I have a set with two vectors in $R^3$. What method should I use to find the vectors orthogonal to both in the original set?
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There are so many different methods I've found on SE and through Matlab, and they're all giving me different results.
Specifically, I have v1 = (1,2,1) and v2 = (2,1,0) in set S. What is the method to find v3 vectors that are orthogonal to both v1 and v2?
I'm preparing for a final and I'm trying to find a flexible method for many cases. The answer I got for above was v3 = 1,-2,3 but different methods are returning different results.
linear-algebra matrices matrix-equations
asked Aug 31 at 8:21
Bryan Walsh
31
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up vote
0
down vote
favorite
There are so many different methods I've found on SE and through Matlab, and they're all giving me different results.
Specifically, I have v1 = (1,2,1) and v2 = (2,1,0) in set S. What is the method to find v3 vectors that are orthogonal to both v1 and v2?
I'm preparing for a final and I'm trying to find a flexible method for many cases. The answer I got for above was v3 = 1,-2,3 but different methods are returning different results.
linear-algebra matrices matrix-equations
asked Aug 31 at 8:21
Bryan Walsh
31
Keep in mind that there is an infinite number of vectors that are perpendicular to two other vectors—in fact they form a subspace of $mathbb R^3$. It shouldn’t be all that surprising that different methods choose a different vector from this space.
– amd
Aug 31 at 20:09
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
There are so many different methods I've found on SE and through Matlab, and they're all giving me different results.
Specifically, I have v1 = (1,2,1) and v2 = (2,1,0) in set S. What is the method to find v3 vectors that are orthogonal to both v1 and v2?
I'm preparing for a final and I'm trying to find a flexible method for many cases. The answer I got for above was v3 = 1,-2,3 but different methods are returning different results.
linear-algebra matrices matrix-equations
asked Aug 31 at 8:21
Bryan Walsh
31
There are so many different methods I've found on SE and through Matlab, and they're all giving me different results.
Specifically, I have v1 = (1,2,1) and v2 = (2,1,0) in set S. What is the method to find v3 vectors that are orthogonal to both v1 and v2?
I'm preparing for a final and I'm trying to find a flexible method for many cases. The answer I got for above was v3 = 1,-2,3 but different methods are returning different results.
linear-algebra matrices matrix-equations
linear-algebra matrices matrix-equations
asked Aug 31 at 8:21
Bryan Walsh
31
asked Aug 31 at 8:21
Bryan Walsh
31
asked Aug 31 at 8:21
Bryan Walsh
31
asked Aug 31 at 8:21
Bryan Walsh
31
asked Aug 31 at 8:21
Bryan Walsh
31
31
Keep in mind that there is an infinite number of vectors that are perpendicular to two other vectors—in fact they form a subspace of $mathbb R^3$. It shouldn’t be all that surprising that different methods choose a different vector from this space.
– amd
Aug 31 at 20:09
add a comment |Â
Keep in mind that there is an infinite number of vectors that are perpendicular to two other vectors—in fact they form a subspace of $mathbb R^3$. It shouldn’t be all that surprising that different methods choose a different vector from this space.
– amd
Aug 31 at 20:09
Keep in mind that there is an infinite number of vectors that are perpendicular to two other vectors—in fact they form a subspace of $mathbb R^3$. It shouldn’t be all that surprising that different methods choose a different vector from this space.
– amd
Aug 31 at 20:09
Keep in mind that there is an infinite number of vectors that are perpendicular to two other vectors—in fact they form a subspace of $mathbb R^3$. It shouldn’t be all that surprising that different methods choose a different vector from this space.
– amd
Aug 31 at 20:09
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Guide:
Your answer should be a non-zero scalar multiple of the $v_3$ that you provided since $v_1$ and $v_2$ are not parallel to each other.
Method $1$:
- Compute the cross product of $v_1$ and $v_2$, that will give you a valid solution.
Method $2$:
- Solve the linear system $v_1^Tx=0$ and $v_2^Tx=0$ by reducing the system say to REF. You will obtain multiple solution, of which all of them are scalar multiple of $v_3$.
Either method should be fine.
answered Aug 31 at 8:25
Siong Thye Goh
81.7k1454104
Thank you! The cross product made it all come together. Had some issues with positive/negative values in my method, but I forgot about the +, -, + property of the cross product.
– Bryan Walsh
Aug 31 at 8:50
add a comment |Â
up vote
0
down vote
What I would do:
- Compute the planes orthogonal to both of your vectors.
A plane orthogonal to the vector $(a,b,c)$ has the equation $ax + by + cz + d = 0, forall d in mathbbR$
- Compute the intersection of the two planes by replacing the first plane equation in the second one
answered Aug 31 at 8:28
PackSciences
41414
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
Guide:
Your answer should be a non-zero scalar multiple of the $v_3$ that you provided since $v_1$ and $v_2$ are not parallel to each other.
Method $1$:
- Compute the cross product of $v_1$ and $v_2$, that will give you a valid solution.
Method $2$:
- Solve the linear system $v_1^Tx=0$ and $v_2^Tx=0$ by reducing the system say to REF. You will obtain multiple solution, of which all of them are scalar multiple of $v_3$.
Either method should be fine.
answered Aug 31 at 8:25
Siong Thye Goh
81.7k1454104
Thank you! The cross product made it all come together. Had some issues with positive/negative values in my method, but I forgot about the +, -, + property of the cross product.
– Bryan Walsh
Aug 31 at 8:50
add a comment |Â
up vote
1
down vote
accepted
Guide:
Your answer should be a non-zero scalar multiple of the $v_3$ that you provided since $v_1$ and $v_2$ are not parallel to each other.
Method $1$:
- Compute the cross product of $v_1$ and $v_2$, that will give you a valid solution.
Method $2$:
- Solve the linear system $v_1^Tx=0$ and $v_2^Tx=0$ by reducing the system say to REF. You will obtain multiple solution, of which all of them are scalar multiple of $v_3$.
Either method should be fine.
answered Aug 31 at 8:25
Siong Thye Goh
81.7k1454104
Thank you! The cross product made it all come together. Had some issues with positive/negative values in my method, but I forgot about the +, -, + property of the cross product.
– Bryan Walsh
Aug 31 at 8:50
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Guide:
Your answer should be a non-zero scalar multiple of the $v_3$ that you provided since $v_1$ and $v_2$ are not parallel to each other.
Method $1$:
- Compute the cross product of $v_1$ and $v_2$, that will give you a valid solution.
Method $2$:
- Solve the linear system $v_1^Tx=0$ and $v_2^Tx=0$ by reducing the system say to REF. You will obtain multiple solution, of which all of them are scalar multiple of $v_3$.
Either method should be fine.
answered Aug 31 at 8:25
Siong Thye Goh
81.7k1454104
Guide:
Your answer should be a non-zero scalar multiple of the $v_3$ that you provided since $v_1$ and $v_2$ are not parallel to each other.
Method $1$:
- Compute the cross product of $v_1$ and $v_2$, that will give you a valid solution.
Method $2$:
- Solve the linear system $v_1^Tx=0$ and $v_2^Tx=0$ by reducing the system say to REF. You will obtain multiple solution, of which all of them are scalar multiple of $v_3$.
Either method should be fine.
answered Aug 31 at 8:25
Siong Thye Goh
81.7k1454104
answered Aug 31 at 8:25
Siong Thye Goh
81.7k1454104
answered Aug 31 at 8:25
Siong Thye Goh
81.7k1454104
answered Aug 31 at 8:25
Siong Thye Goh
81.7k1454104
81.7k1454104
Thank you! The cross product made it all come together. Had some issues with positive/negative values in my method, but I forgot about the +, -, + property of the cross product.
– Bryan Walsh
Aug 31 at 8:50
add a comment |Â
Thank you! The cross product made it all come together. Had some issues with positive/negative values in my method, but I forgot about the +, -, + property of the cross product.
– Bryan Walsh
Aug 31 at 8:50
Thank you! The cross product made it all come together. Had some issues with positive/negative values in my method, but I forgot about the +, -, + property of the cross product.
– Bryan Walsh
Aug 31 at 8:50
Thank you! The cross product made it all come together. Had some issues with positive/negative values in my method, but I forgot about the +, -, + property of the cross product.
– Bryan Walsh
Aug 31 at 8:50
add a comment |Â
up vote
0
down vote
What I would do:
- Compute the planes orthogonal to both of your vectors.
A plane orthogonal to the vector $(a,b,c)$ has the equation $ax + by + cz + d = 0, forall d in mathbbR$
- Compute the intersection of the two planes by replacing the first plane equation in the second one
answered Aug 31 at 8:28
PackSciences
41414
add a comment |Â
up vote
0
down vote
What I would do:
- Compute the planes orthogonal to both of your vectors.
A plane orthogonal to the vector $(a,b,c)$ has the equation $ax + by + cz + d = 0, forall d in mathbbR$
- Compute the intersection of the two planes by replacing the first plane equation in the second one
answered Aug 31 at 8:28
PackSciences
41414
add a comment |Â
up vote
0
down vote
up vote
0
down vote
What I would do:
- Compute the planes orthogonal to both of your vectors.
A plane orthogonal to the vector $(a,b,c)$ has the equation $ax + by + cz + d = 0, forall d in mathbbR$
- Compute the intersection of the two planes by replacing the first plane equation in the second one
answered Aug 31 at 8:28
PackSciences
41414
What I would do:
- Compute the planes orthogonal to both of your vectors.
A plane orthogonal to the vector $(a,b,c)$ has the equation $ax + by + cz + d = 0, forall d in mathbbR$
- Compute the intersection of the two planes by replacing the first plane equation in the second one
answered Aug 31 at 8:28
PackSciences
41414
answered Aug 31 at 8:28
PackSciences
41414
answered Aug 31 at 8:28
PackSciences
41414
answered Aug 31 at 8:28
PackSciences
41414
41414
add a comment |Â
add a comment |Â
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Keep in mind that there is an infinite number of vectors that are perpendicular to two other vectors—in fact they form a subspace of $mathbb R^3$. It shouldn’t be all that surprising that different methods choose a different vector from this space.
– amd
Aug 31 at 20:09