Mixing
Given a vector and angle find new vector
$begingroup$
Given a vector and an angle, how can i find an vector that the angle between the two vector is exactly the given angle?
Edit:
We are in the n-dimensional space and the new vector has a fixed given length.
vector-spaces
asked Sep 2 '15 at 11:30
testytesty
11
$endgroup$
add a comment |
$begingroup$
Given a vector and an angle, how can i find an vector that the angle between the two vector is exactly the given angle?
Edit:
We are in the n-dimensional space and the new vector has a fixed given length.
vector-spaces
asked Sep 2 '15 at 11:30
testytesty
11
$endgroup$
$begingroup$
Unless $n=2$, there exist infinitely many vectors that satisfy your requirement.
$endgroup$
– 5xum
Sep 2 '15 at 11:40
$begingroup$
@5xum is it possible to describe all vectors through linear combinations of finite many vectors?
$endgroup$
– testy
Sep 2 '15 at 11:44
add a comment |
$begingroup$
Given a vector and an angle, how can i find an vector that the angle between the two vector is exactly the given angle?
Edit:
We are in the n-dimensional space and the new vector has a fixed given length.
vector-spaces
asked Sep 2 '15 at 11:30
testytesty
11
$endgroup$
Given a vector and an angle, how can i find an vector that the angle between the two vector is exactly the given angle?
Edit:
We are in the n-dimensional space and the new vector has a fixed given length.
vector-spaces
vector-spaces
asked Sep 2 '15 at 11:30
testytesty
11
asked Sep 2 '15 at 11:30
testytesty
11
edited Sep 2 '15 at 11:39
testy
asked Sep 2 '15 at 11:30
testytesty
11
asked Sep 2 '15 at 11:30
testytesty
11
asked Sep 2 '15 at 11:30
testytesty
11
11
$begingroup$
Unless $n=2$, there exist infinitely many vectors that satisfy your requirement.
$endgroup$
– 5xum
Sep 2 '15 at 11:40
$begingroup$
@5xum is it possible to describe all vectors through linear combinations of finite many vectors?
$endgroup$
– testy
Sep 2 '15 at 11:44
add a comment |
$begingroup$
Unless $n=2$, there exist infinitely many vectors that satisfy your requirement.
$endgroup$
– 5xum
Sep 2 '15 at 11:40
$begingroup$
@5xum is it possible to describe all vectors through linear combinations of finite many vectors?
$endgroup$
– testy
Sep 2 '15 at 11:44
$begingroup$
Unless $n=2$, there exist infinitely many vectors that satisfy your requirement.
$endgroup$
– 5xum
Sep 2 '15 at 11:40
$begingroup$
Unless $n=2$, there exist infinitely many vectors that satisfy your requirement.
$endgroup$
– 5xum
Sep 2 '15 at 11:40
$begingroup$
@5xum is it possible to describe all vectors through linear combinations of finite many vectors?
$endgroup$
– testy
Sep 2 '15 at 11:44
$begingroup$
@5xum is it possible to describe all vectors through linear combinations of finite many vectors?
$endgroup$
– testy
Sep 2 '15 at 11:44
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Assuming you are in $mathbb R^2$ take the rotation matrix
$$begin{bmatrix}
cos theta & -sin theta \
sin theta & cos theta \
end{bmatrix}$$
where $theta$ is your angle and multiply it with your vector.
For a vector $vinmathbb R^n$ with $n>2$ you should first decide in which plain you want to apply the rotation. Find an orthonormal basis $(b_i)_{1le ile n}$ of $mathbb R^n$ such that $b_1$ and $b_2$ span your rotation plain and $vinoperatorname{span}(b_1,b_2)$. Now you can translate the rotation from above into that plain. Actually you don't have to care about $b_i$ for $i>2$ since nothing will change there.
answered Sep 2 '15 at 11:32
principal-ideal-domainprincipal-ideal-domain
2,736521
$endgroup$
$begingroup$
we are in the n-dimensional space
$endgroup$
– testy
Sep 2 '15 at 11:32
$begingroup$
@testy See my edit.
$endgroup$
– principal-ideal-domain
Sep 2 '15 at 11:40
add a comment |
$begingroup$
Find another vector that is not linearly dependent on the first (should be easy if you have more than one dimension), then you use that to construct a normal vector and last you combine the first and the normal vector to construct your final vector.
Let's say your original vector is $u$ and then select one of the basis vectors $e_j$ (you may have to select carefully if $u$ is parallel to some of the axes). Now $v = e_j - (e_jcdot u)u/u^2$ is perpendicular to $u$:
$vcdot u = e_jcdot u - (e_jcdot u)u^2/u^2 = 0$
Then we just combine $u$ and $v$:
$w = ucos(alpha) + vsin(alpha)$
In the special case where there is two dimensions one could immediately select $v$ as $(-u_y, u_x)$ resulting in just applying a rotation matrix:
$$begin{bmatrix}
cos alpha & -sin alpha \
sin alpha & cos alpha \
end{bmatrix}$$
answered Sep 2 '15 at 11:35
skykingskyking
14.3k1929
$endgroup$
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
$begingroup$
Assuming you are in $mathbb R^2$ take the rotation matrix
$$begin{bmatrix}
cos theta & -sin theta \
sin theta & cos theta \
end{bmatrix}$$
where $theta$ is your angle and multiply it with your vector.
For a vector $vinmathbb R^n$ with $n>2$ you should first decide in which plain you want to apply the rotation. Find an orthonormal basis $(b_i)_{1le ile n}$ of $mathbb R^n$ such that $b_1$ and $b_2$ span your rotation plain and $vinoperatorname{span}(b_1,b_2)$. Now you can translate the rotation from above into that plain. Actually you don't have to care about $b_i$ for $i>2$ since nothing will change there.
answered Sep 2 '15 at 11:32
principal-ideal-domainprincipal-ideal-domain
2,736521
$endgroup$
$begingroup$
we are in the n-dimensional space
$endgroup$
– testy
Sep 2 '15 at 11:32
$begingroup$
@testy See my edit.
$endgroup$
– principal-ideal-domain
Sep 2 '15 at 11:40
add a comment |
$begingroup$
Assuming you are in $mathbb R^2$ take the rotation matrix
$$begin{bmatrix}
cos theta & -sin theta \
sin theta & cos theta \
end{bmatrix}$$
where $theta$ is your angle and multiply it with your vector.
For a vector $vinmathbb R^n$ with $n>2$ you should first decide in which plain you want to apply the rotation. Find an orthonormal basis $(b_i)_{1le ile n}$ of $mathbb R^n$ such that $b_1$ and $b_2$ span your rotation plain and $vinoperatorname{span}(b_1,b_2)$. Now you can translate the rotation from above into that plain. Actually you don't have to care about $b_i$ for $i>2$ since nothing will change there.
answered Sep 2 '15 at 11:32
principal-ideal-domainprincipal-ideal-domain
2,736521
$endgroup$
$begingroup$
we are in the n-dimensional space
$endgroup$
– testy
Sep 2 '15 at 11:32
$begingroup$
@testy See my edit.
$endgroup$
– principal-ideal-domain
Sep 2 '15 at 11:40
add a comment |
$begingroup$
Assuming you are in $mathbb R^2$ take the rotation matrix
$$begin{bmatrix}
cos theta & -sin theta \
sin theta & cos theta \
end{bmatrix}$$
where $theta$ is your angle and multiply it with your vector.
For a vector $vinmathbb R^n$ with $n>2$ you should first decide in which plain you want to apply the rotation. Find an orthonormal basis $(b_i)_{1le ile n}$ of $mathbb R^n$ such that $b_1$ and $b_2$ span your rotation plain and $vinoperatorname{span}(b_1,b_2)$. Now you can translate the rotation from above into that plain. Actually you don't have to care about $b_i$ for $i>2$ since nothing will change there.
answered Sep 2 '15 at 11:32
principal-ideal-domainprincipal-ideal-domain
2,736521
$endgroup$
Assuming you are in $mathbb R^2$ take the rotation matrix
$$begin{bmatrix}
cos theta & -sin theta \
sin theta & cos theta \
end{bmatrix}$$
where $theta$ is your angle and multiply it with your vector.
For a vector $vinmathbb R^n$ with $n>2$ you should first decide in which plain you want to apply the rotation. Find an orthonormal basis $(b_i)_{1le ile n}$ of $mathbb R^n$ such that $b_1$ and $b_2$ span your rotation plain and $vinoperatorname{span}(b_1,b_2)$. Now you can translate the rotation from above into that plain. Actually you don't have to care about $b_i$ for $i>2$ since nothing will change there.
answered Sep 2 '15 at 11:32
principal-ideal-domainprincipal-ideal-domain
2,736521
edited Sep 2 '15 at 11:40
answered Sep 2 '15 at 11:32
principal-ideal-domainprincipal-ideal-domain
2,736521
answered Sep 2 '15 at 11:32
principal-ideal-domainprincipal-ideal-domain
2,736521
answered Sep 2 '15 at 11:32
principal-ideal-domainprincipal-ideal-domain
2,736521
2,736521
$begingroup$
we are in the n-dimensional space
$endgroup$
– testy
Sep 2 '15 at 11:32
$begingroup$
@testy See my edit.
$endgroup$
– principal-ideal-domain
Sep 2 '15 at 11:40
add a comment |
$begingroup$
we are in the n-dimensional space
$endgroup$
– testy
Sep 2 '15 at 11:32
$begingroup$
@testy See my edit.
$endgroup$
– principal-ideal-domain
Sep 2 '15 at 11:40
$begingroup$
we are in the n-dimensional space
$endgroup$
– testy
Sep 2 '15 at 11:32
$begingroup$
we are in the n-dimensional space
$endgroup$
– testy
Sep 2 '15 at 11:32
$begingroup$
@testy See my edit.
$endgroup$
– principal-ideal-domain
Sep 2 '15 at 11:40
$begingroup$
@testy See my edit.
$endgroup$
– principal-ideal-domain
Sep 2 '15 at 11:40
add a comment |
$begingroup$
Find another vector that is not linearly dependent on the first (should be easy if you have more than one dimension), then you use that to construct a normal vector and last you combine the first and the normal vector to construct your final vector.
Let's say your original vector is $u$ and then select one of the basis vectors $e_j$ (you may have to select carefully if $u$ is parallel to some of the axes). Now $v = e_j - (e_jcdot u)u/u^2$ is perpendicular to $u$:
$vcdot u = e_jcdot u - (e_jcdot u)u^2/u^2 = 0$
Then we just combine $u$ and $v$:
$w = ucos(alpha) + vsin(alpha)$
In the special case where there is two dimensions one could immediately select $v$ as $(-u_y, u_x)$ resulting in just applying a rotation matrix:
$$begin{bmatrix}
cos alpha & -sin alpha \
sin alpha & cos alpha \
end{bmatrix}$$
answered Sep 2 '15 at 11:35
skykingskyking
14.3k1929
$endgroup$
add a comment |
$begingroup$
Find another vector that is not linearly dependent on the first (should be easy if you have more than one dimension), then you use that to construct a normal vector and last you combine the first and the normal vector to construct your final vector.
Let's say your original vector is $u$ and then select one of the basis vectors $e_j$ (you may have to select carefully if $u$ is parallel to some of the axes). Now $v = e_j - (e_jcdot u)u/u^2$ is perpendicular to $u$:
$vcdot u = e_jcdot u - (e_jcdot u)u^2/u^2 = 0$
Then we just combine $u$ and $v$:
$w = ucos(alpha) + vsin(alpha)$
In the special case where there is two dimensions one could immediately select $v$ as $(-u_y, u_x)$ resulting in just applying a rotation matrix:
$$begin{bmatrix}
cos alpha & -sin alpha \
sin alpha & cos alpha \
end{bmatrix}$$
answered Sep 2 '15 at 11:35
skykingskyking
14.3k1929
$endgroup$
add a comment |
$begingroup$
Find another vector that is not linearly dependent on the first (should be easy if you have more than one dimension), then you use that to construct a normal vector and last you combine the first and the normal vector to construct your final vector.
Let's say your original vector is $u$ and then select one of the basis vectors $e_j$ (you may have to select carefully if $u$ is parallel to some of the axes). Now $v = e_j - (e_jcdot u)u/u^2$ is perpendicular to $u$:
$vcdot u = e_jcdot u - (e_jcdot u)u^2/u^2 = 0$
Then we just combine $u$ and $v$:
$w = ucos(alpha) + vsin(alpha)$
In the special case where there is two dimensions one could immediately select $v$ as $(-u_y, u_x)$ resulting in just applying a rotation matrix:
$$begin{bmatrix}
cos alpha & -sin alpha \
sin alpha & cos alpha \
end{bmatrix}$$
answered Sep 2 '15 at 11:35
skykingskyking
14.3k1929
$endgroup$
Find another vector that is not linearly dependent on the first (should be easy if you have more than one dimension), then you use that to construct a normal vector and last you combine the first and the normal vector to construct your final vector.
Let's say your original vector is $u$ and then select one of the basis vectors $e_j$ (you may have to select carefully if $u$ is parallel to some of the axes). Now $v = e_j - (e_jcdot u)u/u^2$ is perpendicular to $u$:
$vcdot u = e_jcdot u - (e_jcdot u)u^2/u^2 = 0$
Then we just combine $u$ and $v$:
$w = ucos(alpha) + vsin(alpha)$
In the special case where there is two dimensions one could immediately select $v$ as $(-u_y, u_x)$ resulting in just applying a rotation matrix:
$$begin{bmatrix}
cos alpha & -sin alpha \
sin alpha & cos alpha \
end{bmatrix}$$
answered Sep 2 '15 at 11:35
skykingskyking
14.3k1929
edited Sep 2 '15 at 11:43
answered Sep 2 '15 at 11:35
skykingskyking
14.3k1929
answered Sep 2 '15 at 11:35
skykingskyking
14.3k1929
answered Sep 2 '15 at 11:35
skykingskyking
14.3k1929
14.3k1929
add a comment |
add a comment |
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$begingroup$
Unless $n=2$, there exist infinitely many vectors that satisfy your requirement.
$endgroup$
– 5xum
Sep 2 '15 at 11:40
$begingroup$
@5xum is it possible to describe all vectors through linear combinations of finite many vectors?
$endgroup$
– testy
Sep 2 '15 at 11:44