Maximum angle between vectors
$begingroup$
Consider 3 vectors $textbf{v},textbf{v}',textbf{u}$ related by
$$textbf{v}=textbf{u}+textbf{v}'$$
Let $theta$ be the angle between $textbf{v}$ and $textbf{u}$ and let $phi$ be the angle between $textbf{v}$ and $-textbf{v}'$.
For what angle $theta$ is the angle $phi$ a maximum? The magnitudes $v,u$ are given and you can express the angle in terms of them.
I started by doing the dot product of the equation with itself getting
$$v^{2}=vucostheta+vv'cosphi=u^{2}+v^{'2}+2uv'cos (pi/2-phi-theta)$$
I thought I would then find the derivative of this expression with respect to $theta$ and set $dphi/dtheta=0$. That gives
$$-vusintheta=2uv'sin(pi/2-phi-theta)$$ or
$$vsintheta=2v'sin(phi+theta-pi/2)=-2v'cos(phi+theta)$$
which seems overdetermined.
calculus vectors
asked Dec 25 '18 at 23:49
user2175783user2175783
1876
$endgroup$
add a comment |
$begingroup$
Consider 3 vectors $textbf{v},textbf{v}',textbf{u}$ related by
$$textbf{v}=textbf{u}+textbf{v}'$$
Let $theta$ be the angle between $textbf{v}$ and $textbf{u}$ and let $phi$ be the angle between $textbf{v}$ and $-textbf{v}'$.
For what angle $theta$ is the angle $phi$ a maximum? The magnitudes $v,u$ are given and you can express the angle in terms of them.
I started by doing the dot product of the equation with itself getting
$$v^{2}=vucostheta+vv'cosphi=u^{2}+v^{'2}+2uv'cos (pi/2-phi-theta)$$
I thought I would then find the derivative of this expression with respect to $theta$ and set $dphi/dtheta=0$. That gives
$$-vusintheta=2uv'sin(pi/2-phi-theta)$$ or
$$vsintheta=2v'sin(phi+theta-pi/2)=-2v'cos(phi+theta)$$
which seems overdetermined.
calculus vectors
asked Dec 25 '18 at 23:49
user2175783user2175783
1876
$endgroup$
1
$begingroup$
Lenghts are given?
$endgroup$
– greedoid
Dec 25 '18 at 23:51
$begingroup$
On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
$endgroup$
– Henning Makholm
Dec 25 '18 at 23:56
2
$begingroup$
If all three magnitudes are given, then both $theta$ and $phi$ can be computed by standard triangle solving. There's no choice and no maxima or minima.
$endgroup$
– Henning Makholm
Dec 25 '18 at 23:57
$begingroup$
@HenningMakholm Only v,u are given (not v'). I edited the question.
$endgroup$
– user2175783
Dec 26 '18 at 0:02
add a comment |
$begingroup$
Consider 3 vectors $textbf{v},textbf{v}',textbf{u}$ related by
$$textbf{v}=textbf{u}+textbf{v}'$$
Let $theta$ be the angle between $textbf{v}$ and $textbf{u}$ and let $phi$ be the angle between $textbf{v}$ and $-textbf{v}'$.
For what angle $theta$ is the angle $phi$ a maximum? The magnitudes $v,u$ are given and you can express the angle in terms of them.
I started by doing the dot product of the equation with itself getting
$$v^{2}=vucostheta+vv'cosphi=u^{2}+v^{'2}+2uv'cos (pi/2-phi-theta)$$
I thought I would then find the derivative of this expression with respect to $theta$ and set $dphi/dtheta=0$. That gives
$$-vusintheta=2uv'sin(pi/2-phi-theta)$$ or
$$vsintheta=2v'sin(phi+theta-pi/2)=-2v'cos(phi+theta)$$
which seems overdetermined.
calculus vectors
asked Dec 25 '18 at 23:49
user2175783user2175783
1876
$endgroup$
Consider 3 vectors $textbf{v},textbf{v}',textbf{u}$ related by
$$textbf{v}=textbf{u}+textbf{v}'$$
Let $theta$ be the angle between $textbf{v}$ and $textbf{u}$ and let $phi$ be the angle between $textbf{v}$ and $-textbf{v}'$.
For what angle $theta$ is the angle $phi$ a maximum? The magnitudes $v,u$ are given and you can express the angle in terms of them.
I started by doing the dot product of the equation with itself getting
$$v^{2}=vucostheta+vv'cosphi=u^{2}+v^{'2}+2uv'cos (pi/2-phi-theta)$$
I thought I would then find the derivative of this expression with respect to $theta$ and set $dphi/dtheta=0$. That gives
$$-vusintheta=2uv'sin(pi/2-phi-theta)$$ or
$$vsintheta=2v'sin(phi+theta-pi/2)=-2v'cos(phi+theta)$$
which seems overdetermined.
calculus vectors
calculus vectors
asked Dec 25 '18 at 23:49
user2175783user2175783
1876
asked Dec 25 '18 at 23:49
user2175783user2175783
1876
edited Dec 26 '18 at 0:01
user2175783
asked Dec 25 '18 at 23:49
user2175783user2175783
1876
asked Dec 25 '18 at 23:49
user2175783user2175783
1876
asked Dec 25 '18 at 23:49
user2175783user2175783
1876
1876
1
$begingroup$
Lenghts are given?
$endgroup$
– greedoid
Dec 25 '18 at 23:51
$begingroup$
On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
$endgroup$
– Henning Makholm
Dec 25 '18 at 23:56
2
$begingroup$
If all three magnitudes are given, then both $theta$ and $phi$ can be computed by standard triangle solving. There's no choice and no maxima or minima.
$endgroup$
– Henning Makholm
Dec 25 '18 at 23:57
$begingroup$
@HenningMakholm Only v,u are given (not v'). I edited the question.
$endgroup$
– user2175783
Dec 26 '18 at 0:02
add a comment |
1
$begingroup$
Lenghts are given?
$endgroup$
– greedoid
Dec 25 '18 at 23:51
$begingroup$
On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
$endgroup$
– Henning Makholm
Dec 25 '18 at 23:56
2
$begingroup$
If all three magnitudes are given, then both $theta$ and $phi$ can be computed by standard triangle solving. There's no choice and no maxima or minima.
$endgroup$
– Henning Makholm
Dec 25 '18 at 23:57
$begingroup$
@HenningMakholm Only v,u are given (not v'). I edited the question.
$endgroup$
– user2175783
Dec 26 '18 at 0:02
1
1
$begingroup$
Lenghts are given?
$endgroup$
– greedoid
Dec 25 '18 at 23:51
$begingroup$
Lenghts are given?
$endgroup$
– greedoid
Dec 25 '18 at 23:51
$begingroup$
On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
$endgroup$
– Henning Makholm
Dec 25 '18 at 23:56
$begingroup$
On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
$endgroup$
– Henning Makholm
Dec 25 '18 at 23:56
2
2
$begingroup$
If all three magnitudes are given, then both $theta$ and $phi$ can be computed by standard triangle solving. There's no choice and no maxima or minima.
$endgroup$
– Henning Makholm
Dec 25 '18 at 23:57
$begingroup$
If all three magnitudes are given, then both $theta$ and $phi$ can be computed by standard triangle solving. There's no choice and no maxima or minima.
$endgroup$
– Henning Makholm
Dec 25 '18 at 23:57
$begingroup$
@HenningMakholm Only v,u are given (not v'). I edited the question.
$endgroup$
– user2175783
Dec 26 '18 at 0:02
$begingroup$
@HenningMakholm Only v,u are given (not v'). I edited the question.
$endgroup$
– user2175783
Dec 26 '18 at 0:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Unless you're working under peculiar constraints, doing this algebraically is almost certainly a tactical error. Instead, draw a diagram:
A
/
u/ v
|/ |
B------>C
v'
The angles are $phi$ at C and $theta$ at A.
$phi$ will certainly be a maximum if we can make it $pi$ which happens if $|u|>|v|$ and $theta=0$.
Then consider the case $|u|<|v|$. If we have already decided on the positions of A and C, the possible positions of B is a circle with center A. Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle -- which are the ones where the angle is B is right!
Basic trigonometry then gives us $|u|=|v|costheta$.
answered Dec 26 '18 at 0:16
Henning MakholmHenning Makholm
239k17304541
$endgroup$
$begingroup$
"Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
$endgroup$
– user35202
Dec 26 '18 at 1:37
1
$begingroup$
@user35202: Geometrically, the tangent to the circle through C separates the rays that intersect the circle (which are the ones that represent possible $phi$s) from those that don't intersect the circle (and therefor cannot be possible $phi$s).
$endgroup$
– Henning Makholm
Dec 26 '18 at 1:42
add a comment |
$begingroup$
$$
vec v = vec u -(-vec v') = vec v = vec u -vec w
$$
so now
$$
vec ucdotvec v = ||vec u||^2-vec ucdotvec w
$$
or
$$
||vec u||||vec v||costheta=||vec u|| - ||vec u||||vec w||cosphi
$$
and then assuming that $vec ucdotvec w ne 0$
$$
phi = arccosleft(a+bcosthetaright)
$$
and now deriving
$$
frac{dphi}{dtheta} = frac{b sin (theta )}{sqrt{1-(a+b cos (theta ))^2}} = 0
$$
which gives $theta = 0 + kpi$
answered Dec 26 '18 at 0:30
CesareoCesareo
8,6693516
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Unless you're working under peculiar constraints, doing this algebraically is almost certainly a tactical error. Instead, draw a diagram:
A
/
u/ v
|/ |
B------>C
v'
The angles are $phi$ at C and $theta$ at A.
$phi$ will certainly be a maximum if we can make it $pi$ which happens if $|u|>|v|$ and $theta=0$.
Then consider the case $|u|<|v|$. If we have already decided on the positions of A and C, the possible positions of B is a circle with center A. Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle -- which are the ones where the angle is B is right!
Basic trigonometry then gives us $|u|=|v|costheta$.
answered Dec 26 '18 at 0:16
Henning MakholmHenning Makholm
239k17304541
$endgroup$
$begingroup$
"Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
$endgroup$
– user35202
Dec 26 '18 at 1:37
1
$begingroup$
@user35202: Geometrically, the tangent to the circle through C separates the rays that intersect the circle (which are the ones that represent possible $phi$s) from those that don't intersect the circle (and therefor cannot be possible $phi$s).
$endgroup$
– Henning Makholm
Dec 26 '18 at 1:42
add a comment |
$begingroup$
Unless you're working under peculiar constraints, doing this algebraically is almost certainly a tactical error. Instead, draw a diagram:
A
/
u/ v
|/ |
B------>C
v'
The angles are $phi$ at C and $theta$ at A.
$phi$ will certainly be a maximum if we can make it $pi$ which happens if $|u|>|v|$ and $theta=0$.
Then consider the case $|u|<|v|$. If we have already decided on the positions of A and C, the possible positions of B is a circle with center A. Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle -- which are the ones where the angle is B is right!
Basic trigonometry then gives us $|u|=|v|costheta$.
answered Dec 26 '18 at 0:16
Henning MakholmHenning Makholm
239k17304541
$endgroup$
$begingroup$
"Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
$endgroup$
– user35202
Dec 26 '18 at 1:37
1
$begingroup$
@user35202: Geometrically, the tangent to the circle through C separates the rays that intersect the circle (which are the ones that represent possible $phi$s) from those that don't intersect the circle (and therefor cannot be possible $phi$s).
$endgroup$
– Henning Makholm
Dec 26 '18 at 1:42
add a comment |
$begingroup$
Unless you're working under peculiar constraints, doing this algebraically is almost certainly a tactical error. Instead, draw a diagram:
A
/
u/ v
|/ |
B------>C
v'
The angles are $phi$ at C and $theta$ at A.
$phi$ will certainly be a maximum if we can make it $pi$ which happens if $|u|>|v|$ and $theta=0$.
Then consider the case $|u|<|v|$. If we have already decided on the positions of A and C, the possible positions of B is a circle with center A. Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle -- which are the ones where the angle is B is right!
Basic trigonometry then gives us $|u|=|v|costheta$.
answered Dec 26 '18 at 0:16
Henning MakholmHenning Makholm
239k17304541
$endgroup$
Unless you're working under peculiar constraints, doing this algebraically is almost certainly a tactical error. Instead, draw a diagram:
A
/
u/ v
|/ |
B------>C
v'
The angles are $phi$ at C and $theta$ at A.
$phi$ will certainly be a maximum if we can make it $pi$ which happens if $|u|>|v|$ and $theta=0$.
Then consider the case $|u|<|v|$. If we have already decided on the positions of A and C, the possible positions of B is a circle with center A. Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle -- which are the ones where the angle is B is right!
Basic trigonometry then gives us $|u|=|v|costheta$.
answered Dec 26 '18 at 0:16
Henning MakholmHenning Makholm
239k17304541
answered Dec 26 '18 at 0:16
Henning MakholmHenning Makholm
239k17304541
answered Dec 26 '18 at 0:16
Henning MakholmHenning Makholm
239k17304541
answered Dec 26 '18 at 0:16
Henning MakholmHenning Makholm
239k17304541
239k17304541
$begingroup$
"Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
$endgroup$
– user35202
Dec 26 '18 at 1:37
1
$begingroup$
@user35202: Geometrically, the tangent to the circle through C separates the rays that intersect the circle (which are the ones that represent possible $phi$s) from those that don't intersect the circle (and therefor cannot be possible $phi$s).
$endgroup$
– Henning Makholm
Dec 26 '18 at 1:42
add a comment |
$begingroup$
"Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
$endgroup$
– user35202
Dec 26 '18 at 1:37
1
$begingroup$
@user35202: Geometrically, the tangent to the circle through C separates the rays that intersect the circle (which are the ones that represent possible $phi$s) from those that don't intersect the circle (and therefor cannot be possible $phi$s).
$endgroup$
– Henning Makholm
Dec 26 '18 at 1:42
$begingroup$
"Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
$endgroup$
– user35202
Dec 26 '18 at 1:37
$begingroup$
"Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
$endgroup$
– user35202
Dec 26 '18 at 1:37
1
1
$begingroup$
@user35202: Geometrically, the tangent to the circle through C separates the rays that intersect the circle (which are the ones that represent possible $phi$s) from those that don't intersect the circle (and therefor cannot be possible $phi$s).
$endgroup$
– Henning Makholm
Dec 26 '18 at 1:42
$begingroup$
@user35202: Geometrically, the tangent to the circle through C separates the rays that intersect the circle (which are the ones that represent possible $phi$s) from those that don't intersect the circle (and therefor cannot be possible $phi$s).
$endgroup$
– Henning Makholm
Dec 26 '18 at 1:42
add a comment |
$begingroup$
$$
vec v = vec u -(-vec v') = vec v = vec u -vec w
$$
so now
$$
vec ucdotvec v = ||vec u||^2-vec ucdotvec w
$$
or
$$
||vec u||||vec v||costheta=||vec u|| - ||vec u||||vec w||cosphi
$$
and then assuming that $vec ucdotvec w ne 0$
$$
phi = arccosleft(a+bcosthetaright)
$$
and now deriving
$$
frac{dphi}{dtheta} = frac{b sin (theta )}{sqrt{1-(a+b cos (theta ))^2}} = 0
$$
which gives $theta = 0 + kpi$
answered Dec 26 '18 at 0:30
CesareoCesareo
8,6693516
$endgroup$
add a comment |
$begingroup$
$$
vec v = vec u -(-vec v') = vec v = vec u -vec w
$$
so now
$$
vec ucdotvec v = ||vec u||^2-vec ucdotvec w
$$
or
$$
||vec u||||vec v||costheta=||vec u|| - ||vec u||||vec w||cosphi
$$
and then assuming that $vec ucdotvec w ne 0$
$$
phi = arccosleft(a+bcosthetaright)
$$
and now deriving
$$
frac{dphi}{dtheta} = frac{b sin (theta )}{sqrt{1-(a+b cos (theta ))^2}} = 0
$$
which gives $theta = 0 + kpi$
answered Dec 26 '18 at 0:30
CesareoCesareo
8,6693516
$endgroup$
add a comment |
$begingroup$
$$
vec v = vec u -(-vec v') = vec v = vec u -vec w
$$
so now
$$
vec ucdotvec v = ||vec u||^2-vec ucdotvec w
$$
or
$$
||vec u||||vec v||costheta=||vec u|| - ||vec u||||vec w||cosphi
$$
and then assuming that $vec ucdotvec w ne 0$
$$
phi = arccosleft(a+bcosthetaright)
$$
and now deriving
$$
frac{dphi}{dtheta} = frac{b sin (theta )}{sqrt{1-(a+b cos (theta ))^2}} = 0
$$
which gives $theta = 0 + kpi$
answered Dec 26 '18 at 0:30
CesareoCesareo
8,6693516
$endgroup$
$$
vec v = vec u -(-vec v') = vec v = vec u -vec w
$$
so now
$$
vec ucdotvec v = ||vec u||^2-vec ucdotvec w
$$
or
$$
||vec u||||vec v||costheta=||vec u|| - ||vec u||||vec w||cosphi
$$
and then assuming that $vec ucdotvec w ne 0$
$$
phi = arccosleft(a+bcosthetaright)
$$
and now deriving
$$
frac{dphi}{dtheta} = frac{b sin (theta )}{sqrt{1-(a+b cos (theta ))^2}} = 0
$$
which gives $theta = 0 + kpi$
answered Dec 26 '18 at 0:30
CesareoCesareo
8,6693516
edited Dec 26 '18 at 9:12
answered Dec 26 '18 at 0:30
CesareoCesareo
8,6693516
answered Dec 26 '18 at 0:30
CesareoCesareo
8,6693516
answered Dec 26 '18 at 0:30
CesareoCesareo
8,6693516
8,6693516
add a comment |
add a comment |
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1
$begingroup$
Lenghts are given?
$endgroup$
– greedoid
Dec 25 '18 at 23:51
$begingroup$
On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
$endgroup$
– Henning Makholm
Dec 25 '18 at 23:56
2
$begingroup$
If all three magnitudes are given, then both $theta$ and $phi$ can be computed by standard triangle solving. There's no choice and no maxima or minima.
$endgroup$
– Henning Makholm
Dec 25 '18 at 23:57
$begingroup$
@HenningMakholm Only v,u are given (not v'). I edited the question.
$endgroup$
– user2175783
Dec 26 '18 at 0:02