Getting point-coordinates after a rotation
$begingroup$
I have two points $p_1$ and $p_2$ on a 2-dimensional graph, each having an $x$-coordinate and a $y$-coordinate. I want to rotate $p_2$ by $60^circ$ around $p_1$, such that $p_1$ is fixed in its position. So, how to know the new $p_2$ coordinates (after rotation)? I think there's some relation between the line that connects $p_1$ to $p_2$ and the angle $60^circ$, but can't figure what is.
geometry rotations
edited Jul 5 '15 at 9:52
Harish Chandra Rajpoot
29.7k103772
asked Sep 30 '12 at 8:22
FindOutIslamNowFindOutIslamNow
1176
$endgroup$
add a comment |
$begingroup$
I have two points $p_1$ and $p_2$ on a 2-dimensional graph, each having an $x$-coordinate and a $y$-coordinate. I want to rotate $p_2$ by $60^circ$ around $p_1$, such that $p_1$ is fixed in its position. So, how to know the new $p_2$ coordinates (after rotation)? I think there's some relation between the line that connects $p_1$ to $p_2$ and the angle $60^circ$, but can't figure what is.
geometry rotations
edited Jul 5 '15 at 9:52
Harish Chandra Rajpoot
29.7k103772
asked Sep 30 '12 at 8:22
FindOutIslamNowFindOutIslamNow
1176
$endgroup$
add a comment |
$begingroup$
I have two points $p_1$ and $p_2$ on a 2-dimensional graph, each having an $x$-coordinate and a $y$-coordinate. I want to rotate $p_2$ by $60^circ$ around $p_1$, such that $p_1$ is fixed in its position. So, how to know the new $p_2$ coordinates (after rotation)? I think there's some relation between the line that connects $p_1$ to $p_2$ and the angle $60^circ$, but can't figure what is.
geometry rotations
edited Jul 5 '15 at 9:52
Harish Chandra Rajpoot
29.7k103772
asked Sep 30 '12 at 8:22
FindOutIslamNowFindOutIslamNow
1176
$endgroup$
I have two points $p_1$ and $p_2$ on a 2-dimensional graph, each having an $x$-coordinate and a $y$-coordinate. I want to rotate $p_2$ by $60^circ$ around $p_1$, such that $p_1$ is fixed in its position. So, how to know the new $p_2$ coordinates (after rotation)? I think there's some relation between the line that connects $p_1$ to $p_2$ and the angle $60^circ$, but can't figure what is.
geometry rotations
geometry rotations
edited Jul 5 '15 at 9:52
Harish Chandra Rajpoot
29.7k103772
asked Sep 30 '12 at 8:22
FindOutIslamNowFindOutIslamNow
1176
edited Jul 5 '15 at 9:52
Harish Chandra Rajpoot
29.7k103772
asked Sep 30 '12 at 8:22
FindOutIslamNowFindOutIslamNow
1176
edited Jul 5 '15 at 9:52
Harish Chandra Rajpoot
29.7k103772
edited Jul 5 '15 at 9:52
Harish Chandra Rajpoot
29.7k103772
edited Jul 5 '15 at 9:52
Harish Chandra Rajpoot
29.7k103772
29.7k103772
asked Sep 30 '12 at 8:22
FindOutIslamNowFindOutIslamNow
1176
asked Sep 30 '12 at 8:22
FindOutIslamNowFindOutIslamNow
1176
asked Sep 30 '12 at 8:22
FindOutIslamNowFindOutIslamNow
1176
1176
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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Write the points in column vectors: $begin{pmatrix} x\y end{pmatrix}$. Then
$P_2':=begin{pmatrix} cosvarphi & -sinvarphi \ sinvarphi & cosvarphi end{pmatrix}cdot (P_2-P_1) + P_1$
where choose $varphi=pm 60^circ$ according which direction you want to rotate.
answered Sep 30 '12 at 8:34
BerciBerci
62.1k23776
$endgroup$
$begingroup$
Thanks. But why do we have a matrix of derivatives? Is this matrix multiplication?
$endgroup$
– FindOutIslamNow
Sep 30 '12 at 8:38
$begingroup$
Yes, matrix multiplication, I could have expanded it. First reduced the problem where $P_1$ is the origo (shifted by $-P_1$ before rotation and $+P_1$ after), and the matrix multiplication gives the rotated vector.
$endgroup$
– Berci
Sep 30 '12 at 9:22
add a comment |
$begingroup$
In the complex number plane, a rotation of a point, $z$, through an angle, $theta$, is accomplished by $z to z e^{i theta}$.
If $z = x + iy$, where $x,y in mathbb R$, and $theta = 60^circ$, then
$x + iy
to (x + i y)(cos 60^circ + i sin 60^circ)
to (x + iy)(1 + isqrt 3)/2
to (x-frac 12ysqrt 3) + i(y +frac 12xsqrt 3)$
In $mathbb R^2$, this translates into
$(x,y) to (x-frac 12ysqrt 3, frac 12xsqrt 3 + y)$
answered Dec 31 '18 at 16:24
steven gregorysteven gregory
18.5k32359
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Write the points in column vectors: $begin{pmatrix} x\y end{pmatrix}$. Then
$P_2':=begin{pmatrix} cosvarphi & -sinvarphi \ sinvarphi & cosvarphi end{pmatrix}cdot (P_2-P_1) + P_1$
where choose $varphi=pm 60^circ$ according which direction you want to rotate.
answered Sep 30 '12 at 8:34
BerciBerci
62.1k23776
$endgroup$
$begingroup$
Thanks. But why do we have a matrix of derivatives? Is this matrix multiplication?
$endgroup$
– FindOutIslamNow
Sep 30 '12 at 8:38
$begingroup$
Yes, matrix multiplication, I could have expanded it. First reduced the problem where $P_1$ is the origo (shifted by $-P_1$ before rotation and $+P_1$ after), and the matrix multiplication gives the rotated vector.
$endgroup$
– Berci
Sep 30 '12 at 9:22
add a comment |
$begingroup$
Write the points in column vectors: $begin{pmatrix} x\y end{pmatrix}$. Then
$P_2':=begin{pmatrix} cosvarphi & -sinvarphi \ sinvarphi & cosvarphi end{pmatrix}cdot (P_2-P_1) + P_1$
where choose $varphi=pm 60^circ$ according which direction you want to rotate.
answered Sep 30 '12 at 8:34
BerciBerci
62.1k23776
$endgroup$
$begingroup$
Thanks. But why do we have a matrix of derivatives? Is this matrix multiplication?
$endgroup$
– FindOutIslamNow
Sep 30 '12 at 8:38
$begingroup$
Yes, matrix multiplication, I could have expanded it. First reduced the problem where $P_1$ is the origo (shifted by $-P_1$ before rotation and $+P_1$ after), and the matrix multiplication gives the rotated vector.
$endgroup$
– Berci
Sep 30 '12 at 9:22
add a comment |
$begingroup$
Write the points in column vectors: $begin{pmatrix} x\y end{pmatrix}$. Then
$P_2':=begin{pmatrix} cosvarphi & -sinvarphi \ sinvarphi & cosvarphi end{pmatrix}cdot (P_2-P_1) + P_1$
where choose $varphi=pm 60^circ$ according which direction you want to rotate.
answered Sep 30 '12 at 8:34
BerciBerci
62.1k23776
$endgroup$
Write the points in column vectors: $begin{pmatrix} x\y end{pmatrix}$. Then
$P_2':=begin{pmatrix} cosvarphi & -sinvarphi \ sinvarphi & cosvarphi end{pmatrix}cdot (P_2-P_1) + P_1$
where choose $varphi=pm 60^circ$ according which direction you want to rotate.
answered Sep 30 '12 at 8:34
BerciBerci
62.1k23776
answered Sep 30 '12 at 8:34
BerciBerci
62.1k23776
answered Sep 30 '12 at 8:34
BerciBerci
62.1k23776
answered Sep 30 '12 at 8:34
BerciBerci
62.1k23776
62.1k23776
$begingroup$
Thanks. But why do we have a matrix of derivatives? Is this matrix multiplication?
$endgroup$
– FindOutIslamNow
Sep 30 '12 at 8:38
$begingroup$
Yes, matrix multiplication, I could have expanded it. First reduced the problem where $P_1$ is the origo (shifted by $-P_1$ before rotation and $+P_1$ after), and the matrix multiplication gives the rotated vector.
$endgroup$
– Berci
Sep 30 '12 at 9:22
add a comment |
$begingroup$
Thanks. But why do we have a matrix of derivatives? Is this matrix multiplication?
$endgroup$
– FindOutIslamNow
Sep 30 '12 at 8:38
$begingroup$
Yes, matrix multiplication, I could have expanded it. First reduced the problem where $P_1$ is the origo (shifted by $-P_1$ before rotation and $+P_1$ after), and the matrix multiplication gives the rotated vector.
$endgroup$
– Berci
Sep 30 '12 at 9:22
$begingroup$
Thanks. But why do we have a matrix of derivatives? Is this matrix multiplication?
$endgroup$
– FindOutIslamNow
Sep 30 '12 at 8:38
$begingroup$
Thanks. But why do we have a matrix of derivatives? Is this matrix multiplication?
$endgroup$
– FindOutIslamNow
Sep 30 '12 at 8:38
$begingroup$
Yes, matrix multiplication, I could have expanded it. First reduced the problem where $P_1$ is the origo (shifted by $-P_1$ before rotation and $+P_1$ after), and the matrix multiplication gives the rotated vector.
$endgroup$
– Berci
Sep 30 '12 at 9:22
$begingroup$
Yes, matrix multiplication, I could have expanded it. First reduced the problem where $P_1$ is the origo (shifted by $-P_1$ before rotation and $+P_1$ after), and the matrix multiplication gives the rotated vector.
$endgroup$
– Berci
Sep 30 '12 at 9:22
add a comment |
$begingroup$
In the complex number plane, a rotation of a point, $z$, through an angle, $theta$, is accomplished by $z to z e^{i theta}$.
If $z = x + iy$, where $x,y in mathbb R$, and $theta = 60^circ$, then
$x + iy
to (x + i y)(cos 60^circ + i sin 60^circ)
to (x + iy)(1 + isqrt 3)/2
to (x-frac 12ysqrt 3) + i(y +frac 12xsqrt 3)$
In $mathbb R^2$, this translates into
$(x,y) to (x-frac 12ysqrt 3, frac 12xsqrt 3 + y)$
answered Dec 31 '18 at 16:24
steven gregorysteven gregory
18.5k32359
$endgroup$
add a comment |
$begingroup$
In the complex number plane, a rotation of a point, $z$, through an angle, $theta$, is accomplished by $z to z e^{i theta}$.
If $z = x + iy$, where $x,y in mathbb R$, and $theta = 60^circ$, then
$x + iy
to (x + i y)(cos 60^circ + i sin 60^circ)
to (x + iy)(1 + isqrt 3)/2
to (x-frac 12ysqrt 3) + i(y +frac 12xsqrt 3)$
In $mathbb R^2$, this translates into
$(x,y) to (x-frac 12ysqrt 3, frac 12xsqrt 3 + y)$
answered Dec 31 '18 at 16:24
steven gregorysteven gregory
18.5k32359
$endgroup$
add a comment |
$begingroup$
In the complex number plane, a rotation of a point, $z$, through an angle, $theta$, is accomplished by $z to z e^{i theta}$.
If $z = x + iy$, where $x,y in mathbb R$, and $theta = 60^circ$, then
$x + iy
to (x + i y)(cos 60^circ + i sin 60^circ)
to (x + iy)(1 + isqrt 3)/2
to (x-frac 12ysqrt 3) + i(y +frac 12xsqrt 3)$
In $mathbb R^2$, this translates into
$(x,y) to (x-frac 12ysqrt 3, frac 12xsqrt 3 + y)$
answered Dec 31 '18 at 16:24
steven gregorysteven gregory
18.5k32359
$endgroup$
In the complex number plane, a rotation of a point, $z$, through an angle, $theta$, is accomplished by $z to z e^{i theta}$.
If $z = x + iy$, where $x,y in mathbb R$, and $theta = 60^circ$, then
$x + iy
to (x + i y)(cos 60^circ + i sin 60^circ)
to (x + iy)(1 + isqrt 3)/2
to (x-frac 12ysqrt 3) + i(y +frac 12xsqrt 3)$
In $mathbb R^2$, this translates into
$(x,y) to (x-frac 12ysqrt 3, frac 12xsqrt 3 + y)$
answered Dec 31 '18 at 16:24
steven gregorysteven gregory
18.5k32359
answered Dec 31 '18 at 16:24
steven gregorysteven gregory
18.5k32359
answered Dec 31 '18 at 16:24
steven gregorysteven gregory
18.5k32359
answered Dec 31 '18 at 16:24
steven gregorysteven gregory
18.5k32359
18.5k32359
add a comment |
add a comment |
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