Orthogonal unit vector in spherical coordinates
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A particle is traveling in the direction of the positive $z$ axis, until eventually, it is deflected. The new direction is given by an azimuthal angle (w.r.t. the positive $z$ axis) $theta$ and a polar angle (within the $x$-$y$ plane) $phi$. The new direction is given by $vec u = (cos phi sin theta, sin phi sin theta, cos theta)$, having length 1.
I am looking for an orthogonal direction $vec v$ to $vec u$, having the same length 1.
Is there a direct expression in $theta$ and $phi$ that yields a normalized $vec v$, without having to compute some intermediate vector and diving all entries of said vector by its norm?
vectors angle
asked Nov 28 '18 at 12:41
bersbers
260214
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add a comment |
$begingroup$
A particle is traveling in the direction of the positive $z$ axis, until eventually, it is deflected. The new direction is given by an azimuthal angle (w.r.t. the positive $z$ axis) $theta$ and a polar angle (within the $x$-$y$ plane) $phi$. The new direction is given by $vec u = (cos phi sin theta, sin phi sin theta, cos theta)$, having length 1.
I am looking for an orthogonal direction $vec v$ to $vec u$, having the same length 1.
Is there a direct expression in $theta$ and $phi$ that yields a normalized $vec v$, without having to compute some intermediate vector and diving all entries of said vector by its norm?
vectors angle
asked Nov 28 '18 at 12:41
bersbers
260214
$endgroup$
$begingroup$
There are infinitely many. Do you care which one?
$endgroup$
– Randall
Nov 28 '18 at 12:48
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@Randall no, any one is fine.
$endgroup$
– bers
Nov 28 '18 at 12:52
$begingroup$
I feel $(sin phi, -cosphi, 0)$ is an easy answer. Or am I missing some edge case there, such as $theta in {0, pi}$?
$endgroup$
– bers
Nov 28 '18 at 13:12
add a comment |
$begingroup$
A particle is traveling in the direction of the positive $z$ axis, until eventually, it is deflected. The new direction is given by an azimuthal angle (w.r.t. the positive $z$ axis) $theta$ and a polar angle (within the $x$-$y$ plane) $phi$. The new direction is given by $vec u = (cos phi sin theta, sin phi sin theta, cos theta)$, having length 1.
I am looking for an orthogonal direction $vec v$ to $vec u$, having the same length 1.
Is there a direct expression in $theta$ and $phi$ that yields a normalized $vec v$, without having to compute some intermediate vector and diving all entries of said vector by its norm?
vectors angle
asked Nov 28 '18 at 12:41
bersbers
260214
$endgroup$
A particle is traveling in the direction of the positive $z$ axis, until eventually, it is deflected. The new direction is given by an azimuthal angle (w.r.t. the positive $z$ axis) $theta$ and a polar angle (within the $x$-$y$ plane) $phi$. The new direction is given by $vec u = (cos phi sin theta, sin phi sin theta, cos theta)$, having length 1.
I am looking for an orthogonal direction $vec v$ to $vec u$, having the same length 1.
Is there a direct expression in $theta$ and $phi$ that yields a normalized $vec v$, without having to compute some intermediate vector and diving all entries of said vector by its norm?
vectors angle
vectors angle
asked Nov 28 '18 at 12:41
bersbers
260214
asked Nov 28 '18 at 12:41
bersbers
260214
asked Nov 28 '18 at 12:41
bersbers
260214
asked Nov 28 '18 at 12:41
bersbers
260214
asked Nov 28 '18 at 12:41
bersbers
260214
260214
$begingroup$
There are infinitely many. Do you care which one?
$endgroup$
– Randall
Nov 28 '18 at 12:48
$begingroup$
@Randall no, any one is fine.
$endgroup$
– bers
Nov 28 '18 at 12:52
$begingroup$
I feel $(sin phi, -cosphi, 0)$ is an easy answer. Or am I missing some edge case there, such as $theta in {0, pi}$?
$endgroup$
– bers
Nov 28 '18 at 13:12
add a comment |
$begingroup$
There are infinitely many. Do you care which one?
$endgroup$
– Randall
Nov 28 '18 at 12:48
$begingroup$
@Randall no, any one is fine.
$endgroup$
– bers
Nov 28 '18 at 12:52
$begingroup$
I feel $(sin phi, -cosphi, 0)$ is an easy answer. Or am I missing some edge case there, such as $theta in {0, pi}$?
$endgroup$
– bers
Nov 28 '18 at 13:12
$begingroup$
There are infinitely many. Do you care which one?
$endgroup$
– Randall
Nov 28 '18 at 12:48
$begingroup$
There are infinitely many. Do you care which one?
$endgroup$
– Randall
Nov 28 '18 at 12:48
$begingroup$
@Randall no, any one is fine.
$endgroup$
– bers
Nov 28 '18 at 12:52
$begingroup$
@Randall no, any one is fine.
$endgroup$
– bers
Nov 28 '18 at 12:52
$begingroup$
I feel $(sin phi, -cosphi, 0)$ is an easy answer. Or am I missing some edge case there, such as $theta in {0, pi}$?
$endgroup$
– bers
Nov 28 '18 at 13:12
$begingroup$
I feel $(sin phi, -cosphi, 0)$ is an easy answer. Or am I missing some edge case there, such as $theta in {0, pi}$?
$endgroup$
– bers
Nov 28 '18 at 13:12
add a comment |
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$begingroup$
There are infinitely many. Do you care which one?
$endgroup$
– Randall
Nov 28 '18 at 12:48
$begingroup$
@Randall no, any one is fine.
$endgroup$
– bers
Nov 28 '18 at 12:52
$begingroup$
I feel $(sin phi, -cosphi, 0)$ is an easy answer. Or am I missing some edge case there, such as $theta in {0, pi}$?
$endgroup$
– bers
Nov 28 '18 at 13:12