If $4x^2 + 9y^2 + z^2 = 108$ and $6xy + frac32yz + 2xz = 108$, then find $x^2 + y^2 + z^2$ [on hold]
Multi tool use
A CBSE class 9 question:
If
$$4x^2 + 9y^2 + z^2 = 108quadtext{and}quad 6xy + frac32yz + 2xz = 108$$ then
$$x^2 + y^2 + z^2 =,text{?}$$
algebra-precalculus
edited 2 days ago
Blue
47.7k870151
asked 2 days ago
Aditya SinghAditya Singh
973
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put on hold as off-topic by Peter, amWhy, Davide Giraudo, Henrik, Ali Caglayan 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Peter, amWhy, Davide Giraudo, Henrik, Ali Caglayan
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
A CBSE class 9 question:
If
$$4x^2 + 9y^2 + z^2 = 108quadtext{and}quad 6xy + frac32yz + 2xz = 108$$ then
$$x^2 + y^2 + z^2 =,text{?}$$
algebra-precalculus
edited 2 days ago
Blue
47.7k870151
asked 2 days ago
Aditya SinghAditya Singh
973
New contributor
Aditya Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Peter, amWhy, Davide Giraudo, Henrik, Ali Caglayan 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Peter, amWhy, Davide Giraudo, Henrik, Ali Caglayan
If this question can be reworded to fit the rules in the help center, please edit the question.
Welcome to math.SE!What have you tried?
– Thomas Shelby
2 days ago
1
Note that you have only two equations in three unknowns. About the only hope for a single solution is if you can make a sum of squares be zero, which suggests subtracting the two equations. Have you tried that, then writing the left side as a sum of squares? Does it work? Alpha finds a mess of complex solutions
– Ross Millikan
2 days ago
The closest solution to this I could find by trial and error is the tuple $[3,2,6]$, but if the second equation were $6xy + 3yz + 2xz = 108$, the tuple would work and $x^2+y^2+z^2 = 3^2 + 2^2 + 6^2 = 9 + 4 + 36 = 49$. Otherwise, the current equations do not have a solution.
– bjcolby15
2 days ago
add a comment |
A CBSE class 9 question:
If
$$4x^2 + 9y^2 + z^2 = 108quadtext{and}quad 6xy + frac32yz + 2xz = 108$$ then
$$x^2 + y^2 + z^2 =,text{?}$$
algebra-precalculus
edited 2 days ago
Blue
47.7k870151
asked 2 days ago
Aditya SinghAditya Singh
973
New contributor
Aditya Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
A CBSE class 9 question:
If
$$4x^2 + 9y^2 + z^2 = 108quadtext{and}quad 6xy + frac32yz + 2xz = 108$$ then
$$x^2 + y^2 + z^2 =,text{?}$$
algebra-precalculus
algebra-precalculus
edited 2 days ago
Blue
47.7k870151
asked 2 days ago
Aditya SinghAditya Singh
973
New contributor
Aditya Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Blue
47.7k870151
asked 2 days ago
Aditya SinghAditya Singh
973
New contributor
Aditya Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Blue
47.7k870151
edited 2 days ago
Blue
47.7k870151
edited 2 days ago
Blue
47.7k870151
47.7k870151
asked 2 days ago
Aditya SinghAditya Singh
973
New contributor
Aditya Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Aditya SinghAditya Singh
973
asked 2 days ago
Aditya SinghAditya Singh
973
973
New contributor
Aditya Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Aditya Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Aditya Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Peter, amWhy, Davide Giraudo, Henrik, Ali Caglayan 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Peter, amWhy, Davide Giraudo, Henrik, Ali Caglayan
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Peter, amWhy, Davide Giraudo, Henrik, Ali Caglayan 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Peter, amWhy, Davide Giraudo, Henrik, Ali Caglayan
If this question can be reworded to fit the rules in the help center, please edit the question.
Welcome to math.SE!What have you tried?
– Thomas Shelby
2 days ago
1
Note that you have only two equations in three unknowns. About the only hope for a single solution is if you can make a sum of squares be zero, which suggests subtracting the two equations. Have you tried that, then writing the left side as a sum of squares? Does it work? Alpha finds a mess of complex solutions
– Ross Millikan
2 days ago
The closest solution to this I could find by trial and error is the tuple $[3,2,6]$, but if the second equation were $6xy + 3yz + 2xz = 108$, the tuple would work and $x^2+y^2+z^2 = 3^2 + 2^2 + 6^2 = 9 + 4 + 36 = 49$. Otherwise, the current equations do not have a solution.
– bjcolby15
2 days ago
add a comment |
Welcome to math.SE!What have you tried?
– Thomas Shelby
2 days ago
1
Note that you have only two equations in three unknowns. About the only hope for a single solution is if you can make a sum of squares be zero, which suggests subtracting the two equations. Have you tried that, then writing the left side as a sum of squares? Does it work? Alpha finds a mess of complex solutions
– Ross Millikan
2 days ago
The closest solution to this I could find by trial and error is the tuple $[3,2,6]$, but if the second equation were $6xy + 3yz + 2xz = 108$, the tuple would work and $x^2+y^2+z^2 = 3^2 + 2^2 + 6^2 = 9 + 4 + 36 = 49$. Otherwise, the current equations do not have a solution.
– bjcolby15
2 days ago
Welcome to math.SE!What have you tried?
– Thomas Shelby
2 days ago
Welcome to math.SE!What have you tried?
– Thomas Shelby
2 days ago
1
1
Note that you have only two equations in three unknowns. About the only hope for a single solution is if you can make a sum of squares be zero, which suggests subtracting the two equations. Have you tried that, then writing the left side as a sum of squares? Does it work? Alpha finds a mess of complex solutions
– Ross Millikan
2 days ago
Note that you have only two equations in three unknowns. About the only hope for a single solution is if you can make a sum of squares be zero, which suggests subtracting the two equations. Have you tried that, then writing the left side as a sum of squares? Does it work? Alpha finds a mess of complex solutions
– Ross Millikan
2 days ago
The closest solution to this I could find by trial and error is the tuple $[3,2,6]$, but if the second equation were $6xy + 3yz + 2xz = 108$, the tuple would work and $x^2+y^2+z^2 = 3^2 + 2^2 + 6^2 = 9 + 4 + 36 = 49$. Otherwise, the current equations do not have a solution.
– bjcolby15
2 days ago
The closest solution to this I could find by trial and error is the tuple $[3,2,6]$, but if the second equation were $6xy + 3yz + 2xz = 108$, the tuple would work and $x^2+y^2+z^2 = 3^2 + 2^2 + 6^2 = 9 + 4 + 36 = 49$. Otherwise, the current equations do not have a solution.
– bjcolby15
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
There is no actual intersection of the two surfaces. Take the two quadratic forms, the Hessian matrix of twice the difference is positive definite, the only (real) point it is zero is the origin, which is not part of either original surface.
$$ H = left(
begin{array}{rrr}
16 & - 12 & - 4 \
- 12 & 36 & - 3 \
- 4 & - 3 & 4 \
end{array}
right)
$$
$$ D_0 = H $$
$$ E_j^T D_{j-1} E_j = D_j $$
$$ P_{j-1} E_j = P_j $$
$$ E_j^{-1} Q_{j-1} = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = left(
begin{array}{rrr}
16 & - 12 & - 4 \
- 12 & 36 & - 3 \
- 4 & - 3 & 4 \
end{array}
right)
$$
==============================================
$$ E_{1} = left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{1} = left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{1} = left(
begin{array}{rrr}
1 & - frac{ 3 }{ 4 } & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{1} = left(
begin{array}{rrr}
16 & 0 & - 4 \
0 & 27 & - 6 \
- 4 & - 6 & 4 \
end{array}
right)
$$
==============================================
$$ E_{2} = left(
begin{array}{rrr}
1 & 0 & frac{ 1 }{ 4 } \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{2} = left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & frac{ 1 }{ 4 } \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{2} = left(
begin{array}{rrr}
1 & - frac{ 3 }{ 4 } & - frac{ 1 }{ 4 } \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{2} = left(
begin{array}{rrr}
16 & 0 & 0 \
0 & 27 & - 6 \
0 & - 6 & 3 \
end{array}
right)
$$
==============================================
$$ E_{3} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & 1 & frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{3} = left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & frac{ 5 }{ 12 } \
0 & 1 & frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{3} = left(
begin{array}{rrr}
1 & - frac{ 3 }{ 4 } & - frac{ 1 }{ 4 } \
0 & 1 & - frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{3} = left(
begin{array}{rrr}
16 & 0 & 0 \
0 & 27 & 0 \
0 & 0 & frac{ 5 }{ 3 } \
end{array}
right)
$$
==============================================
$$ P^T H P = D $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
frac{ 3 }{ 4 } & 1 & 0 \
frac{ 5 }{ 12 } & frac{ 2 }{ 9 } & 1 \
end{array}
right)
left(
begin{array}{rrr}
16 & - 12 & - 4 \
- 12 & 36 & - 3 \
- 4 & - 3 & 4 \
end{array}
right)
left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & frac{ 5 }{ 12 } \
0 & 1 & frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
16 & 0 & 0 \
0 & 27 & 0 \
0 & 0 & frac{ 5 }{ 3 } \
end{array}
right)
$$
$$ Q^T D Q = H $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
- frac{ 3 }{ 4 } & 1 & 0 \
- frac{ 1 }{ 4 } & - frac{ 2 }{ 9 } & 1 \
end{array}
right)
left(
begin{array}{rrr}
16 & 0 & 0 \
0 & 27 & 0 \
0 & 0 & frac{ 5 }{ 3 } \
end{array}
right)
left(
begin{array}{rrr}
1 & - frac{ 3 }{ 4 } & - frac{ 1 }{ 4 } \
0 & 1 & - frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
16 & - 12 & - 4 \
- 12 & 36 & - 3 \
- 4 & - 3 & 4 \
end{array}
right)
$$
answered 2 days ago
Will JagyWill Jagy
102k599199
add a comment |
Perhaps the lazy way to do this:
Thus, there is no intersection.
answered 2 days ago
glowstonetreesglowstonetrees
2,305318
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
There is no actual intersection of the two surfaces. Take the two quadratic forms, the Hessian matrix of twice the difference is positive definite, the only (real) point it is zero is the origin, which is not part of either original surface.
$$ H = left(
begin{array}{rrr}
16 & - 12 & - 4 \
- 12 & 36 & - 3 \
- 4 & - 3 & 4 \
end{array}
right)
$$
$$ D_0 = H $$
$$ E_j^T D_{j-1} E_j = D_j $$
$$ P_{j-1} E_j = P_j $$
$$ E_j^{-1} Q_{j-1} = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = left(
begin{array}{rrr}
16 & - 12 & - 4 \
- 12 & 36 & - 3 \
- 4 & - 3 & 4 \
end{array}
right)
$$
==============================================
$$ E_{1} = left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{1} = left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{1} = left(
begin{array}{rrr}
1 & - frac{ 3 }{ 4 } & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{1} = left(
begin{array}{rrr}
16 & 0 & - 4 \
0 & 27 & - 6 \
- 4 & - 6 & 4 \
end{array}
right)
$$
==============================================
$$ E_{2} = left(
begin{array}{rrr}
1 & 0 & frac{ 1 }{ 4 } \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{2} = left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & frac{ 1 }{ 4 } \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{2} = left(
begin{array}{rrr}
1 & - frac{ 3 }{ 4 } & - frac{ 1 }{ 4 } \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{2} = left(
begin{array}{rrr}
16 & 0 & 0 \
0 & 27 & - 6 \
0 & - 6 & 3 \
end{array}
right)
$$
==============================================
$$ E_{3} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & 1 & frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{3} = left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & frac{ 5 }{ 12 } \
0 & 1 & frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{3} = left(
begin{array}{rrr}
1 & - frac{ 3 }{ 4 } & - frac{ 1 }{ 4 } \
0 & 1 & - frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{3} = left(
begin{array}{rrr}
16 & 0 & 0 \
0 & 27 & 0 \
0 & 0 & frac{ 5 }{ 3 } \
end{array}
right)
$$
==============================================
$$ P^T H P = D $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
frac{ 3 }{ 4 } & 1 & 0 \
frac{ 5 }{ 12 } & frac{ 2 }{ 9 } & 1 \
end{array}
right)
left(
begin{array}{rrr}
16 & - 12 & - 4 \
- 12 & 36 & - 3 \
- 4 & - 3 & 4 \
end{array}
right)
left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & frac{ 5 }{ 12 } \
0 & 1 & frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
16 & 0 & 0 \
0 & 27 & 0 \
0 & 0 & frac{ 5 }{ 3 } \
end{array}
right)
$$
$$ Q^T D Q = H $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
- frac{ 3 }{ 4 } & 1 & 0 \
- frac{ 1 }{ 4 } & - frac{ 2 }{ 9 } & 1 \
end{array}
right)
left(
begin{array}{rrr}
16 & 0 & 0 \
0 & 27 & 0 \
0 & 0 & frac{ 5 }{ 3 } \
end{array}
right)
left(
begin{array}{rrr}
1 & - frac{ 3 }{ 4 } & - frac{ 1 }{ 4 } \
0 & 1 & - frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
16 & - 12 & - 4 \
- 12 & 36 & - 3 \
- 4 & - 3 & 4 \
end{array}
right)
$$
answered 2 days ago
Will JagyWill Jagy
102k599199
add a comment |
There is no actual intersection of the two surfaces. Take the two quadratic forms, the Hessian matrix of twice the difference is positive definite, the only (real) point it is zero is the origin, which is not part of either original surface.
$$ H = left(
begin{array}{rrr}
16 & - 12 & - 4 \
- 12 & 36 & - 3 \
- 4 & - 3 & 4 \
end{array}
right)
$$
$$ D_0 = H $$
$$ E_j^T D_{j-1} E_j = D_j $$
$$ P_{j-1} E_j = P_j $$
$$ E_j^{-1} Q_{j-1} = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = left(
begin{array}{rrr}
16 & - 12 & - 4 \
- 12 & 36 & - 3 \
- 4 & - 3 & 4 \
end{array}
right)
$$
==============================================
$$ E_{1} = left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{1} = left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{1} = left(
begin{array}{rrr}
1 & - frac{ 3 }{ 4 } & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{1} = left(
begin{array}{rrr}
16 & 0 & - 4 \
0 & 27 & - 6 \
- 4 & - 6 & 4 \
end{array}
right)
$$
==============================================
$$ E_{2} = left(
begin{array}{rrr}
1 & 0 & frac{ 1 }{ 4 } \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{2} = left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & frac{ 1 }{ 4 } \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{2} = left(
begin{array}{rrr}
1 & - frac{ 3 }{ 4 } & - frac{ 1 }{ 4 } \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{2} = left(
begin{array}{rrr}
16 & 0 & 0 \
0 & 27 & - 6 \
0 & - 6 & 3 \
end{array}
right)
$$
==============================================
$$ E_{3} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & 1 & frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{3} = left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & frac{ 5 }{ 12 } \
0 & 1 & frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{3} = left(
begin{array}{rrr}
1 & - frac{ 3 }{ 4 } & - frac{ 1 }{ 4 } \
0 & 1 & - frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{3} = left(
begin{array}{rrr}
16 & 0 & 0 \
0 & 27 & 0 \
0 & 0 & frac{ 5 }{ 3 } \
end{array}
right)
$$
==============================================
$$ P^T H P = D $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
frac{ 3 }{ 4 } & 1 & 0 \
frac{ 5 }{ 12 } & frac{ 2 }{ 9 } & 1 \
end{array}
right)
left(
begin{array}{rrr}
16 & - 12 & - 4 \
- 12 & 36 & - 3 \
- 4 & - 3 & 4 \
end{array}
right)
left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & frac{ 5 }{ 12 } \
0 & 1 & frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
16 & 0 & 0 \
0 & 27 & 0 \
0 & 0 & frac{ 5 }{ 3 } \
end{array}
right)
$$
$$ Q^T D Q = H $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
- frac{ 3 }{ 4 } & 1 & 0 \
- frac{ 1 }{ 4 } & - frac{ 2 }{ 9 } & 1 \
end{array}
right)
left(
begin{array}{rrr}
16 & 0 & 0 \
0 & 27 & 0 \
0 & 0 & frac{ 5 }{ 3 } \
end{array}
right)
left(
begin{array}{rrr}
1 & - frac{ 3 }{ 4 } & - frac{ 1 }{ 4 } \
0 & 1 & - frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
16 & - 12 & - 4 \
- 12 & 36 & - 3 \
- 4 & - 3 & 4 \
end{array}
right)
$$
answered 2 days ago
Will JagyWill Jagy
102k599199
add a comment |
There is no actual intersection of the two surfaces. Take the two quadratic forms, the Hessian matrix of twice the difference is positive definite, the only (real) point it is zero is the origin, which is not part of either original surface.
$$ H = left(
begin{array}{rrr}
16 & - 12 & - 4 \
- 12 & 36 & - 3 \
- 4 & - 3 & 4 \
end{array}
right)
$$
$$ D_0 = H $$
$$ E_j^T D_{j-1} E_j = D_j $$
$$ P_{j-1} E_j = P_j $$
$$ E_j^{-1} Q_{j-1} = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = left(
begin{array}{rrr}
16 & - 12 & - 4 \
- 12 & 36 & - 3 \
- 4 & - 3 & 4 \
end{array}
right)
$$
==============================================
$$ E_{1} = left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{1} = left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{1} = left(
begin{array}{rrr}
1 & - frac{ 3 }{ 4 } & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{1} = left(
begin{array}{rrr}
16 & 0 & - 4 \
0 & 27 & - 6 \
- 4 & - 6 & 4 \
end{array}
right)
$$
==============================================
$$ E_{2} = left(
begin{array}{rrr}
1 & 0 & frac{ 1 }{ 4 } \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{2} = left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & frac{ 1 }{ 4 } \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{2} = left(
begin{array}{rrr}
1 & - frac{ 3 }{ 4 } & - frac{ 1 }{ 4 } \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{2} = left(
begin{array}{rrr}
16 & 0 & 0 \
0 & 27 & - 6 \
0 & - 6 & 3 \
end{array}
right)
$$
==============================================
$$ E_{3} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & 1 & frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{3} = left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & frac{ 5 }{ 12 } \
0 & 1 & frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{3} = left(
begin{array}{rrr}
1 & - frac{ 3 }{ 4 } & - frac{ 1 }{ 4 } \
0 & 1 & - frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{3} = left(
begin{array}{rrr}
16 & 0 & 0 \
0 & 27 & 0 \
0 & 0 & frac{ 5 }{ 3 } \
end{array}
right)
$$
==============================================
$$ P^T H P = D $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
frac{ 3 }{ 4 } & 1 & 0 \
frac{ 5 }{ 12 } & frac{ 2 }{ 9 } & 1 \
end{array}
right)
left(
begin{array}{rrr}
16 & - 12 & - 4 \
- 12 & 36 & - 3 \
- 4 & - 3 & 4 \
end{array}
right)
left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & frac{ 5 }{ 12 } \
0 & 1 & frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
16 & 0 & 0 \
0 & 27 & 0 \
0 & 0 & frac{ 5 }{ 3 } \
end{array}
right)
$$
$$ Q^T D Q = H $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
- frac{ 3 }{ 4 } & 1 & 0 \
- frac{ 1 }{ 4 } & - frac{ 2 }{ 9 } & 1 \
end{array}
right)
left(
begin{array}{rrr}
16 & 0 & 0 \
0 & 27 & 0 \
0 & 0 & frac{ 5 }{ 3 } \
end{array}
right)
left(
begin{array}{rrr}
1 & - frac{ 3 }{ 4 } & - frac{ 1 }{ 4 } \
0 & 1 & - frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
16 & - 12 & - 4 \
- 12 & 36 & - 3 \
- 4 & - 3 & 4 \
end{array}
right)
$$
answered 2 days ago
Will JagyWill Jagy
102k599199
There is no actual intersection of the two surfaces. Take the two quadratic forms, the Hessian matrix of twice the difference is positive definite, the only (real) point it is zero is the origin, which is not part of either original surface.
$$ H = left(
begin{array}{rrr}
16 & - 12 & - 4 \
- 12 & 36 & - 3 \
- 4 & - 3 & 4 \
end{array}
right)
$$
$$ D_0 = H $$
$$ E_j^T D_{j-1} E_j = D_j $$
$$ P_{j-1} E_j = P_j $$
$$ E_j^{-1} Q_{j-1} = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = left(
begin{array}{rrr}
16 & - 12 & - 4 \
- 12 & 36 & - 3 \
- 4 & - 3 & 4 \
end{array}
right)
$$
==============================================
$$ E_{1} = left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{1} = left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{1} = left(
begin{array}{rrr}
1 & - frac{ 3 }{ 4 } & 0 \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{1} = left(
begin{array}{rrr}
16 & 0 & - 4 \
0 & 27 & - 6 \
- 4 & - 6 & 4 \
end{array}
right)
$$
==============================================
$$ E_{2} = left(
begin{array}{rrr}
1 & 0 & frac{ 1 }{ 4 } \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{2} = left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & frac{ 1 }{ 4 } \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{2} = left(
begin{array}{rrr}
1 & - frac{ 3 }{ 4 } & - frac{ 1 }{ 4 } \
0 & 1 & 0 \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{2} = left(
begin{array}{rrr}
16 & 0 & 0 \
0 & 27 & - 6 \
0 & - 6 & 3 \
end{array}
right)
$$
==============================================
$$ E_{3} = left(
begin{array}{rrr}
1 & 0 & 0 \
0 & 1 & frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
$$
$$ P_{3} = left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & frac{ 5 }{ 12 } \
0 & 1 & frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
, ; ; ; Q_{3} = left(
begin{array}{rrr}
1 & - frac{ 3 }{ 4 } & - frac{ 1 }{ 4 } \
0 & 1 & - frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
, ; ; ; D_{3} = left(
begin{array}{rrr}
16 & 0 & 0 \
0 & 27 & 0 \
0 & 0 & frac{ 5 }{ 3 } \
end{array}
right)
$$
==============================================
$$ P^T H P = D $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
frac{ 3 }{ 4 } & 1 & 0 \
frac{ 5 }{ 12 } & frac{ 2 }{ 9 } & 1 \
end{array}
right)
left(
begin{array}{rrr}
16 & - 12 & - 4 \
- 12 & 36 & - 3 \
- 4 & - 3 & 4 \
end{array}
right)
left(
begin{array}{rrr}
1 & frac{ 3 }{ 4 } & frac{ 5 }{ 12 } \
0 & 1 & frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
16 & 0 & 0 \
0 & 27 & 0 \
0 & 0 & frac{ 5 }{ 3 } \
end{array}
right)
$$
$$ Q^T D Q = H $$
$$left(
begin{array}{rrr}
1 & 0 & 0 \
- frac{ 3 }{ 4 } & 1 & 0 \
- frac{ 1 }{ 4 } & - frac{ 2 }{ 9 } & 1 \
end{array}
right)
left(
begin{array}{rrr}
16 & 0 & 0 \
0 & 27 & 0 \
0 & 0 & frac{ 5 }{ 3 } \
end{array}
right)
left(
begin{array}{rrr}
1 & - frac{ 3 }{ 4 } & - frac{ 1 }{ 4 } \
0 & 1 & - frac{ 2 }{ 9 } \
0 & 0 & 1 \
end{array}
right)
= left(
begin{array}{rrr}
16 & - 12 & - 4 \
- 12 & 36 & - 3 \
- 4 & - 3 & 4 \
end{array}
right)
$$
answered 2 days ago
Will JagyWill Jagy
102k599199
edited 2 days ago
answered 2 days ago
Will JagyWill Jagy
102k599199
answered 2 days ago
Will JagyWill Jagy
102k599199
answered 2 days ago
Will JagyWill Jagy
102k599199
102k599199
add a comment |
add a comment |
Perhaps the lazy way to do this:
Thus, there is no intersection.
answered 2 days ago
glowstonetreesglowstonetrees
2,305318
add a comment |
Perhaps the lazy way to do this:
Thus, there is no intersection.
answered 2 days ago
glowstonetreesglowstonetrees
2,305318
add a comment |
Perhaps the lazy way to do this:
Thus, there is no intersection.
answered 2 days ago
glowstonetreesglowstonetrees
2,305318
Perhaps the lazy way to do this:
Thus, there is no intersection.
answered 2 days ago
glowstonetreesglowstonetrees
2,305318
answered 2 days ago
glowstonetreesglowstonetrees
2,305318
answered 2 days ago
glowstonetreesglowstonetrees
2,305318
answered 2 days ago
glowstonetreesglowstonetrees
2,305318
2,305318
add a comment |
add a comment |
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Welcome to math.SE!What have you tried?
– Thomas Shelby
2 days ago
1
Note that you have only two equations in three unknowns. About the only hope for a single solution is if you can make a sum of squares be zero, which suggests subtracting the two equations. Have you tried that, then writing the left side as a sum of squares? Does it work? Alpha finds a mess of complex solutions
– Ross Millikan
2 days ago
The closest solution to this I could find by trial and error is the tuple $[3,2,6]$, but if the second equation were $6xy + 3yz + 2xz = 108$, the tuple would work and $x^2+y^2+z^2 = 3^2 + 2^2 + 6^2 = 9 + 4 + 36 = 49$. Otherwise, the current equations do not have a solution.
– bjcolby15
2 days ago