How to get vector a from known inner product?
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up vote
3
down vote
favorite
Assuming that a.b=z
,where b=b1,b2,b3
and z=x b1+y b2+z b3
, obviously a=x,y,z
, how to realize it by mathematica?
vector products coefficients
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up vote
3
down vote
favorite
Assuming that a.b=z
,where b=b1,b2,b3
and z=x b1+y b2+z b3
, obviously a=x,y,z
, how to realize it by mathematica?
vector products coefficients
dnvin, welcome to mma.se. Our standard welcome message: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) If you haven't already done so please take the tour! ...
â kglr
Sep 6 at 10:16
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â kglr
Sep 6 at 10:19
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Assuming that a.b=z
,where b=b1,b2,b3
and z=x b1+y b2+z b3
, obviously a=x,y,z
, how to realize it by mathematica?
vector products coefficients
Assuming that a.b=z
,where b=b1,b2,b3
and z=x b1+y b2+z b3
, obviously a=x,y,z
, how to realize it by mathematica?
vector products coefficients
vector products coefficients
edited Sep 6 at 9:18
kglr
161k8185384
161k8185384
asked Sep 6 at 8:08
dnvin
182
182
dnvin, welcome to mma.se. Our standard welcome message: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) If you haven't already done so please take the tour! ...
â kglr
Sep 6 at 10:16
... 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
â kglr
Sep 6 at 10:19
add a comment |Â
dnvin, welcome to mma.se. Our standard welcome message: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) If you haven't already done so please take the tour! ...
â kglr
Sep 6 at 10:16
... 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
â kglr
Sep 6 at 10:19
dnvin, welcome to mma.se. Our standard welcome message: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) If you haven't already done so please take the tour! ...
â kglr
Sep 6 at 10:16
dnvin, welcome to mma.se. Our standard welcome message: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) If you haven't already done so please take the tour! ...
â kglr
Sep 6 at 10:16
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â kglr
Sep 6 at 10:19
... 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
â kglr
Sep 6 at 10:19
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2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
If a.b1,b2,b3 == x b1+y b2+z b3
for all b1,b2,b3
:
Reduce[ForAll[b1, b2, b3, Dot[a1, a2, a3, b1, b2, b3] == x b1 + y b2 + z b3]]
a3 == z && a2 == y && a1 == x
or
SolveAlways[ Dot[a1, a2, a3, b1, b2, b3] == x b1 + y b2 + z b3, b1, b2, b3]
a1 -> x, a2 -> y, a3 -> z
Thanks a lot! This is what I want.
â dnvin
Sep 6 at 10:09
@dnvin, my pleasure. Welcome mma.se.
â kglr
Sep 6 at 10:13
add a comment |Â
up vote
3
down vote
If I understand correctly, you have a single linear equation with 3 unknowns.
It's not possible to back out the 3 unknown values. You have an equation for a 3D hyperplane with an infinite number of solutions.
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
If a.b1,b2,b3 == x b1+y b2+z b3
for all b1,b2,b3
:
Reduce[ForAll[b1, b2, b3, Dot[a1, a2, a3, b1, b2, b3] == x b1 + y b2 + z b3]]
a3 == z && a2 == y && a1 == x
or
SolveAlways[ Dot[a1, a2, a3, b1, b2, b3] == x b1 + y b2 + z b3, b1, b2, b3]
a1 -> x, a2 -> y, a3 -> z
Thanks a lot! This is what I want.
â dnvin
Sep 6 at 10:09
@dnvin, my pleasure. Welcome mma.se.
â kglr
Sep 6 at 10:13
add a comment |Â
up vote
4
down vote
accepted
If a.b1,b2,b3 == x b1+y b2+z b3
for all b1,b2,b3
:
Reduce[ForAll[b1, b2, b3, Dot[a1, a2, a3, b1, b2, b3] == x b1 + y b2 + z b3]]
a3 == z && a2 == y && a1 == x
or
SolveAlways[ Dot[a1, a2, a3, b1, b2, b3] == x b1 + y b2 + z b3, b1, b2, b3]
a1 -> x, a2 -> y, a3 -> z
Thanks a lot! This is what I want.
â dnvin
Sep 6 at 10:09
@dnvin, my pleasure. Welcome mma.se.
â kglr
Sep 6 at 10:13
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
If a.b1,b2,b3 == x b1+y b2+z b3
for all b1,b2,b3
:
Reduce[ForAll[b1, b2, b3, Dot[a1, a2, a3, b1, b2, b3] == x b1 + y b2 + z b3]]
a3 == z && a2 == y && a1 == x
or
SolveAlways[ Dot[a1, a2, a3, b1, b2, b3] == x b1 + y b2 + z b3, b1, b2, b3]
a1 -> x, a2 -> y, a3 -> z
If a.b1,b2,b3 == x b1+y b2+z b3
for all b1,b2,b3
:
Reduce[ForAll[b1, b2, b3, Dot[a1, a2, a3, b1, b2, b3] == x b1 + y b2 + z b3]]
a3 == z && a2 == y && a1 == x
or
SolveAlways[ Dot[a1, a2, a3, b1, b2, b3] == x b1 + y b2 + z b3, b1, b2, b3]
a1 -> x, a2 -> y, a3 -> z
answered Sep 6 at 9:19
kglr
161k8185384
161k8185384
Thanks a lot! This is what I want.
â dnvin
Sep 6 at 10:09
@dnvin, my pleasure. Welcome mma.se.
â kglr
Sep 6 at 10:13
add a comment |Â
Thanks a lot! This is what I want.
â dnvin
Sep 6 at 10:09
@dnvin, my pleasure. Welcome mma.se.
â kglr
Sep 6 at 10:13
Thanks a lot! This is what I want.
â dnvin
Sep 6 at 10:09
Thanks a lot! This is what I want.
â dnvin
Sep 6 at 10:09
@dnvin, my pleasure. Welcome mma.se.
â kglr
Sep 6 at 10:13
@dnvin, my pleasure. Welcome mma.se.
â kglr
Sep 6 at 10:13
add a comment |Â
up vote
3
down vote
If I understand correctly, you have a single linear equation with 3 unknowns.
It's not possible to back out the 3 unknown values. You have an equation for a 3D hyperplane with an infinite number of solutions.
add a comment |Â
up vote
3
down vote
If I understand correctly, you have a single linear equation with 3 unknowns.
It's not possible to back out the 3 unknown values. You have an equation for a 3D hyperplane with an infinite number of solutions.
add a comment |Â
up vote
3
down vote
up vote
3
down vote
If I understand correctly, you have a single linear equation with 3 unknowns.
It's not possible to back out the 3 unknown values. You have an equation for a 3D hyperplane with an infinite number of solutions.
If I understand correctly, you have a single linear equation with 3 unknowns.
It's not possible to back out the 3 unknown values. You have an equation for a 3D hyperplane with an infinite number of solutions.
answered Sep 6 at 8:16
GerardF123
986
986
add a comment |Â
add a comment |Â
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dnvin, welcome to mma.se. Our standard welcome message: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) If you haven't already done so please take the tour! ...
â kglr
Sep 6 at 10:16
... 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
â kglr
Sep 6 at 10:19