Signed angle in plane
What is the formula to compute the signed angle between two vectors $u, vinmathbb{R}^2$ where positive angle is equivalent to a counter-clockwise rotation on the plane
In other words I would like to create a function $angle(u, v)$ where
$$angle(u, v) = -angle(v, u)$$
$$angle(u, v) = -angle(u, -v)$$
$$angle(u, v) = angle(-u, -v)$$
$$angle(u, v) = pi + angle(-u, v)$$
and I want that $angle(e_1, e_2)=frac{pi}{2}$
linear-algebra angle
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What is the formula to compute the signed angle between two vectors $u, vinmathbb{R}^2$ where positive angle is equivalent to a counter-clockwise rotation on the plane
In other words I would like to create a function $angle(u, v)$ where
$$angle(u, v) = -angle(v, u)$$
$$angle(u, v) = -angle(u, -v)$$
$$angle(u, v) = angle(-u, -v)$$
$$angle(u, v) = pi + angle(-u, v)$$
and I want that $angle(e_1, e_2)=frac{pi}{2}$
linear-algebra angle
add a comment |
What is the formula to compute the signed angle between two vectors $u, vinmathbb{R}^2$ where positive angle is equivalent to a counter-clockwise rotation on the plane
In other words I would like to create a function $angle(u, v)$ where
$$angle(u, v) = -angle(v, u)$$
$$angle(u, v) = -angle(u, -v)$$
$$angle(u, v) = angle(-u, -v)$$
$$angle(u, v) = pi + angle(-u, v)$$
and I want that $angle(e_1, e_2)=frac{pi}{2}$
linear-algebra angle
What is the formula to compute the signed angle between two vectors $u, vinmathbb{R}^2$ where positive angle is equivalent to a counter-clockwise rotation on the plane
In other words I would like to create a function $angle(u, v)$ where
$$angle(u, v) = -angle(v, u)$$
$$angle(u, v) = -angle(u, -v)$$
$$angle(u, v) = angle(-u, -v)$$
$$angle(u, v) = pi + angle(-u, v)$$
and I want that $angle(e_1, e_2)=frac{pi}{2}$
linear-algebra angle
linear-algebra angle
edited Nov 30 at 14:03
Andrei
11k21025
11k21025
asked Nov 30 at 13:33
gota
404315
404315
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1 Answer
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Just a small observation first, you will have some minor inconsistencies when the angle is either $pi$ or $0$.
To calculate the angle, you can use the $sin$ and $cos$ function, at the same time. Using both will allow you to uniquely define the angle between $-pi$ and $pi$. The cosine of the angle is given by the scalar product:$$ucdot v=|u||v|costheta$$In $mathbb R^2$ you can write the scalar product as $ucdot v=u_xv_x+u_yv_y$.
You get the sine of the angle using cross product $$utimes v=|u||v|sintheta$$
In $mathbb R^2$ you can write the cross product as $utimes v=u_xv_y-u_yv_x$.
Note that the ratio of the cross product and scalar product is the tangent of the angle. But the range of the $arctan$ function is from $-pi/2$ to $pi/2$. Note however that there is a function $mathrm{arctan2}$ or $mathrm{atan2}$ in some programming languages that take the sine and cosine and give you the correct angle. See for example the definition on wikipedia
Then $$angle(u,v)=mathrm{arctan2}(utimes v,ucdot v)$$
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Just a small observation first, you will have some minor inconsistencies when the angle is either $pi$ or $0$.
To calculate the angle, you can use the $sin$ and $cos$ function, at the same time. Using both will allow you to uniquely define the angle between $-pi$ and $pi$. The cosine of the angle is given by the scalar product:$$ucdot v=|u||v|costheta$$In $mathbb R^2$ you can write the scalar product as $ucdot v=u_xv_x+u_yv_y$.
You get the sine of the angle using cross product $$utimes v=|u||v|sintheta$$
In $mathbb R^2$ you can write the cross product as $utimes v=u_xv_y-u_yv_x$.
Note that the ratio of the cross product and scalar product is the tangent of the angle. But the range of the $arctan$ function is from $-pi/2$ to $pi/2$. Note however that there is a function $mathrm{arctan2}$ or $mathrm{atan2}$ in some programming languages that take the sine and cosine and give you the correct angle. See for example the definition on wikipedia
Then $$angle(u,v)=mathrm{arctan2}(utimes v,ucdot v)$$
add a comment |
Just a small observation first, you will have some minor inconsistencies when the angle is either $pi$ or $0$.
To calculate the angle, you can use the $sin$ and $cos$ function, at the same time. Using both will allow you to uniquely define the angle between $-pi$ and $pi$. The cosine of the angle is given by the scalar product:$$ucdot v=|u||v|costheta$$In $mathbb R^2$ you can write the scalar product as $ucdot v=u_xv_x+u_yv_y$.
You get the sine of the angle using cross product $$utimes v=|u||v|sintheta$$
In $mathbb R^2$ you can write the cross product as $utimes v=u_xv_y-u_yv_x$.
Note that the ratio of the cross product and scalar product is the tangent of the angle. But the range of the $arctan$ function is from $-pi/2$ to $pi/2$. Note however that there is a function $mathrm{arctan2}$ or $mathrm{atan2}$ in some programming languages that take the sine and cosine and give you the correct angle. See for example the definition on wikipedia
Then $$angle(u,v)=mathrm{arctan2}(utimes v,ucdot v)$$
add a comment |
Just a small observation first, you will have some minor inconsistencies when the angle is either $pi$ or $0$.
To calculate the angle, you can use the $sin$ and $cos$ function, at the same time. Using both will allow you to uniquely define the angle between $-pi$ and $pi$. The cosine of the angle is given by the scalar product:$$ucdot v=|u||v|costheta$$In $mathbb R^2$ you can write the scalar product as $ucdot v=u_xv_x+u_yv_y$.
You get the sine of the angle using cross product $$utimes v=|u||v|sintheta$$
In $mathbb R^2$ you can write the cross product as $utimes v=u_xv_y-u_yv_x$.
Note that the ratio of the cross product and scalar product is the tangent of the angle. But the range of the $arctan$ function is from $-pi/2$ to $pi/2$. Note however that there is a function $mathrm{arctan2}$ or $mathrm{atan2}$ in some programming languages that take the sine and cosine and give you the correct angle. See for example the definition on wikipedia
Then $$angle(u,v)=mathrm{arctan2}(utimes v,ucdot v)$$
Just a small observation first, you will have some minor inconsistencies when the angle is either $pi$ or $0$.
To calculate the angle, you can use the $sin$ and $cos$ function, at the same time. Using both will allow you to uniquely define the angle between $-pi$ and $pi$. The cosine of the angle is given by the scalar product:$$ucdot v=|u||v|costheta$$In $mathbb R^2$ you can write the scalar product as $ucdot v=u_xv_x+u_yv_y$.
You get the sine of the angle using cross product $$utimes v=|u||v|sintheta$$
In $mathbb R^2$ you can write the cross product as $utimes v=u_xv_y-u_yv_x$.
Note that the ratio of the cross product and scalar product is the tangent of the angle. But the range of the $arctan$ function is from $-pi/2$ to $pi/2$. Note however that there is a function $mathrm{arctan2}$ or $mathrm{atan2}$ in some programming languages that take the sine and cosine and give you the correct angle. See for example the definition on wikipedia
Then $$angle(u,v)=mathrm{arctan2}(utimes v,ucdot v)$$
answered Nov 30 at 14:22
Andrei
11k21025
11k21025
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