How can I represent the rotation of a point in $mathbb{R}^2$ graphically?
$begingroup$
Let $P = (x, y)$ be a point in the plane; now if I want to rotate it by an angle $theta$ the point $P$ will be moved from $(x, y)$ to $(X,Y)$ where:
$$X = x(cos theta) - y(sin theta)$$
$$Y = x(sin theta) + y(cos theta)$$
Why is this true, how can I represent it graphically?
Can somebody explain me how to get the representative matrix of the rotation function?
linear-algebra rotations
edited Dec 1 '18 at 21:15
Hans Hüttel
3,1972921
asked Dec 1 '18 at 20:29
RykRyk
234
$endgroup$
add a comment |
$begingroup$
Let $P = (x, y)$ be a point in the plane; now if I want to rotate it by an angle $theta$ the point $P$ will be moved from $(x, y)$ to $(X,Y)$ where:
$$X = x(cos theta) - y(sin theta)$$
$$Y = x(sin theta) + y(cos theta)$$
Why is this true, how can I represent it graphically?
Can somebody explain me how to get the representative matrix of the rotation function?
linear-algebra rotations
edited Dec 1 '18 at 21:15
Hans Hüttel
3,1972921
asked Dec 1 '18 at 20:29
RykRyk
234
$endgroup$
$begingroup$
Sure. Draw a point on a paper and rotate it some angle. Then try to write the new point in terms of the old. You should be able to get it.
$endgroup$
– Dog_69
Dec 1 '18 at 21:08
$begingroup$
@Dog_69 I don't find intuitive at all what you said there, sorry.
$endgroup$
– Ryk
Dec 1 '18 at 21:24
add a comment |
$begingroup$
Let $P = (x, y)$ be a point in the plane; now if I want to rotate it by an angle $theta$ the point $P$ will be moved from $(x, y)$ to $(X,Y)$ where:
$$X = x(cos theta) - y(sin theta)$$
$$Y = x(sin theta) + y(cos theta)$$
Why is this true, how can I represent it graphically?
Can somebody explain me how to get the representative matrix of the rotation function?
linear-algebra rotations
edited Dec 1 '18 at 21:15
Hans Hüttel
3,1972921
asked Dec 1 '18 at 20:29
RykRyk
234
$endgroup$
Let $P = (x, y)$ be a point in the plane; now if I want to rotate it by an angle $theta$ the point $P$ will be moved from $(x, y)$ to $(X,Y)$ where:
$$X = x(cos theta) - y(sin theta)$$
$$Y = x(sin theta) + y(cos theta)$$
Why is this true, how can I represent it graphically?
Can somebody explain me how to get the representative matrix of the rotation function?
linear-algebra rotations
linear-algebra rotations
edited Dec 1 '18 at 21:15
Hans Hüttel
3,1972921
asked Dec 1 '18 at 20:29
RykRyk
234
edited Dec 1 '18 at 21:15
Hans Hüttel
3,1972921
asked Dec 1 '18 at 20:29
RykRyk
234
edited Dec 1 '18 at 21:15
Hans Hüttel
3,1972921
edited Dec 1 '18 at 21:15
Hans Hüttel
3,1972921
edited Dec 1 '18 at 21:15
Hans Hüttel
3,1972921
3,1972921
asked Dec 1 '18 at 20:29
RykRyk
234
asked Dec 1 '18 at 20:29
RykRyk
234
asked Dec 1 '18 at 20:29
RykRyk
234
234
$begingroup$
Sure. Draw a point on a paper and rotate it some angle. Then try to write the new point in terms of the old. You should be able to get it.
$endgroup$
– Dog_69
Dec 1 '18 at 21:08
$begingroup$
@Dog_69 I don't find intuitive at all what you said there, sorry.
$endgroup$
– Ryk
Dec 1 '18 at 21:24
add a comment |
$begingroup$
Sure. Draw a point on a paper and rotate it some angle. Then try to write the new point in terms of the old. You should be able to get it.
$endgroup$
– Dog_69
Dec 1 '18 at 21:08
$begingroup$
@Dog_69 I don't find intuitive at all what you said there, sorry.
$endgroup$
– Ryk
Dec 1 '18 at 21:24
$begingroup$
Sure. Draw a point on a paper and rotate it some angle. Then try to write the new point in terms of the old. You should be able to get it.
$endgroup$
– Dog_69
Dec 1 '18 at 21:08
$begingroup$
Sure. Draw a point on a paper and rotate it some angle. Then try to write the new point in terms of the old. You should be able to get it.
$endgroup$
– Dog_69
Dec 1 '18 at 21:08
$begingroup$
@Dog_69 I don't find intuitive at all what you said there, sorry.
$endgroup$
– Ryk
Dec 1 '18 at 21:24
$begingroup$
@Dog_69 I don't find intuitive at all what you said there, sorry.
$endgroup$
– Ryk
Dec 1 '18 at 21:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For any linear transformation $T: V_1 rightarrow V_2$, where $V_1$ and $V_2$ are vector spaces, its matrix representation consists of the images under $T$ of the basis vectors of $V_2$. Since $V_1 = mathbb{R}^2$, we have that the matrix for a counterclockwise rotation of $theta$ is
$$left[begin{array}{ll} cos theta & - sin theta \
sin theta & cos theta end{array}right]$$
answered Dec 1 '18 at 21:20
Hans HüttelHans Hüttel
3,1972921
$endgroup$
$begingroup$
Well I know that from a booklet but how can you actually reach that matrix, what's the proof behind it?
$endgroup$
– Ryk
Dec 1 '18 at 21:22
add a comment |
$begingroup$
Prove the parallelogram rule: if $(x,y)$ and $(x',y')$ are any two points, then the four points $(0,0)$, $(x,y)$, $(x',y')$ and $(x+x',y+y')$ form a parallelogram.
Now: Where does $(1,0)$ go when rotated by $theta$? Where does $(0,1)$ go? (Use the definitions of sine and cosine.)
Where does $(x,0)$ go when rotated by $theta$? Where does $(0,y)$ go? (Use the results of the last paragraph.)
Where does $(x,y)$ go when rotated by $theta$? (Hint: use the parallelogram rule on the points $(x,0)$ and $(0,y)$. Notice that when you rotate a parallelogram, you get another parallelogram. Also notice that a rectangle is a special type of parallelogram.)
answered Dec 1 '18 at 21:30
Akiva WeinbergerAkiva Weinberger
13.8k12167
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
For any linear transformation $T: V_1 rightarrow V_2$, where $V_1$ and $V_2$ are vector spaces, its matrix representation consists of the images under $T$ of the basis vectors of $V_2$. Since $V_1 = mathbb{R}^2$, we have that the matrix for a counterclockwise rotation of $theta$ is
$$left[begin{array}{ll} cos theta & - sin theta \
sin theta & cos theta end{array}right]$$
answered Dec 1 '18 at 21:20
Hans HüttelHans Hüttel
3,1972921
$endgroup$
$begingroup$
Well I know that from a booklet but how can you actually reach that matrix, what's the proof behind it?
$endgroup$
– Ryk
Dec 1 '18 at 21:22
add a comment |
$begingroup$
For any linear transformation $T: V_1 rightarrow V_2$, where $V_1$ and $V_2$ are vector spaces, its matrix representation consists of the images under $T$ of the basis vectors of $V_2$. Since $V_1 = mathbb{R}^2$, we have that the matrix for a counterclockwise rotation of $theta$ is
$$left[begin{array}{ll} cos theta & - sin theta \
sin theta & cos theta end{array}right]$$
answered Dec 1 '18 at 21:20
Hans HüttelHans Hüttel
3,1972921
$endgroup$
$begingroup$
Well I know that from a booklet but how can you actually reach that matrix, what's the proof behind it?
$endgroup$
– Ryk
Dec 1 '18 at 21:22
add a comment |
$begingroup$
For any linear transformation $T: V_1 rightarrow V_2$, where $V_1$ and $V_2$ are vector spaces, its matrix representation consists of the images under $T$ of the basis vectors of $V_2$. Since $V_1 = mathbb{R}^2$, we have that the matrix for a counterclockwise rotation of $theta$ is
$$left[begin{array}{ll} cos theta & - sin theta \
sin theta & cos theta end{array}right]$$
answered Dec 1 '18 at 21:20
Hans HüttelHans Hüttel
3,1972921
$endgroup$
For any linear transformation $T: V_1 rightarrow V_2$, where $V_1$ and $V_2$ are vector spaces, its matrix representation consists of the images under $T$ of the basis vectors of $V_2$. Since $V_1 = mathbb{R}^2$, we have that the matrix for a counterclockwise rotation of $theta$ is
$$left[begin{array}{ll} cos theta & - sin theta \
sin theta & cos theta end{array}right]$$
answered Dec 1 '18 at 21:20
Hans HüttelHans Hüttel
3,1972921
answered Dec 1 '18 at 21:20
Hans HüttelHans Hüttel
3,1972921
answered Dec 1 '18 at 21:20
Hans HüttelHans Hüttel
3,1972921
answered Dec 1 '18 at 21:20
Hans HüttelHans Hüttel
3,1972921
3,1972921
$begingroup$
Well I know that from a booklet but how can you actually reach that matrix, what's the proof behind it?
$endgroup$
– Ryk
Dec 1 '18 at 21:22
add a comment |
$begingroup$
Well I know that from a booklet but how can you actually reach that matrix, what's the proof behind it?
$endgroup$
– Ryk
Dec 1 '18 at 21:22
$begingroup$
Well I know that from a booklet but how can you actually reach that matrix, what's the proof behind it?
$endgroup$
– Ryk
Dec 1 '18 at 21:22
$begingroup$
Well I know that from a booklet but how can you actually reach that matrix, what's the proof behind it?
$endgroup$
– Ryk
Dec 1 '18 at 21:22
add a comment |
$begingroup$
Prove the parallelogram rule: if $(x,y)$ and $(x',y')$ are any two points, then the four points $(0,0)$, $(x,y)$, $(x',y')$ and $(x+x',y+y')$ form a parallelogram.
Now: Where does $(1,0)$ go when rotated by $theta$? Where does $(0,1)$ go? (Use the definitions of sine and cosine.)
Where does $(x,0)$ go when rotated by $theta$? Where does $(0,y)$ go? (Use the results of the last paragraph.)
Where does $(x,y)$ go when rotated by $theta$? (Hint: use the parallelogram rule on the points $(x,0)$ and $(0,y)$. Notice that when you rotate a parallelogram, you get another parallelogram. Also notice that a rectangle is a special type of parallelogram.)
answered Dec 1 '18 at 21:30
Akiva WeinbergerAkiva Weinberger
13.8k12167
$endgroup$
add a comment |
$begingroup$
Prove the parallelogram rule: if $(x,y)$ and $(x',y')$ are any two points, then the four points $(0,0)$, $(x,y)$, $(x',y')$ and $(x+x',y+y')$ form a parallelogram.
Now: Where does $(1,0)$ go when rotated by $theta$? Where does $(0,1)$ go? (Use the definitions of sine and cosine.)
Where does $(x,0)$ go when rotated by $theta$? Where does $(0,y)$ go? (Use the results of the last paragraph.)
Where does $(x,y)$ go when rotated by $theta$? (Hint: use the parallelogram rule on the points $(x,0)$ and $(0,y)$. Notice that when you rotate a parallelogram, you get another parallelogram. Also notice that a rectangle is a special type of parallelogram.)
answered Dec 1 '18 at 21:30
Akiva WeinbergerAkiva Weinberger
13.8k12167
$endgroup$
add a comment |
$begingroup$
Prove the parallelogram rule: if $(x,y)$ and $(x',y')$ are any two points, then the four points $(0,0)$, $(x,y)$, $(x',y')$ and $(x+x',y+y')$ form a parallelogram.
Now: Where does $(1,0)$ go when rotated by $theta$? Where does $(0,1)$ go? (Use the definitions of sine and cosine.)
Where does $(x,0)$ go when rotated by $theta$? Where does $(0,y)$ go? (Use the results of the last paragraph.)
Where does $(x,y)$ go when rotated by $theta$? (Hint: use the parallelogram rule on the points $(x,0)$ and $(0,y)$. Notice that when you rotate a parallelogram, you get another parallelogram. Also notice that a rectangle is a special type of parallelogram.)
answered Dec 1 '18 at 21:30
Akiva WeinbergerAkiva Weinberger
13.8k12167
$endgroup$
Prove the parallelogram rule: if $(x,y)$ and $(x',y')$ are any two points, then the four points $(0,0)$, $(x,y)$, $(x',y')$ and $(x+x',y+y')$ form a parallelogram.
Now: Where does $(1,0)$ go when rotated by $theta$? Where does $(0,1)$ go? (Use the definitions of sine and cosine.)
Where does $(x,0)$ go when rotated by $theta$? Where does $(0,y)$ go? (Use the results of the last paragraph.)
Where does $(x,y)$ go when rotated by $theta$? (Hint: use the parallelogram rule on the points $(x,0)$ and $(0,y)$. Notice that when you rotate a parallelogram, you get another parallelogram. Also notice that a rectangle is a special type of parallelogram.)
answered Dec 1 '18 at 21:30
Akiva WeinbergerAkiva Weinberger
13.8k12167
answered Dec 1 '18 at 21:30
Akiva WeinbergerAkiva Weinberger
13.8k12167
answered Dec 1 '18 at 21:30
Akiva WeinbergerAkiva Weinberger
13.8k12167
answered Dec 1 '18 at 21:30
Akiva WeinbergerAkiva Weinberger
13.8k12167
13.8k12167
add a comment |
add a comment |
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$begingroup$
Sure. Draw a point on a paper and rotate it some angle. Then try to write the new point in terms of the old. You should be able to get it.
$endgroup$
– Dog_69
Dec 1 '18 at 21:08
$begingroup$
@Dog_69 I don't find intuitive at all what you said there, sorry.
$endgroup$
– Ryk
Dec 1 '18 at 21:24