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Use a matrix to rotate a vector in the direction of another vector
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I have two 3D vectors, say A and B, that are perpendicular to each other in an arbitrary plane in 3D space. A and B share origin point P. I would like to rotate vector A in this plane (pivoting about point P) towards vector B by a certain number of degrees $theta$ (or away from B if $theta$ is negative). I would like to do this via a transformation matrix, but I'm having some trouble crafting one on my own. How would I go about this?
vectors transformation
asked Jul 25 at 21:55
derekantrican
1104
add a comment |Â
up vote
0
down vote
favorite
I have two 3D vectors, say A and B, that are perpendicular to each other in an arbitrary plane in 3D space. A and B share origin point P. I would like to rotate vector A in this plane (pivoting about point P) towards vector B by a certain number of degrees $theta$ (or away from B if $theta$ is negative). I would like to do this via a transformation matrix, but I'm having some trouble crafting one on my own. How would I go about this?
vectors transformation
asked Jul 25 at 21:55
derekantrican
1104
In your idea vectors A and B have an origin in common?
– gimusi
Jul 25 at 22:07
and is the plane through the origin?
– gimusi
Jul 25 at 22:12
@gimusi yes and yes
– derekantrican
Jul 26 at 12:55
ok then I can revise my answer a little bit to simplify that, if you have some numerica example I can try to solve it
– gimusi
Jul 26 at 12:57
@gimusi I have updated my question a little bit. I will try to draw a picture or something
– derekantrican
Jul 26 at 13:00
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have two 3D vectors, say A and B, that are perpendicular to each other in an arbitrary plane in 3D space. A and B share origin point P. I would like to rotate vector A in this plane (pivoting about point P) towards vector B by a certain number of degrees $theta$ (or away from B if $theta$ is negative). I would like to do this via a transformation matrix, but I'm having some trouble crafting one on my own. How would I go about this?
vectors transformation
asked Jul 25 at 21:55
derekantrican
1104
I have two 3D vectors, say A and B, that are perpendicular to each other in an arbitrary plane in 3D space. A and B share origin point P. I would like to rotate vector A in this plane (pivoting about point P) towards vector B by a certain number of degrees $theta$ (or away from B if $theta$ is negative). I would like to do this via a transformation matrix, but I'm having some trouble crafting one on my own. How would I go about this?
vectors transformation
asked Jul 25 at 21:55
derekantrican
1104
edited Jul 26 at 12:59
asked Jul 25 at 21:55
derekantrican
1104
asked Jul 25 at 21:55
derekantrican
1104
asked Jul 25 at 21:55
derekantrican
1104
1104
In your idea vectors A and B have an origin in common?
– gimusi
Jul 25 at 22:07
and is the plane through the origin?
– gimusi
Jul 25 at 22:12
@gimusi yes and yes
– derekantrican
Jul 26 at 12:55
ok then I can revise my answer a little bit to simplify that, if you have some numerica example I can try to solve it
– gimusi
Jul 26 at 12:57
@gimusi I have updated my question a little bit. I will try to draw a picture or something
– derekantrican
Jul 26 at 13:00
add a comment |Â
In your idea vectors A and B have an origin in common?
– gimusi
Jul 25 at 22:07
and is the plane through the origin?
– gimusi
Jul 25 at 22:12
@gimusi yes and yes
– derekantrican
Jul 26 at 12:55
ok then I can revise my answer a little bit to simplify that, if you have some numerica example I can try to solve it
– gimusi
Jul 26 at 12:57
@gimusi I have updated my question a little bit. I will try to draw a picture or something
– derekantrican
Jul 26 at 13:00
In your idea vectors A and B have an origin in common?
– gimusi
Jul 25 at 22:07
In your idea vectors A and B have an origin in common?
– gimusi
Jul 25 at 22:07
and is the plane through the origin?
– gimusi
Jul 25 at 22:12
and is the plane through the origin?
– gimusi
Jul 25 at 22:12
@gimusi yes and yes
– derekantrican
Jul 26 at 12:55
@gimusi yes and yes
– derekantrican
Jul 26 at 12:55
ok then I can revise my answer a little bit to simplify that, if you have some numerica example I can try to solve it
– gimusi
Jul 26 at 12:57
ok then I can revise my answer a little bit to simplify that, if you have some numerica example I can try to solve it
– gimusi
Jul 26 at 12:57
@gimusi I have updated my question a little bit. I will try to draw a picture or something
– derekantrican
Jul 26 at 13:00
@gimusi I have updated my question a little bit. I will try to draw a picture or something
– derekantrican
Jul 26 at 13:00
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
3
down vote
accepted
You don’t really need to construct a matrix to do this. Since $A$ and $B$ are orthogonal, $mathbf u = A/|A|$ and $mathbf v = B/|B|$ form an orthonormal basis for their common plane. If you rotate $mathbf u$ through an angle $theta$ toward $mathbf v$, you get $mathbf ucostheta+mathbf vsintheta$, so to rotate $A$ toward $B$, simply multiply this by $|A|$, i.e., $$Acostheta + B Bsintheta.$$
If you really want a matrix, you can use Rodrigues’ rotation formula to construct one. Let $mathbf n = $, the right-handed normal to the plane of rotation, and designate by $mathbf N$ its “cross-product matrix†$$mathbf N = beginbmatrix0&-n_z&n_y\n_z&0&-n_x\-n_y&n_x&0endbmatrix.$$ The corresponding rotation matrix is then $$mathbf R = mathbf I + (sintheta)mathbf N + (1-costheta)mathbf N^2.$$
answered Jul 26 at 8:14
amd
25.8k2943
We don't need but we can always do that! By the way, very useful answer.
– gimusi
Jul 26 at 13:04
Excellent. Your first solutionAcosθ+(∥A∥/∥B∥)Bsinθended up being what I used. Thank you!
– derekantrican
Jul 26 at 15:52
add a comment |Â
up vote
0
down vote
Hint:
first, move your reference system to be in $x'=A/|A|,y'=B/|B|,z'=(A/|A|) times (B/|B|)$, through the corresponding transformation matrix T;
then apply a rotation matrix $R$ bringing $x'$ towards $y'$ by the required angle
finally multiply by $T^-1$
answered Jul 25 at 22:09
G Cab
15.1k31136
add a comment |Â
up vote
0
down vote
First replace $A$ by $A/|A|$, and similarly for $B$, so that they're unit vectors.
Compute $C = A times B$, the cross product. Let $M$ by the matrix whose columns are $A, B, C$; then $M^-1$ happens to be $M^t$, which is easy to compute -- just put in $A, B, C$ as the rows!
Now let
$$
R = M S M^-1 = M S M^t
$$
where
$$
S = pmatrixc & -s & 0 \
s & c & 0 \
0 & 0 & 1
$$
where $c = cos theta$ and $s = sin theta$, and $R$ is the rotation matrix that you want.
NB: Note that if $theta$ is in degrees, then you need to use $c = cos (pi * theta / 180)$, and similarly for $s$, where $sin$ and $cos$ denote the usual functions that work on angles expressed in radians.
answered Jul 25 at 22:15
John Hughes
59.4k23785
add a comment |Â
up vote
0
down vote
We can proceed as follow
- Fix a basis $mathcalB=v_1,v_2,v_3$ with $v_1=n$ is the normal to the plane, $v_2=A$ and $v_3=B$.
- Then write the rotation matrix $R_mathcalB$ with respect to that basis (we can use the standard rotation matrix).
- Then with respect to the canonical basis, indicating with $M=[v_1,v_2,v_3]$ we have
$$w_mathcalB=R_mathcalBv_mathcalB implies M^-1w=R_mathcalBM^-1vimplies w=MR_mathcalBM^-1v=Rv$$
answered Jul 25 at 22:27
gimusi
65k73583
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You don’t really need to construct a matrix to do this. Since $A$ and $B$ are orthogonal, $mathbf u = A/|A|$ and $mathbf v = B/|B|$ form an orthonormal basis for their common plane. If you rotate $mathbf u$ through an angle $theta$ toward $mathbf v$, you get $mathbf ucostheta+mathbf vsintheta$, so to rotate $A$ toward $B$, simply multiply this by $|A|$, i.e., $$Acostheta + B Bsintheta.$$
If you really want a matrix, you can use Rodrigues’ rotation formula to construct one. Let $mathbf n = $, the right-handed normal to the plane of rotation, and designate by $mathbf N$ its “cross-product matrix†$$mathbf N = beginbmatrix0&-n_z&n_y\n_z&0&-n_x\-n_y&n_x&0endbmatrix.$$ The corresponding rotation matrix is then $$mathbf R = mathbf I + (sintheta)mathbf N + (1-costheta)mathbf N^2.$$
answered Jul 26 at 8:14
amd
25.8k2943
We don't need but we can always do that! By the way, very useful answer.
– gimusi
Jul 26 at 13:04
Excellent. Your first solutionAcosθ+(∥A∥/∥B∥)Bsinθended up being what I used. Thank you!
– derekantrican
Jul 26 at 15:52
add a comment |Â
up vote
3
down vote
accepted
You don’t really need to construct a matrix to do this. Since $A$ and $B$ are orthogonal, $mathbf u = A/|A|$ and $mathbf v = B/|B|$ form an orthonormal basis for their common plane. If you rotate $mathbf u$ through an angle $theta$ toward $mathbf v$, you get $mathbf ucostheta+mathbf vsintheta$, so to rotate $A$ toward $B$, simply multiply this by $|A|$, i.e., $$Acostheta + B Bsintheta.$$
If you really want a matrix, you can use Rodrigues’ rotation formula to construct one. Let $mathbf n = $, the right-handed normal to the plane of rotation, and designate by $mathbf N$ its “cross-product matrix†$$mathbf N = beginbmatrix0&-n_z&n_y\n_z&0&-n_x\-n_y&n_x&0endbmatrix.$$ The corresponding rotation matrix is then $$mathbf R = mathbf I + (sintheta)mathbf N + (1-costheta)mathbf N^2.$$
answered Jul 26 at 8:14
amd
25.8k2943
We don't need but we can always do that! By the way, very useful answer.
– gimusi
Jul 26 at 13:04
Excellent. Your first solutionAcosθ+(∥A∥/∥B∥)Bsinθended up being what I used. Thank you!
– derekantrican
Jul 26 at 15:52
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You don’t really need to construct a matrix to do this. Since $A$ and $B$ are orthogonal, $mathbf u = A/|A|$ and $mathbf v = B/|B|$ form an orthonormal basis for their common plane. If you rotate $mathbf u$ through an angle $theta$ toward $mathbf v$, you get $mathbf ucostheta+mathbf vsintheta$, so to rotate $A$ toward $B$, simply multiply this by $|A|$, i.e., $$Acostheta + B Bsintheta.$$
If you really want a matrix, you can use Rodrigues’ rotation formula to construct one. Let $mathbf n = $, the right-handed normal to the plane of rotation, and designate by $mathbf N$ its “cross-product matrix†$$mathbf N = beginbmatrix0&-n_z&n_y\n_z&0&-n_x\-n_y&n_x&0endbmatrix.$$ The corresponding rotation matrix is then $$mathbf R = mathbf I + (sintheta)mathbf N + (1-costheta)mathbf N^2.$$
answered Jul 26 at 8:14
amd
25.8k2943
You don’t really need to construct a matrix to do this. Since $A$ and $B$ are orthogonal, $mathbf u = A/|A|$ and $mathbf v = B/|B|$ form an orthonormal basis for their common plane. If you rotate $mathbf u$ through an angle $theta$ toward $mathbf v$, you get $mathbf ucostheta+mathbf vsintheta$, so to rotate $A$ toward $B$, simply multiply this by $|A|$, i.e., $$Acostheta + B Bsintheta.$$
If you really want a matrix, you can use Rodrigues’ rotation formula to construct one. Let $mathbf n = $, the right-handed normal to the plane of rotation, and designate by $mathbf N$ its “cross-product matrix†$$mathbf N = beginbmatrix0&-n_z&n_y\n_z&0&-n_x\-n_y&n_x&0endbmatrix.$$ The corresponding rotation matrix is then $$mathbf R = mathbf I + (sintheta)mathbf N + (1-costheta)mathbf N^2.$$
answered Jul 26 at 8:14
amd
25.8k2943
edited Jul 26 at 8:33
answered Jul 26 at 8:14
amd
25.8k2943
answered Jul 26 at 8:14
amd
25.8k2943
answered Jul 26 at 8:14
amd
25.8k2943
25.8k2943
We don't need but we can always do that! By the way, very useful answer.
– gimusi
Jul 26 at 13:04
Excellent. Your first solutionAcosθ+(∥A∥/∥B∥)Bsinθended up being what I used. Thank you!
– derekantrican
Jul 26 at 15:52
add a comment |Â
We don't need but we can always do that! By the way, very useful answer.
– gimusi
Jul 26 at 13:04
Excellent. Your first solutionAcosθ+(∥A∥/∥B∥)Bsinθended up being what I used. Thank you!
– derekantrican
Jul 26 at 15:52
We don't need but we can always do that! By the way, very useful answer.
– gimusi
Jul 26 at 13:04
We don't need but we can always do that! By the way, very useful answer.
– gimusi
Jul 26 at 13:04
Excellent. Your first solution
Acosθ+(∥A∥/∥B∥)Bsinθ ended up being what I used. Thank you!– derekantrican
Jul 26 at 15:52
Excellent. Your first solution
Acosθ+(∥A∥/∥B∥)Bsinθ ended up being what I used. Thank you!– derekantrican
Jul 26 at 15:52
add a comment |Â
up vote
0
down vote
Hint:
first, move your reference system to be in $x'=A/|A|,y'=B/|B|,z'=(A/|A|) times (B/|B|)$, through the corresponding transformation matrix T;
then apply a rotation matrix $R$ bringing $x'$ towards $y'$ by the required angle
finally multiply by $T^-1$
answered Jul 25 at 22:09
G Cab
15.1k31136
add a comment |Â
up vote
0
down vote
Hint:
first, move your reference system to be in $x'=A/|A|,y'=B/|B|,z'=(A/|A|) times (B/|B|)$, through the corresponding transformation matrix T;
then apply a rotation matrix $R$ bringing $x'$ towards $y'$ by the required angle
finally multiply by $T^-1$
answered Jul 25 at 22:09
G Cab
15.1k31136
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint:
first, move your reference system to be in $x'=A/|A|,y'=B/|B|,z'=(A/|A|) times (B/|B|)$, through the corresponding transformation matrix T;
then apply a rotation matrix $R$ bringing $x'$ towards $y'$ by the required angle
finally multiply by $T^-1$
answered Jul 25 at 22:09
G Cab
15.1k31136
Hint:
first, move your reference system to be in $x'=A/|A|,y'=B/|B|,z'=(A/|A|) times (B/|B|)$, through the corresponding transformation matrix T;
then apply a rotation matrix $R$ bringing $x'$ towards $y'$ by the required angle
finally multiply by $T^-1$
answered Jul 25 at 22:09
G Cab
15.1k31136
answered Jul 25 at 22:09
G Cab
15.1k31136
answered Jul 25 at 22:09
G Cab
15.1k31136
answered Jul 25 at 22:09
G Cab
15.1k31136
15.1k31136
add a comment |Â
add a comment |Â
up vote
0
down vote
First replace $A$ by $A/|A|$, and similarly for $B$, so that they're unit vectors.
Compute $C = A times B$, the cross product. Let $M$ by the matrix whose columns are $A, B, C$; then $M^-1$ happens to be $M^t$, which is easy to compute -- just put in $A, B, C$ as the rows!
Now let
$$
R = M S M^-1 = M S M^t
$$
where
$$
S = pmatrixc & -s & 0 \
s & c & 0 \
0 & 0 & 1
$$
where $c = cos theta$ and $s = sin theta$, and $R$ is the rotation matrix that you want.
NB: Note that if $theta$ is in degrees, then you need to use $c = cos (pi * theta / 180)$, and similarly for $s$, where $sin$ and $cos$ denote the usual functions that work on angles expressed in radians.
answered Jul 25 at 22:15
John Hughes
59.4k23785
add a comment |Â
up vote
0
down vote
First replace $A$ by $A/|A|$, and similarly for $B$, so that they're unit vectors.
Compute $C = A times B$, the cross product. Let $M$ by the matrix whose columns are $A, B, C$; then $M^-1$ happens to be $M^t$, which is easy to compute -- just put in $A, B, C$ as the rows!
Now let
$$
R = M S M^-1 = M S M^t
$$
where
$$
S = pmatrixc & -s & 0 \
s & c & 0 \
0 & 0 & 1
$$
where $c = cos theta$ and $s = sin theta$, and $R$ is the rotation matrix that you want.
NB: Note that if $theta$ is in degrees, then you need to use $c = cos (pi * theta / 180)$, and similarly for $s$, where $sin$ and $cos$ denote the usual functions that work on angles expressed in radians.
answered Jul 25 at 22:15
John Hughes
59.4k23785
add a comment |Â
up vote
0
down vote
up vote
0
down vote
First replace $A$ by $A/|A|$, and similarly for $B$, so that they're unit vectors.
Compute $C = A times B$, the cross product. Let $M$ by the matrix whose columns are $A, B, C$; then $M^-1$ happens to be $M^t$, which is easy to compute -- just put in $A, B, C$ as the rows!
Now let
$$
R = M S M^-1 = M S M^t
$$
where
$$
S = pmatrixc & -s & 0 \
s & c & 0 \
0 & 0 & 1
$$
where $c = cos theta$ and $s = sin theta$, and $R$ is the rotation matrix that you want.
NB: Note that if $theta$ is in degrees, then you need to use $c = cos (pi * theta / 180)$, and similarly for $s$, where $sin$ and $cos$ denote the usual functions that work on angles expressed in radians.
answered Jul 25 at 22:15
John Hughes
59.4k23785
First replace $A$ by $A/|A|$, and similarly for $B$, so that they're unit vectors.
Compute $C = A times B$, the cross product. Let $M$ by the matrix whose columns are $A, B, C$; then $M^-1$ happens to be $M^t$, which is easy to compute -- just put in $A, B, C$ as the rows!
Now let
$$
R = M S M^-1 = M S M^t
$$
where
$$
S = pmatrixc & -s & 0 \
s & c & 0 \
0 & 0 & 1
$$
where $c = cos theta$ and $s = sin theta$, and $R$ is the rotation matrix that you want.
NB: Note that if $theta$ is in degrees, then you need to use $c = cos (pi * theta / 180)$, and similarly for $s$, where $sin$ and $cos$ denote the usual functions that work on angles expressed in radians.
answered Jul 25 at 22:15
John Hughes
59.4k23785
answered Jul 25 at 22:15
John Hughes
59.4k23785
answered Jul 25 at 22:15
John Hughes
59.4k23785
answered Jul 25 at 22:15
John Hughes
59.4k23785
59.4k23785
add a comment |Â
add a comment |Â
up vote
0
down vote
We can proceed as follow
- Fix a basis $mathcalB=v_1,v_2,v_3$ with $v_1=n$ is the normal to the plane, $v_2=A$ and $v_3=B$.
- Then write the rotation matrix $R_mathcalB$ with respect to that basis (we can use the standard rotation matrix).
- Then with respect to the canonical basis, indicating with $M=[v_1,v_2,v_3]$ we have
$$w_mathcalB=R_mathcalBv_mathcalB implies M^-1w=R_mathcalBM^-1vimplies w=MR_mathcalBM^-1v=Rv$$
answered Jul 25 at 22:27
gimusi
65k73583
add a comment |Â
up vote
0
down vote
We can proceed as follow
- Fix a basis $mathcalB=v_1,v_2,v_3$ with $v_1=n$ is the normal to the plane, $v_2=A$ and $v_3=B$.
- Then write the rotation matrix $R_mathcalB$ with respect to that basis (we can use the standard rotation matrix).
- Then with respect to the canonical basis, indicating with $M=[v_1,v_2,v_3]$ we have
$$w_mathcalB=R_mathcalBv_mathcalB implies M^-1w=R_mathcalBM^-1vimplies w=MR_mathcalBM^-1v=Rv$$
answered Jul 25 at 22:27
gimusi
65k73583
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We can proceed as follow
- Fix a basis $mathcalB=v_1,v_2,v_3$ with $v_1=n$ is the normal to the plane, $v_2=A$ and $v_3=B$.
- Then write the rotation matrix $R_mathcalB$ with respect to that basis (we can use the standard rotation matrix).
- Then with respect to the canonical basis, indicating with $M=[v_1,v_2,v_3]$ we have
$$w_mathcalB=R_mathcalBv_mathcalB implies M^-1w=R_mathcalBM^-1vimplies w=MR_mathcalBM^-1v=Rv$$
answered Jul 25 at 22:27
gimusi
65k73583
We can proceed as follow
- Fix a basis $mathcalB=v_1,v_2,v_3$ with $v_1=n$ is the normal to the plane, $v_2=A$ and $v_3=B$.
- Then write the rotation matrix $R_mathcalB$ with respect to that basis (we can use the standard rotation matrix).
- Then with respect to the canonical basis, indicating with $M=[v_1,v_2,v_3]$ we have
$$w_mathcalB=R_mathcalBv_mathcalB implies M^-1w=R_mathcalBM^-1vimplies w=MR_mathcalBM^-1v=Rv$$
answered Jul 25 at 22:27
gimusi
65k73583
edited Jul 26 at 13:03
answered Jul 25 at 22:27
gimusi
65k73583
answered Jul 25 at 22:27
gimusi
65k73583
answered Jul 25 at 22:27
gimusi
65k73583
65k73583
add a comment |Â
add a comment |Â
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In your idea vectors A and B have an origin in common?
– gimusi
Jul 25 at 22:07
and is the plane through the origin?
– gimusi
Jul 25 at 22:12
@gimusi yes and yes
– derekantrican
Jul 26 at 12:55
ok then I can revise my answer a little bit to simplify that, if you have some numerica example I can try to solve it
– gimusi
Jul 26 at 12:57
@gimusi I have updated my question a little bit. I will try to draw a picture or something
– derekantrican
Jul 26 at 13:00