Finding new position of a point in $Bbb R^2$ after rotating it about a vector in $Bbb R^2$.
$begingroup$
Suppose $(x,y) in Bbb R^2$. Suppose we rotate this point about a vector $(a,b) in Bbb R^2$ through an angle $theta$. Find the coordinate of the new point.
What I have done is as follows $:$
I first translate the vector $(a,b)$ suitably so that it becomes the origin in the new coordinate system. Let $(x',y')$ be the new coordinate of $(x,y)$. Then $x'=x-a$ and $y'=y-b$. Now in order to find out the coordinate of the point $(x,y)$ after rotating it about the vector $(a,b)$ through an angle $theta$ we need to rotate $(x',y')$ about the origin in the new coordinate system through the same angle $theta$. After rotating let the coordinate of $(x',y')$ be changed to $(x'',y'')$ in the new coordinate system. Then $x'' = x' cos theta - y' sin theta$ and $y'' = x' sin theta + y' cos theta$. Replacing $x'$ and $y'$ by $x-a$ and $y-b$ respectively we get the coordinate of $(x'',y'')$ in the old coordinate system which is the required coordinate of $(x,y)$ after rotating it about the vector $(a,b)$ through an angle $theta$. So
$$x'' = (x-a) cos theta -(y-b) sin theta.$$
$$y'' = (x-a) sin theta + (y-b) cos theta.$$
is the required coordinate of the new point. But the answer given in my book is not matching with the above one. It is given as
$$x'' = (x-a) cos theta - (y-b) sin theta + a.$$
$$y'' = (x-a) sin theta + (y-b) cos theta + b.$$
Where have I done mistake? Please help me in this regard.
Thank you very much.
analytic-geometry rotations
asked Dec 22 '18 at 7:50
Dbchatto67Dbchatto67
963118
$endgroup$
add a comment |
$begingroup$
Suppose $(x,y) in Bbb R^2$. Suppose we rotate this point about a vector $(a,b) in Bbb R^2$ through an angle $theta$. Find the coordinate of the new point.
What I have done is as follows $:$
I first translate the vector $(a,b)$ suitably so that it becomes the origin in the new coordinate system. Let $(x',y')$ be the new coordinate of $(x,y)$. Then $x'=x-a$ and $y'=y-b$. Now in order to find out the coordinate of the point $(x,y)$ after rotating it about the vector $(a,b)$ through an angle $theta$ we need to rotate $(x',y')$ about the origin in the new coordinate system through the same angle $theta$. After rotating let the coordinate of $(x',y')$ be changed to $(x'',y'')$ in the new coordinate system. Then $x'' = x' cos theta - y' sin theta$ and $y'' = x' sin theta + y' cos theta$. Replacing $x'$ and $y'$ by $x-a$ and $y-b$ respectively we get the coordinate of $(x'',y'')$ in the old coordinate system which is the required coordinate of $(x,y)$ after rotating it about the vector $(a,b)$ through an angle $theta$. So
$$x'' = (x-a) cos theta -(y-b) sin theta.$$
$$y'' = (x-a) sin theta + (y-b) cos theta.$$
is the required coordinate of the new point. But the answer given in my book is not matching with the above one. It is given as
$$x'' = (x-a) cos theta - (y-b) sin theta + a.$$
$$y'' = (x-a) sin theta + (y-b) cos theta + b.$$
Where have I done mistake? Please help me in this regard.
Thank you very much.
analytic-geometry rotations
asked Dec 22 '18 at 7:50
Dbchatto67Dbchatto67
963118
$endgroup$
add a comment |
$begingroup$
Suppose $(x,y) in Bbb R^2$. Suppose we rotate this point about a vector $(a,b) in Bbb R^2$ through an angle $theta$. Find the coordinate of the new point.
What I have done is as follows $:$
I first translate the vector $(a,b)$ suitably so that it becomes the origin in the new coordinate system. Let $(x',y')$ be the new coordinate of $(x,y)$. Then $x'=x-a$ and $y'=y-b$. Now in order to find out the coordinate of the point $(x,y)$ after rotating it about the vector $(a,b)$ through an angle $theta$ we need to rotate $(x',y')$ about the origin in the new coordinate system through the same angle $theta$. After rotating let the coordinate of $(x',y')$ be changed to $(x'',y'')$ in the new coordinate system. Then $x'' = x' cos theta - y' sin theta$ and $y'' = x' sin theta + y' cos theta$. Replacing $x'$ and $y'$ by $x-a$ and $y-b$ respectively we get the coordinate of $(x'',y'')$ in the old coordinate system which is the required coordinate of $(x,y)$ after rotating it about the vector $(a,b)$ through an angle $theta$. So
$$x'' = (x-a) cos theta -(y-b) sin theta.$$
$$y'' = (x-a) sin theta + (y-b) cos theta.$$
is the required coordinate of the new point. But the answer given in my book is not matching with the above one. It is given as
$$x'' = (x-a) cos theta - (y-b) sin theta + a.$$
$$y'' = (x-a) sin theta + (y-b) cos theta + b.$$
Where have I done mistake? Please help me in this regard.
Thank you very much.
analytic-geometry rotations
asked Dec 22 '18 at 7:50
Dbchatto67Dbchatto67
963118
$endgroup$
Suppose $(x,y) in Bbb R^2$. Suppose we rotate this point about a vector $(a,b) in Bbb R^2$ through an angle $theta$. Find the coordinate of the new point.
What I have done is as follows $:$
I first translate the vector $(a,b)$ suitably so that it becomes the origin in the new coordinate system. Let $(x',y')$ be the new coordinate of $(x,y)$. Then $x'=x-a$ and $y'=y-b$. Now in order to find out the coordinate of the point $(x,y)$ after rotating it about the vector $(a,b)$ through an angle $theta$ we need to rotate $(x',y')$ about the origin in the new coordinate system through the same angle $theta$. After rotating let the coordinate of $(x',y')$ be changed to $(x'',y'')$ in the new coordinate system. Then $x'' = x' cos theta - y' sin theta$ and $y'' = x' sin theta + y' cos theta$. Replacing $x'$ and $y'$ by $x-a$ and $y-b$ respectively we get the coordinate of $(x'',y'')$ in the old coordinate system which is the required coordinate of $(x,y)$ after rotating it about the vector $(a,b)$ through an angle $theta$. So
$$x'' = (x-a) cos theta -(y-b) sin theta.$$
$$y'' = (x-a) sin theta + (y-b) cos theta.$$
is the required coordinate of the new point. But the answer given in my book is not matching with the above one. It is given as
$$x'' = (x-a) cos theta - (y-b) sin theta + a.$$
$$y'' = (x-a) sin theta + (y-b) cos theta + b.$$
Where have I done mistake? Please help me in this regard.
Thank you very much.
analytic-geometry rotations
analytic-geometry rotations
asked Dec 22 '18 at 7:50
Dbchatto67Dbchatto67
963118
asked Dec 22 '18 at 7:50
Dbchatto67Dbchatto67
963118
asked Dec 22 '18 at 7:50
Dbchatto67Dbchatto67
963118
asked Dec 22 '18 at 7:50
Dbchatto67Dbchatto67
963118
asked Dec 22 '18 at 7:50
Dbchatto67Dbchatto67
963118
963118
add a comment |
add a comment |
1 Answer
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active
oldest
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$begingroup$
Carefully note that $(x'',y'')$ are co-ordinates of the required point with respect to the translated axis, the new co-ordinate system $(x',y')$. To get the co-ordinates of the point $(p,q)$ in the original co-ordinate system, we have $x''=p-aimplies p=x''+a;y''=q-bimplies q=y''+b$.
answered Dec 22 '18 at 8:00
Shubham JohriShubham Johri
5,177717
$endgroup$
$begingroup$
Yeah you are right @Shubham Johri. I messed up with all those primes. Thanks for pointing this out.
$endgroup$
– Dbchatto67
Dec 22 '18 at 8:07
$begingroup$
I observe another thing from here that rotations and translation do not commute with each other. What I have done here is translation followed by rotation. If I do the same thing in the reverse direction i.e. first rotation and then translation I would have ended up with the coordinate $(x cos theta - y sin theta + a, x sin theta + y cos theta + b)$. Isn't it so? Which actually proves in the group theoritic context that the group $G$ of rigid motoins in a plane is not commutative. In particular the subgroup $H subset G$ of all orientation preserving rigid motoins is not commutative.
$endgroup$
– Dbchatto67
Dec 22 '18 at 8:15
$begingroup$
While I'm not really sure about Group Theory, you are correct about the former. Although, here, the question meant specifically for the rotation to happen about $(a,b)$, and it wouldn't make much sense to rotate about $(0,0)$
$endgroup$
– Shubham Johri
Dec 22 '18 at 8:19
$begingroup$
Ok. It's fine. What is your background @Shubham Johri? Your name suggests that you are an Indian national like me.
$endgroup$
– Dbchatto67
Dec 22 '18 at 8:21
$begingroup$
Yes, I'm an Indian undergraduate in Applied Mathematics.
$endgroup$
– Shubham Johri
Dec 22 '18 at 8:27
add a comment |
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$begingroup$
Carefully note that $(x'',y'')$ are co-ordinates of the required point with respect to the translated axis, the new co-ordinate system $(x',y')$. To get the co-ordinates of the point $(p,q)$ in the original co-ordinate system, we have $x''=p-aimplies p=x''+a;y''=q-bimplies q=y''+b$.
answered Dec 22 '18 at 8:00
Shubham JohriShubham Johri
5,177717
$endgroup$
$begingroup$
Yeah you are right @Shubham Johri. I messed up with all those primes. Thanks for pointing this out.
$endgroup$
– Dbchatto67
Dec 22 '18 at 8:07
$begingroup$
I observe another thing from here that rotations and translation do not commute with each other. What I have done here is translation followed by rotation. If I do the same thing in the reverse direction i.e. first rotation and then translation I would have ended up with the coordinate $(x cos theta - y sin theta + a, x sin theta + y cos theta + b)$. Isn't it so? Which actually proves in the group theoritic context that the group $G$ of rigid motoins in a plane is not commutative. In particular the subgroup $H subset G$ of all orientation preserving rigid motoins is not commutative.
$endgroup$
– Dbchatto67
Dec 22 '18 at 8:15
$begingroup$
While I'm not really sure about Group Theory, you are correct about the former. Although, here, the question meant specifically for the rotation to happen about $(a,b)$, and it wouldn't make much sense to rotate about $(0,0)$
$endgroup$
– Shubham Johri
Dec 22 '18 at 8:19
$begingroup$
Ok. It's fine. What is your background @Shubham Johri? Your name suggests that you are an Indian national like me.
$endgroup$
– Dbchatto67
Dec 22 '18 at 8:21
$begingroup$
Yes, I'm an Indian undergraduate in Applied Mathematics.
$endgroup$
– Shubham Johri
Dec 22 '18 at 8:27
add a comment |
$begingroup$
Carefully note that $(x'',y'')$ are co-ordinates of the required point with respect to the translated axis, the new co-ordinate system $(x',y')$. To get the co-ordinates of the point $(p,q)$ in the original co-ordinate system, we have $x''=p-aimplies p=x''+a;y''=q-bimplies q=y''+b$.
answered Dec 22 '18 at 8:00
Shubham JohriShubham Johri
5,177717
$endgroup$
$begingroup$
Yeah you are right @Shubham Johri. I messed up with all those primes. Thanks for pointing this out.
$endgroup$
– Dbchatto67
Dec 22 '18 at 8:07
$begingroup$
I observe another thing from here that rotations and translation do not commute with each other. What I have done here is translation followed by rotation. If I do the same thing in the reverse direction i.e. first rotation and then translation I would have ended up with the coordinate $(x cos theta - y sin theta + a, x sin theta + y cos theta + b)$. Isn't it so? Which actually proves in the group theoritic context that the group $G$ of rigid motoins in a plane is not commutative. In particular the subgroup $H subset G$ of all orientation preserving rigid motoins is not commutative.
$endgroup$
– Dbchatto67
Dec 22 '18 at 8:15
$begingroup$
While I'm not really sure about Group Theory, you are correct about the former. Although, here, the question meant specifically for the rotation to happen about $(a,b)$, and it wouldn't make much sense to rotate about $(0,0)$
$endgroup$
– Shubham Johri
Dec 22 '18 at 8:19
$begingroup$
Ok. It's fine. What is your background @Shubham Johri? Your name suggests that you are an Indian national like me.
$endgroup$
– Dbchatto67
Dec 22 '18 at 8:21
$begingroup$
Yes, I'm an Indian undergraduate in Applied Mathematics.
$endgroup$
– Shubham Johri
Dec 22 '18 at 8:27
add a comment |
$begingroup$
Carefully note that $(x'',y'')$ are co-ordinates of the required point with respect to the translated axis, the new co-ordinate system $(x',y')$. To get the co-ordinates of the point $(p,q)$ in the original co-ordinate system, we have $x''=p-aimplies p=x''+a;y''=q-bimplies q=y''+b$.
answered Dec 22 '18 at 8:00
Shubham JohriShubham Johri
5,177717
$endgroup$
Carefully note that $(x'',y'')$ are co-ordinates of the required point with respect to the translated axis, the new co-ordinate system $(x',y')$. To get the co-ordinates of the point $(p,q)$ in the original co-ordinate system, we have $x''=p-aimplies p=x''+a;y''=q-bimplies q=y''+b$.
answered Dec 22 '18 at 8:00
Shubham JohriShubham Johri
5,177717
answered Dec 22 '18 at 8:00
Shubham JohriShubham Johri
5,177717
answered Dec 22 '18 at 8:00
Shubham JohriShubham Johri
5,177717
answered Dec 22 '18 at 8:00
Shubham JohriShubham Johri
5,177717
5,177717
$begingroup$
Yeah you are right @Shubham Johri. I messed up with all those primes. Thanks for pointing this out.
$endgroup$
– Dbchatto67
Dec 22 '18 at 8:07
$begingroup$
I observe another thing from here that rotations and translation do not commute with each other. What I have done here is translation followed by rotation. If I do the same thing in the reverse direction i.e. first rotation and then translation I would have ended up with the coordinate $(x cos theta - y sin theta + a, x sin theta + y cos theta + b)$. Isn't it so? Which actually proves in the group theoritic context that the group $G$ of rigid motoins in a plane is not commutative. In particular the subgroup $H subset G$ of all orientation preserving rigid motoins is not commutative.
$endgroup$
– Dbchatto67
Dec 22 '18 at 8:15
$begingroup$
While I'm not really sure about Group Theory, you are correct about the former. Although, here, the question meant specifically for the rotation to happen about $(a,b)$, and it wouldn't make much sense to rotate about $(0,0)$
$endgroup$
– Shubham Johri
Dec 22 '18 at 8:19
$begingroup$
Ok. It's fine. What is your background @Shubham Johri? Your name suggests that you are an Indian national like me.
$endgroup$
– Dbchatto67
Dec 22 '18 at 8:21
$begingroup$
Yes, I'm an Indian undergraduate in Applied Mathematics.
$endgroup$
– Shubham Johri
Dec 22 '18 at 8:27
add a comment |
$begingroup$
Yeah you are right @Shubham Johri. I messed up with all those primes. Thanks for pointing this out.
$endgroup$
– Dbchatto67
Dec 22 '18 at 8:07
$begingroup$
I observe another thing from here that rotations and translation do not commute with each other. What I have done here is translation followed by rotation. If I do the same thing in the reverse direction i.e. first rotation and then translation I would have ended up with the coordinate $(x cos theta - y sin theta + a, x sin theta + y cos theta + b)$. Isn't it so? Which actually proves in the group theoritic context that the group $G$ of rigid motoins in a plane is not commutative. In particular the subgroup $H subset G$ of all orientation preserving rigid motoins is not commutative.
$endgroup$
– Dbchatto67
Dec 22 '18 at 8:15
$begingroup$
While I'm not really sure about Group Theory, you are correct about the former. Although, here, the question meant specifically for the rotation to happen about $(a,b)$, and it wouldn't make much sense to rotate about $(0,0)$
$endgroup$
– Shubham Johri
Dec 22 '18 at 8:19
$begingroup$
Ok. It's fine. What is your background @Shubham Johri? Your name suggests that you are an Indian national like me.
$endgroup$
– Dbchatto67
Dec 22 '18 at 8:21
$begingroup$
Yes, I'm an Indian undergraduate in Applied Mathematics.
$endgroup$
– Shubham Johri
Dec 22 '18 at 8:27
$begingroup$
Yeah you are right @Shubham Johri. I messed up with all those primes. Thanks for pointing this out.
$endgroup$
– Dbchatto67
Dec 22 '18 at 8:07
$begingroup$
Yeah you are right @Shubham Johri. I messed up with all those primes. Thanks for pointing this out.
$endgroup$
– Dbchatto67
Dec 22 '18 at 8:07
$begingroup$
I observe another thing from here that rotations and translation do not commute with each other. What I have done here is translation followed by rotation. If I do the same thing in the reverse direction i.e. first rotation and then translation I would have ended up with the coordinate $(x cos theta - y sin theta + a, x sin theta + y cos theta + b)$. Isn't it so? Which actually proves in the group theoritic context that the group $G$ of rigid motoins in a plane is not commutative. In particular the subgroup $H subset G$ of all orientation preserving rigid motoins is not commutative.
$endgroup$
– Dbchatto67
Dec 22 '18 at 8:15
$begingroup$
I observe another thing from here that rotations and translation do not commute with each other. What I have done here is translation followed by rotation. If I do the same thing in the reverse direction i.e. first rotation and then translation I would have ended up with the coordinate $(x cos theta - y sin theta + a, x sin theta + y cos theta + b)$. Isn't it so? Which actually proves in the group theoritic context that the group $G$ of rigid motoins in a plane is not commutative. In particular the subgroup $H subset G$ of all orientation preserving rigid motoins is not commutative.
$endgroup$
– Dbchatto67
Dec 22 '18 at 8:15
$begingroup$
While I'm not really sure about Group Theory, you are correct about the former. Although, here, the question meant specifically for the rotation to happen about $(a,b)$, and it wouldn't make much sense to rotate about $(0,0)$
$endgroup$
– Shubham Johri
Dec 22 '18 at 8:19
$begingroup$
While I'm not really sure about Group Theory, you are correct about the former. Although, here, the question meant specifically for the rotation to happen about $(a,b)$, and it wouldn't make much sense to rotate about $(0,0)$
$endgroup$
– Shubham Johri
Dec 22 '18 at 8:19
$begingroup$
Ok. It's fine. What is your background @Shubham Johri? Your name suggests that you are an Indian national like me.
$endgroup$
– Dbchatto67
Dec 22 '18 at 8:21
$begingroup$
Ok. It's fine. What is your background @Shubham Johri? Your name suggests that you are an Indian national like me.
$endgroup$
– Dbchatto67
Dec 22 '18 at 8:21
$begingroup$
Yes, I'm an Indian undergraduate in Applied Mathematics.
$endgroup$
– Shubham Johri
Dec 22 '18 at 8:27
$begingroup$
Yes, I'm an Indian undergraduate in Applied Mathematics.
$endgroup$
– Shubham Johri
Dec 22 '18 at 8:27
add a comment |
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