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Degree between vector and point
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$begingroup$
I have a vector and a point $(x, y)$. The vector starts from $(0, 0)$ and goes to $(x_1, y_1)$. $x$, $y$, $x_1$, $y_1$ are known. How can I get the degree that vector should rotate clockwise to face this point? Scheme
trigonometry vectors
edited Mar 21 at 15:09
Javi
3,15321032
asked Mar 21 at 14:58
Grigory PankoGrigory Panko
31
$endgroup$
add a comment |
$begingroup$
I have a vector and a point $(x, y)$. The vector starts from $(0, 0)$ and goes to $(x_1, y_1)$. $x$, $y$, $x_1$, $y_1$ are known. How can I get the degree that vector should rotate clockwise to face this point? Scheme
trigonometry vectors
edited Mar 21 at 15:09
Javi
3,15321032
asked Mar 21 at 14:58
Grigory PankoGrigory Panko
31
$endgroup$
$begingroup$
Trigonometry; apply it to vector from $(0,0)$ to $(x,y)$ to compute $text {cos}(theta)$ and find the angle $theta$. The same for the other one, finding $theta_1$.
$endgroup$
– Mauro ALLEGRANZA
Mar 21 at 15:05
add a comment |
$begingroup$
I have a vector and a point $(x, y)$. The vector starts from $(0, 0)$ and goes to $(x_1, y_1)$. $x$, $y$, $x_1$, $y_1$ are known. How can I get the degree that vector should rotate clockwise to face this point? Scheme
trigonometry vectors
edited Mar 21 at 15:09
Javi
3,15321032
asked Mar 21 at 14:58
Grigory PankoGrigory Panko
31
$endgroup$
I have a vector and a point $(x, y)$. The vector starts from $(0, 0)$ and goes to $(x_1, y_1)$. $x$, $y$, $x_1$, $y_1$ are known. How can I get the degree that vector should rotate clockwise to face this point? Scheme
trigonometry vectors
trigonometry vectors
edited Mar 21 at 15:09
Javi
3,15321032
asked Mar 21 at 14:58
Grigory PankoGrigory Panko
31
edited Mar 21 at 15:09
Javi
3,15321032
asked Mar 21 at 14:58
Grigory PankoGrigory Panko
31
edited Mar 21 at 15:09
Javi
3,15321032
edited Mar 21 at 15:09
Javi
3,15321032
edited Mar 21 at 15:09
Javi
3,15321032
3,15321032
asked Mar 21 at 14:58
Grigory PankoGrigory Panko
31
asked Mar 21 at 14:58
Grigory PankoGrigory Panko
31
asked Mar 21 at 14:58
Grigory PankoGrigory Panko
31
31
$begingroup$
Trigonometry; apply it to vector from $(0,0)$ to $(x,y)$ to compute $text {cos}(theta)$ and find the angle $theta$. The same for the other one, finding $theta_1$.
$endgroup$
– Mauro ALLEGRANZA
Mar 21 at 15:05
add a comment |
$begingroup$
Trigonometry; apply it to vector from $(0,0)$ to $(x,y)$ to compute $text {cos}(theta)$ and find the angle $theta$. The same for the other one, finding $theta_1$.
$endgroup$
– Mauro ALLEGRANZA
Mar 21 at 15:05
$begingroup$
Trigonometry; apply it to vector from $(0,0)$ to $(x,y)$ to compute $text {cos}(theta)$ and find the angle $theta$. The same for the other one, finding $theta_1$.
$endgroup$
– Mauro ALLEGRANZA
Mar 21 at 15:05
$begingroup$
Trigonometry; apply it to vector from $(0,0)$ to $(x,y)$ to compute $text {cos}(theta)$ and find the angle $theta$. The same for the other one, finding $theta_1$.
$endgroup$
– Mauro ALLEGRANZA
Mar 21 at 15:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have an another vector $vec A$ which joins $(0,0)$ and $(x,y)$. So now simply calculate the angle between this factor and your vector using dot product.
You have:
$ cos theta = frac{xx_1+yy_1}{sqrt{x^2+y^2}sqrt{x_1^2+y_1^2}}$
Where $theta$ is angle between the two vectors
answered Mar 21 at 15:08
TojrahTojrah
4036
$endgroup$
$begingroup$
This returns cosine of angle in range [0°; 180°] (smallest angle of two possible). How can I get only clockwise angle (in range [0°; 360°] or [-180°; 180°])?
$endgroup$
– Grigory Panko
Mar 21 at 17:12
$begingroup$
This can be determined on the basis of in which quadrant do $(x,y)$ and $(x_1,y_1)$ lie. So based on their relative position(you will have various cases), you can get the clockwise angle in terms of $theta$ :)
$endgroup$
– Tojrah
Mar 21 at 17:21
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have an another vector $vec A$ which joins $(0,0)$ and $(x,y)$. So now simply calculate the angle between this factor and your vector using dot product.
You have:
$ cos theta = frac{xx_1+yy_1}{sqrt{x^2+y^2}sqrt{x_1^2+y_1^2}}$
Where $theta$ is angle between the two vectors
answered Mar 21 at 15:08
TojrahTojrah
4036
$endgroup$
$begingroup$
This returns cosine of angle in range [0°; 180°] (smallest angle of two possible). How can I get only clockwise angle (in range [0°; 360°] or [-180°; 180°])?
$endgroup$
– Grigory Panko
Mar 21 at 17:12
$begingroup$
This can be determined on the basis of in which quadrant do $(x,y)$ and $(x_1,y_1)$ lie. So based on their relative position(you will have various cases), you can get the clockwise angle in terms of $theta$ :)
$endgroup$
– Tojrah
Mar 21 at 17:21
add a comment |
$begingroup$
You have an another vector $vec A$ which joins $(0,0)$ and $(x,y)$. So now simply calculate the angle between this factor and your vector using dot product.
You have:
$ cos theta = frac{xx_1+yy_1}{sqrt{x^2+y^2}sqrt{x_1^2+y_1^2}}$
Where $theta$ is angle between the two vectors
answered Mar 21 at 15:08
TojrahTojrah
4036
$endgroup$
$begingroup$
This returns cosine of angle in range [0°; 180°] (smallest angle of two possible). How can I get only clockwise angle (in range [0°; 360°] or [-180°; 180°])?
$endgroup$
– Grigory Panko
Mar 21 at 17:12
$begingroup$
This can be determined on the basis of in which quadrant do $(x,y)$ and $(x_1,y_1)$ lie. So based on their relative position(you will have various cases), you can get the clockwise angle in terms of $theta$ :)
$endgroup$
– Tojrah
Mar 21 at 17:21
add a comment |
$begingroup$
You have an another vector $vec A$ which joins $(0,0)$ and $(x,y)$. So now simply calculate the angle between this factor and your vector using dot product.
You have:
$ cos theta = frac{xx_1+yy_1}{sqrt{x^2+y^2}sqrt{x_1^2+y_1^2}}$
Where $theta$ is angle between the two vectors
answered Mar 21 at 15:08
TojrahTojrah
4036
$endgroup$
You have an another vector $vec A$ which joins $(0,0)$ and $(x,y)$. So now simply calculate the angle between this factor and your vector using dot product.
You have:
$ cos theta = frac{xx_1+yy_1}{sqrt{x^2+y^2}sqrt{x_1^2+y_1^2}}$
Where $theta$ is angle between the two vectors
answered Mar 21 at 15:08
TojrahTojrah
4036
edited Mar 21 at 15:13
answered Mar 21 at 15:08
TojrahTojrah
4036
answered Mar 21 at 15:08
TojrahTojrah
4036
answered Mar 21 at 15:08
TojrahTojrah
4036
4036
$begingroup$
This returns cosine of angle in range [0°; 180°] (smallest angle of two possible). How can I get only clockwise angle (in range [0°; 360°] or [-180°; 180°])?
$endgroup$
– Grigory Panko
Mar 21 at 17:12
$begingroup$
This can be determined on the basis of in which quadrant do $(x,y)$ and $(x_1,y_1)$ lie. So based on their relative position(you will have various cases), you can get the clockwise angle in terms of $theta$ :)
$endgroup$
– Tojrah
Mar 21 at 17:21
add a comment |
$begingroup$
This returns cosine of angle in range [0°; 180°] (smallest angle of two possible). How can I get only clockwise angle (in range [0°; 360°] or [-180°; 180°])?
$endgroup$
– Grigory Panko
Mar 21 at 17:12
$begingroup$
This can be determined on the basis of in which quadrant do $(x,y)$ and $(x_1,y_1)$ lie. So based on their relative position(you will have various cases), you can get the clockwise angle in terms of $theta$ :)
$endgroup$
– Tojrah
Mar 21 at 17:21
$begingroup$
This returns cosine of angle in range [0°; 180°] (smallest angle of two possible). How can I get only clockwise angle (in range [0°; 360°] or [-180°; 180°])?
$endgroup$
– Grigory Panko
Mar 21 at 17:12
$begingroup$
This returns cosine of angle in range [0°; 180°] (smallest angle of two possible). How can I get only clockwise angle (in range [0°; 360°] or [-180°; 180°])?
$endgroup$
– Grigory Panko
Mar 21 at 17:12
$begingroup$
This can be determined on the basis of in which quadrant do $(x,y)$ and $(x_1,y_1)$ lie. So based on their relative position(you will have various cases), you can get the clockwise angle in terms of $theta$ :)
$endgroup$
– Tojrah
Mar 21 at 17:21
$begingroup$
This can be determined on the basis of in which quadrant do $(x,y)$ and $(x_1,y_1)$ lie. So based on their relative position(you will have various cases), you can get the clockwise angle in terms of $theta$ :)
$endgroup$
– Tojrah
Mar 21 at 17:21
add a comment |
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$begingroup$
Trigonometry; apply it to vector from $(0,0)$ to $(x,y)$ to compute $text {cos}(theta)$ and find the angle $theta$. The same for the other one, finding $theta_1$.
$endgroup$
– Mauro ALLEGRANZA
Mar 21 at 15:05