Maximum angle between vectors
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 10 hours ago
user2175783
1575
add a comment |
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 10 hours ago
user2175783
1575
1
Lenghts are given?
– greedoid
10 hours ago
On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
– Henning Makholm
10 hours ago
2
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.
– Henning Makholm
10 hours ago
@HenningMakholm Only v,u are given (not v'). I edited the question.
– user2175783
10 hours ago
add a comment |
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 10 hours ago
user2175783
1575
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 10 hours ago
user2175783
1575
asked 10 hours ago
user2175783
1575
edited 10 hours ago
asked 10 hours ago
user2175783
1575
asked 10 hours ago
user2175783
1575
asked 10 hours ago
user2175783
1575
1575
1
Lenghts are given?
– greedoid
10 hours ago
On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
– Henning Makholm
10 hours ago
2
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.
– Henning Makholm
10 hours ago
@HenningMakholm Only v,u are given (not v'). I edited the question.
– user2175783
10 hours ago
add a comment |
1
Lenghts are given?
– greedoid
10 hours ago
On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
– Henning Makholm
10 hours ago
2
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.
– Henning Makholm
10 hours ago
@HenningMakholm Only v,u are given (not v'). I edited the question.
– user2175783
10 hours ago
1
1
Lenghts are given?
– greedoid
10 hours ago
Lenghts are given?
– greedoid
10 hours ago
On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
– Henning Makholm
10 hours ago
On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
– Henning Makholm
10 hours ago
2
2
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.
– Henning Makholm
10 hours ago
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.
– Henning Makholm
10 hours ago
@HenningMakholm Only v,u are given (not v'). I edited the question.
– user2175783
10 hours ago
@HenningMakholm Only v,u are given (not v'). I edited the question.
– user2175783
10 hours ago
add a comment |
2 Answers
2
active
oldest
votes
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 10 hours ago
Henning Makholm
237k16301536
"Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
– user35202
9 hours ago
1
@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).
– Henning Makholm
8 hours ago
add a comment |
$$
vec v = vec u -(-vec v') = vec v = vec u -vec w
$$
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 10 hours ago
Cesareo
8,0883516
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
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 10 hours ago
Henning Makholm
237k16301536
"Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
– user35202
9 hours ago
1
@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).
– Henning Makholm
8 hours ago
add a comment |
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 10 hours ago
Henning Makholm
237k16301536
"Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
– user35202
9 hours ago
1
@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).
– Henning Makholm
8 hours ago
add a comment |
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 10 hours ago
Henning Makholm
237k16301536
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 10 hours ago
Henning Makholm
237k16301536
answered 10 hours ago
Henning Makholm
237k16301536
answered 10 hours ago
Henning Makholm
237k16301536
answered 10 hours ago
Henning Makholm
237k16301536
237k16301536
"Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
– user35202
9 hours ago
1
@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).
– Henning Makholm
8 hours ago
add a comment |
"Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
– user35202
9 hours ago
1
@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).
– Henning Makholm
8 hours ago
"Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
– user35202
9 hours ago
"Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle". Why is that?
– user35202
9 hours ago
1
1
@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).
– Henning Makholm
8 hours ago
@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).
– Henning Makholm
8 hours ago
add a comment |
$$
vec v = vec u -(-vec v') = vec v = vec u -vec w
$$
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 10 hours ago
Cesareo
8,0883516
add a comment |
$$
vec v = vec u -(-vec v') = vec v = vec u -vec w
$$
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 10 hours ago
Cesareo
8,0883516
add a comment |
$$
vec v = vec u -(-vec v') = vec v = vec u -vec w
$$
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 10 hours ago
Cesareo
8,0883516
$$
vec v = vec u -(-vec v') = vec v = vec u -vec w
$$
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 10 hours ago
Cesareo
8,0883516
edited 1 hour ago
answered 10 hours ago
Cesareo
8,0883516
answered 10 hours ago
Cesareo
8,0883516
answered 10 hours ago
Cesareo
8,0883516
8,0883516
add a comment |
add a comment |
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1
Lenghts are given?
– greedoid
10 hours ago
On geometric grounds, if $|u|<|v'|$, then $theta$ should be $0$ or $pi$ such that $v$ and $v'$ are opposite ...
– Henning Makholm
10 hours ago
2
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.
– Henning Makholm
10 hours ago
@HenningMakholm Only v,u are given (not v'). I edited the question.
– user2175783
10 hours ago