Mixing
How do I find the 3 possible 4th points when given 3 unnamed vertices of a parallelogram?
$begingroup$
So, I got have this task:
"The points (3, -4, 5), (1, 0, 5) and (3, 1, -2) are three of four vertices of parallelogram ABCD. Explain why there are three possibilities for the location of the fourth vertex, and find the three points."
Now, the points are unnamed, so I am to assign A, B and C to them myself. I know this because if they were named, point D could only be in one place due to alphabetical order. But since I have no idea which point is/can be which, I'm not sure where to start.
I'm not looking for an answer to the whole problem, just a hint as to where I would start in terms of determining the possible combinations of ABC these 3 points could be.
geometry vectors
asked Jan 6 '14 at 20:27
ThreethumbThreethumb
43631631
$endgroup$
add a comment |
$begingroup$
So, I got have this task:
"The points (3, -4, 5), (1, 0, 5) and (3, 1, -2) are three of four vertices of parallelogram ABCD. Explain why there are three possibilities for the location of the fourth vertex, and find the three points."
Now, the points are unnamed, so I am to assign A, B and C to them myself. I know this because if they were named, point D could only be in one place due to alphabetical order. But since I have no idea which point is/can be which, I'm not sure where to start.
I'm not looking for an answer to the whole problem, just a hint as to where I would start in terms of determining the possible combinations of ABC these 3 points could be.
geometry vectors
asked Jan 6 '14 at 20:27
ThreethumbThreethumb
43631631
$endgroup$
add a comment |
$begingroup$
So, I got have this task:
"The points (3, -4, 5), (1, 0, 5) and (3, 1, -2) are three of four vertices of parallelogram ABCD. Explain why there are three possibilities for the location of the fourth vertex, and find the three points."
Now, the points are unnamed, so I am to assign A, B and C to them myself. I know this because if they were named, point D could only be in one place due to alphabetical order. But since I have no idea which point is/can be which, I'm not sure where to start.
I'm not looking for an answer to the whole problem, just a hint as to where I would start in terms of determining the possible combinations of ABC these 3 points could be.
geometry vectors
asked Jan 6 '14 at 20:27
ThreethumbThreethumb
43631631
$endgroup$
So, I got have this task:
"The points (3, -4, 5), (1, 0, 5) and (3, 1, -2) are three of four vertices of parallelogram ABCD. Explain why there are three possibilities for the location of the fourth vertex, and find the three points."
Now, the points are unnamed, so I am to assign A, B and C to them myself. I know this because if they were named, point D could only be in one place due to alphabetical order. But since I have no idea which point is/can be which, I'm not sure where to start.
I'm not looking for an answer to the whole problem, just a hint as to where I would start in terms of determining the possible combinations of ABC these 3 points could be.
geometry vectors
geometry vectors
asked Jan 6 '14 at 20:27
ThreethumbThreethumb
43631631
asked Jan 6 '14 at 20:27
ThreethumbThreethumb
43631631
asked Jan 6 '14 at 20:27
ThreethumbThreethumb
43631631
asked Jan 6 '14 at 20:27
ThreethumbThreethumb
43631631
asked Jan 6 '14 at 20:27
ThreethumbThreethumb
43631631
43631631
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Everything happens in a plane. Perhaps you can figure out the equation of the plane. Anyway, after that, the relationships are simple.
answered Jan 6 '14 at 20:44
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
Oh wow, that illustration really helps. I understand how to explain it and do the calculations now. Thanks!
$endgroup$
– Threethumb
Jan 6 '14 at 20:53
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I found the points (5, -3, -2), (1, -5, 12) and (1, 5, -2). Does this look right to you? When I tried plotting them into some software, two of them looked like they formed parallelograms, but (5, -3, -2) almost looked like a rectangle.
$endgroup$
– Threethumb
Jan 6 '14 at 22:30
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@Treethumb: A rectangle is a parallelogram.
$endgroup$
– TonyK
Jan 7 '14 at 0:01
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@Threethumb, I see, three thumbs. unusual. If you calculate the lengths of the three edges of the original triangle, that is enough information to find out the angles at the given vertices. Also instructive.
$endgroup$
– Will Jagy
Jan 7 '14 at 0:05
$begingroup$
@TonyK, it turns out to be THreethumb. No idea why.
$endgroup$
– Will Jagy
Jan 7 '14 at 0:06
add a comment |
$begingroup$
Two of the given points form a diagonal of the parallelogram. You must find the fourth point which forms the other diagonal with the third point.
There are three ways you can choose the first diagonal, so there are three ways to form the other.
To arrive at the fourth point, start at one end of the diagonal you know. Go to the off diagonal point you know by displacing the point by $vec{A}$. From that point, go the other end of the diagonal you know by a displacement of $vec{B}$. Finally, go the fourth point by displacing that point by $-vec{A}.$ You can compute the components of $vec{A}$ and $vec{B}$ easily for each of your three cases.
answered Jan 6 '14 at 20:33
JohnJohn
22.8k32550
$endgroup$
$begingroup$
I also think so.
$endgroup$
– kmitov
Jan 6 '14 at 20:35
add a comment |
$begingroup$
The reason there are three possible loci for the fourth point is — that there are three existing points. For each point,
1. Select it $(x_0|y_0|z_0)$ and consider it to be a known vertex of the parallelogram.
2. Consider the other two points, $(x_1|y_1|z_1)$ and $(x_2|y_2|z_2)$, as endpoints of a diagonal of the parallelogram.
3. The sought-after opposite vertex is $(x_1+x_2-x_0|y_1+y_2-y_0|z_1+z_2-z_0)$.
answered Jan 6 '14 at 23:28
Senex Ægypti ParviSenex Ægypti Parvi
2,2231816
$endgroup$
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
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$begingroup$
Everything happens in a plane. Perhaps you can figure out the equation of the plane. Anyway, after that, the relationships are simple.
answered Jan 6 '14 at 20:44
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
Oh wow, that illustration really helps. I understand how to explain it and do the calculations now. Thanks!
$endgroup$
– Threethumb
Jan 6 '14 at 20:53
$begingroup$
I found the points (5, -3, -2), (1, -5, 12) and (1, 5, -2). Does this look right to you? When I tried plotting them into some software, two of them looked like they formed parallelograms, but (5, -3, -2) almost looked like a rectangle.
$endgroup$
– Threethumb
Jan 6 '14 at 22:30
$begingroup$
@Treethumb: A rectangle is a parallelogram.
$endgroup$
– TonyK
Jan 7 '14 at 0:01
$begingroup$
@Threethumb, I see, three thumbs. unusual. If you calculate the lengths of the three edges of the original triangle, that is enough information to find out the angles at the given vertices. Also instructive.
$endgroup$
– Will Jagy
Jan 7 '14 at 0:05
$begingroup$
@TonyK, it turns out to be THreethumb. No idea why.
$endgroup$
– Will Jagy
Jan 7 '14 at 0:06
add a comment |
$begingroup$
Everything happens in a plane. Perhaps you can figure out the equation of the plane. Anyway, after that, the relationships are simple.
answered Jan 6 '14 at 20:44
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
Oh wow, that illustration really helps. I understand how to explain it and do the calculations now. Thanks!
$endgroup$
– Threethumb
Jan 6 '14 at 20:53
$begingroup$
I found the points (5, -3, -2), (1, -5, 12) and (1, 5, -2). Does this look right to you? When I tried plotting them into some software, two of them looked like they formed parallelograms, but (5, -3, -2) almost looked like a rectangle.
$endgroup$
– Threethumb
Jan 6 '14 at 22:30
$begingroup$
@Treethumb: A rectangle is a parallelogram.
$endgroup$
– TonyK
Jan 7 '14 at 0:01
$begingroup$
@Threethumb, I see, three thumbs. unusual. If you calculate the lengths of the three edges of the original triangle, that is enough information to find out the angles at the given vertices. Also instructive.
$endgroup$
– Will Jagy
Jan 7 '14 at 0:05
$begingroup$
@TonyK, it turns out to be THreethumb. No idea why.
$endgroup$
– Will Jagy
Jan 7 '14 at 0:06
add a comment |
$begingroup$
Everything happens in a plane. Perhaps you can figure out the equation of the plane. Anyway, after that, the relationships are simple.
answered Jan 6 '14 at 20:44
Will JagyWill Jagy
104k5102201
$endgroup$
Everything happens in a plane. Perhaps you can figure out the equation of the plane. Anyway, after that, the relationships are simple.
answered Jan 6 '14 at 20:44
Will JagyWill Jagy
104k5102201
answered Jan 6 '14 at 20:44
Will JagyWill Jagy
104k5102201
answered Jan 6 '14 at 20:44
Will JagyWill Jagy
104k5102201
answered Jan 6 '14 at 20:44
Will JagyWill Jagy
104k5102201
104k5102201
$begingroup$
Oh wow, that illustration really helps. I understand how to explain it and do the calculations now. Thanks!
$endgroup$
– Threethumb
Jan 6 '14 at 20:53
$begingroup$
I found the points (5, -3, -2), (1, -5, 12) and (1, 5, -2). Does this look right to you? When I tried plotting them into some software, two of them looked like they formed parallelograms, but (5, -3, -2) almost looked like a rectangle.
$endgroup$
– Threethumb
Jan 6 '14 at 22:30
$begingroup$
@Treethumb: A rectangle is a parallelogram.
$endgroup$
– TonyK
Jan 7 '14 at 0:01
$begingroup$
@Threethumb, I see, three thumbs. unusual. If you calculate the lengths of the three edges of the original triangle, that is enough information to find out the angles at the given vertices. Also instructive.
$endgroup$
– Will Jagy
Jan 7 '14 at 0:05
$begingroup$
@TonyK, it turns out to be THreethumb. No idea why.
$endgroup$
– Will Jagy
Jan 7 '14 at 0:06
add a comment |
$begingroup$
Oh wow, that illustration really helps. I understand how to explain it and do the calculations now. Thanks!
$endgroup$
– Threethumb
Jan 6 '14 at 20:53
$begingroup$
I found the points (5, -3, -2), (1, -5, 12) and (1, 5, -2). Does this look right to you? When I tried plotting them into some software, two of them looked like they formed parallelograms, but (5, -3, -2) almost looked like a rectangle.
$endgroup$
– Threethumb
Jan 6 '14 at 22:30
$begingroup$
@Treethumb: A rectangle is a parallelogram.
$endgroup$
– TonyK
Jan 7 '14 at 0:01
$begingroup$
@Threethumb, I see, three thumbs. unusual. If you calculate the lengths of the three edges of the original triangle, that is enough information to find out the angles at the given vertices. Also instructive.
$endgroup$
– Will Jagy
Jan 7 '14 at 0:05
$begingroup$
@TonyK, it turns out to be THreethumb. No idea why.
$endgroup$
– Will Jagy
Jan 7 '14 at 0:06
$begingroup$
Oh wow, that illustration really helps. I understand how to explain it and do the calculations now. Thanks!
$endgroup$
– Threethumb
Jan 6 '14 at 20:53
$begingroup$
Oh wow, that illustration really helps. I understand how to explain it and do the calculations now. Thanks!
$endgroup$
– Threethumb
Jan 6 '14 at 20:53
$begingroup$
I found the points (5, -3, -2), (1, -5, 12) and (1, 5, -2). Does this look right to you? When I tried plotting them into some software, two of them looked like they formed parallelograms, but (5, -3, -2) almost looked like a rectangle.
$endgroup$
– Threethumb
Jan 6 '14 at 22:30
$begingroup$
I found the points (5, -3, -2), (1, -5, 12) and (1, 5, -2). Does this look right to you? When I tried plotting them into some software, two of them looked like they formed parallelograms, but (5, -3, -2) almost looked like a rectangle.
$endgroup$
– Threethumb
Jan 6 '14 at 22:30
$begingroup$
@Treethumb: A rectangle is a parallelogram.
$endgroup$
– TonyK
Jan 7 '14 at 0:01
$begingroup$
@Treethumb: A rectangle is a parallelogram.
$endgroup$
– TonyK
Jan 7 '14 at 0:01
$begingroup$
@Threethumb, I see, three thumbs. unusual. If you calculate the lengths of the three edges of the original triangle, that is enough information to find out the angles at the given vertices. Also instructive.
$endgroup$
– Will Jagy
Jan 7 '14 at 0:05
$begingroup$
@Threethumb, I see, three thumbs. unusual. If you calculate the lengths of the three edges of the original triangle, that is enough information to find out the angles at the given vertices. Also instructive.
$endgroup$
– Will Jagy
Jan 7 '14 at 0:05
$begingroup$
@TonyK, it turns out to be THreethumb. No idea why.
$endgroup$
– Will Jagy
Jan 7 '14 at 0:06
$begingroup$
@TonyK, it turns out to be THreethumb. No idea why.
$endgroup$
– Will Jagy
Jan 7 '14 at 0:06
add a comment |
$begingroup$
Two of the given points form a diagonal of the parallelogram. You must find the fourth point which forms the other diagonal with the third point.
There are three ways you can choose the first diagonal, so there are three ways to form the other.
To arrive at the fourth point, start at one end of the diagonal you know. Go to the off diagonal point you know by displacing the point by $vec{A}$. From that point, go the other end of the diagonal you know by a displacement of $vec{B}$. Finally, go the fourth point by displacing that point by $-vec{A}.$ You can compute the components of $vec{A}$ and $vec{B}$ easily for each of your three cases.
answered Jan 6 '14 at 20:33
JohnJohn
22.8k32550
$endgroup$
$begingroup$
I also think so.
$endgroup$
– kmitov
Jan 6 '14 at 20:35
add a comment |
$begingroup$
Two of the given points form a diagonal of the parallelogram. You must find the fourth point which forms the other diagonal with the third point.
There are three ways you can choose the first diagonal, so there are three ways to form the other.
To arrive at the fourth point, start at one end of the diagonal you know. Go to the off diagonal point you know by displacing the point by $vec{A}$. From that point, go the other end of the diagonal you know by a displacement of $vec{B}$. Finally, go the fourth point by displacing that point by $-vec{A}.$ You can compute the components of $vec{A}$ and $vec{B}$ easily for each of your three cases.
answered Jan 6 '14 at 20:33
JohnJohn
22.8k32550
$endgroup$
$begingroup$
I also think so.
$endgroup$
– kmitov
Jan 6 '14 at 20:35
add a comment |
$begingroup$
Two of the given points form a diagonal of the parallelogram. You must find the fourth point which forms the other diagonal with the third point.
There are three ways you can choose the first diagonal, so there are three ways to form the other.
To arrive at the fourth point, start at one end of the diagonal you know. Go to the off diagonal point you know by displacing the point by $vec{A}$. From that point, go the other end of the diagonal you know by a displacement of $vec{B}$. Finally, go the fourth point by displacing that point by $-vec{A}.$ You can compute the components of $vec{A}$ and $vec{B}$ easily for each of your three cases.
answered Jan 6 '14 at 20:33
JohnJohn
22.8k32550
$endgroup$
Two of the given points form a diagonal of the parallelogram. You must find the fourth point which forms the other diagonal with the third point.
There are three ways you can choose the first diagonal, so there are three ways to form the other.
To arrive at the fourth point, start at one end of the diagonal you know. Go to the off diagonal point you know by displacing the point by $vec{A}$. From that point, go the other end of the diagonal you know by a displacement of $vec{B}$. Finally, go the fourth point by displacing that point by $-vec{A}.$ You can compute the components of $vec{A}$ and $vec{B}$ easily for each of your three cases.
answered Jan 6 '14 at 20:33
JohnJohn
22.8k32550
edited Jan 6 '14 at 20:49
answered Jan 6 '14 at 20:33
JohnJohn
22.8k32550
answered Jan 6 '14 at 20:33
JohnJohn
22.8k32550
answered Jan 6 '14 at 20:33
JohnJohn
22.8k32550
22.8k32550
$begingroup$
I also think so.
$endgroup$
– kmitov
Jan 6 '14 at 20:35
add a comment |
$begingroup$
I also think so.
$endgroup$
– kmitov
Jan 6 '14 at 20:35
$begingroup$
I also think so.
$endgroup$
– kmitov
Jan 6 '14 at 20:35
$begingroup$
I also think so.
$endgroup$
– kmitov
Jan 6 '14 at 20:35
add a comment |
$begingroup$
The reason there are three possible loci for the fourth point is — that there are three existing points. For each point,
1. Select it $(x_0|y_0|z_0)$ and consider it to be a known vertex of the parallelogram.
2. Consider the other two points, $(x_1|y_1|z_1)$ and $(x_2|y_2|z_2)$, as endpoints of a diagonal of the parallelogram.
3. The sought-after opposite vertex is $(x_1+x_2-x_0|y_1+y_2-y_0|z_1+z_2-z_0)$.
answered Jan 6 '14 at 23:28
Senex Ægypti ParviSenex Ægypti Parvi
2,2231816
$endgroup$
add a comment |
$begingroup$
The reason there are three possible loci for the fourth point is — that there are three existing points. For each point,
1. Select it $(x_0|y_0|z_0)$ and consider it to be a known vertex of the parallelogram.
2. Consider the other two points, $(x_1|y_1|z_1)$ and $(x_2|y_2|z_2)$, as endpoints of a diagonal of the parallelogram.
3. The sought-after opposite vertex is $(x_1+x_2-x_0|y_1+y_2-y_0|z_1+z_2-z_0)$.
answered Jan 6 '14 at 23:28
Senex Ægypti ParviSenex Ægypti Parvi
2,2231816
$endgroup$
add a comment |
$begingroup$
The reason there are three possible loci for the fourth point is — that there are three existing points. For each point,
1. Select it $(x_0|y_0|z_0)$ and consider it to be a known vertex of the parallelogram.
2. Consider the other two points, $(x_1|y_1|z_1)$ and $(x_2|y_2|z_2)$, as endpoints of a diagonal of the parallelogram.
3. The sought-after opposite vertex is $(x_1+x_2-x_0|y_1+y_2-y_0|z_1+z_2-z_0)$.
answered Jan 6 '14 at 23:28
Senex Ægypti ParviSenex Ægypti Parvi
2,2231816
$endgroup$
The reason there are three possible loci for the fourth point is — that there are three existing points. For each point,
1. Select it $(x_0|y_0|z_0)$ and consider it to be a known vertex of the parallelogram.
2. Consider the other two points, $(x_1|y_1|z_1)$ and $(x_2|y_2|z_2)$, as endpoints of a diagonal of the parallelogram.
3. The sought-after opposite vertex is $(x_1+x_2-x_0|y_1+y_2-y_0|z_1+z_2-z_0)$.
answered Jan 6 '14 at 23:28
Senex Ægypti ParviSenex Ægypti Parvi
2,2231816
edited Feb 17 '14 at 17:26
answered Jan 6 '14 at 23:28
Senex Ægypti ParviSenex Ægypti Parvi
2,2231816
answered Jan 6 '14 at 23:28
Senex Ægypti ParviSenex Ægypti Parvi
2,2231816
answered Jan 6 '14 at 23:28
Senex Ægypti ParviSenex Ægypti Parvi
2,2231816
2,2231816
add a comment |
add a comment |
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