates of $x_1-x_0, x_2-x_0, x_3-x_0$ has a rank of two.
answered Jul 15 at 3:31
steven gregory
16.5k22055
Let $x_0, x_1, x_2, x_3$ be the points. Then all four points are on the same plane if array whos rows are the coordinates of $x_1-x_0, x_2-x_0, x_3-x_0$ has a rank of two.
answered Jul 15 at 3:31
steven gregory
16.5k22055
edited Jul 15 at 13:04
answered Jul 15 at 3:31
steven gregory
16.5k22055
answered Jul 15 at 3:31
steven gregory
16.5k22055
answered Jul 15 at 3:31
steven gregory
16.5k22055
16.5k22055
add a comment |Â
add a comment |Â
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