Distance between a 3D point and a vector defined by two 3D points
$begingroup$
Drawing of the vector and line
I have vector points A (Xa, Ya, Za) and B (Xb, Yb, Zb)
Point to measure the distance from;
C (Xc, Yc, Zc)
I don't know if it's correct, but this is my equation for distance: D = (|(C-A) * (C-B)|) / (|B-A|)
Can I split that formula to:
Dx = (|(Xc-Xa) * (Xc-Xb)|) / (|Xb-Xa|)
Dy = (|(Yc-Ya ) * (Yc-Yb)|) / (|Yb-Ya|)
Dz = (|(Zc-Za) * (Zc-Zb)|) / (|Zb-Za|)
, to get the X distance, Y distance and Z distance from the point to vector?
If that equation is even correct.
I'm pretty insecure since I'm still in high school and don't know much about 3D math. Would really appreciate if you helped me out or gave me a better solution if there is one. (This is for game programming purposes)
trigonometry 3d
asked Dec 13 '18 at 16:35
Marin KovačMarin Kovač
52
$endgroup$
add a comment |
$begingroup$
Drawing of the vector and line
I have vector points A (Xa, Ya, Za) and B (Xb, Yb, Zb)
Point to measure the distance from;
C (Xc, Yc, Zc)
I don't know if it's correct, but this is my equation for distance: D = (|(C-A) * (C-B)|) / (|B-A|)
Can I split that formula to:
Dx = (|(Xc-Xa) * (Xc-Xb)|) / (|Xb-Xa|)
Dy = (|(Yc-Ya ) * (Yc-Yb)|) / (|Yb-Ya|)
Dz = (|(Zc-Za) * (Zc-Zb)|) / (|Zb-Za|)
, to get the X distance, Y distance and Z distance from the point to vector?
If that equation is even correct.
I'm pretty insecure since I'm still in high school and don't know much about 3D math. Would really appreciate if you helped me out or gave me a better solution if there is one. (This is for game programming purposes)
trigonometry 3d
asked Dec 13 '18 at 16:35
Marin KovačMarin Kovač
52
$endgroup$
$begingroup$
Does*refer to the cross product here? If so, it's still not quite correct. You want the distance from C to the line going through AB?
$endgroup$
– Dylan
Dec 13 '18 at 16:53
$begingroup$
Are you looking for the distance to the line through $A$ and $B$ or to the line segement $overline{AB}$? If the latter, then you also have to deal with the case that the perpendicular from $C$ doesn’t intersect the segment at all.
$endgroup$
– amd
Dec 13 '18 at 18:02
add a comment |
$begingroup$
Drawing of the vector and line
I have vector points A (Xa, Ya, Za) and B (Xb, Yb, Zb)
Point to measure the distance from;
C (Xc, Yc, Zc)
I don't know if it's correct, but this is my equation for distance: D = (|(C-A) * (C-B)|) / (|B-A|)
Can I split that formula to:
Dx = (|(Xc-Xa) * (Xc-Xb)|) / (|Xb-Xa|)
Dy = (|(Yc-Ya ) * (Yc-Yb)|) / (|Yb-Ya|)
Dz = (|(Zc-Za) * (Zc-Zb)|) / (|Zb-Za|)
, to get the X distance, Y distance and Z distance from the point to vector?
If that equation is even correct.
I'm pretty insecure since I'm still in high school and don't know much about 3D math. Would really appreciate if you helped me out or gave me a better solution if there is one. (This is for game programming purposes)
trigonometry 3d
asked Dec 13 '18 at 16:35
Marin KovačMarin Kovač
52
$endgroup$
Drawing of the vector and line
I have vector points A (Xa, Ya, Za) and B (Xb, Yb, Zb)
Point to measure the distance from;
C (Xc, Yc, Zc)
I don't know if it's correct, but this is my equation for distance: D = (|(C-A) * (C-B)|) / (|B-A|)
Can I split that formula to:
Dx = (|(Xc-Xa) * (Xc-Xb)|) / (|Xb-Xa|)
Dy = (|(Yc-Ya ) * (Yc-Yb)|) / (|Yb-Ya|)
Dz = (|(Zc-Za) * (Zc-Zb)|) / (|Zb-Za|)
, to get the X distance, Y distance and Z distance from the point to vector?
If that equation is even correct.
I'm pretty insecure since I'm still in high school and don't know much about 3D math. Would really appreciate if you helped me out or gave me a better solution if there is one. (This is for game programming purposes)
trigonometry 3d
trigonometry 3d
asked Dec 13 '18 at 16:35
Marin KovačMarin Kovač
52
asked Dec 13 '18 at 16:35
Marin KovačMarin Kovač
52
asked Dec 13 '18 at 16:35
Marin KovačMarin Kovač
52
asked Dec 13 '18 at 16:35
Marin KovačMarin Kovač
52
asked Dec 13 '18 at 16:35
Marin KovačMarin Kovač
52
52
$begingroup$
Does*refer to the cross product here? If so, it's still not quite correct. You want the distance from C to the line going through AB?
$endgroup$
– Dylan
Dec 13 '18 at 16:53
$begingroup$
Are you looking for the distance to the line through $A$ and $B$ or to the line segement $overline{AB}$? If the latter, then you also have to deal with the case that the perpendicular from $C$ doesn’t intersect the segment at all.
$endgroup$
– amd
Dec 13 '18 at 18:02
add a comment |
$begingroup$
Does*refer to the cross product here? If so, it's still not quite correct. You want the distance from C to the line going through AB?
$endgroup$
– Dylan
Dec 13 '18 at 16:53
$begingroup$
Are you looking for the distance to the line through $A$ and $B$ or to the line segement $overline{AB}$? If the latter, then you also have to deal with the case that the perpendicular from $C$ doesn’t intersect the segment at all.
$endgroup$
– amd
Dec 13 '18 at 18:02
$begingroup$
Does
* refer to the cross product here? If so, it's still not quite correct. You want the distance from C to the line going through AB?$endgroup$
– Dylan
Dec 13 '18 at 16:53
$begingroup$
Does
* refer to the cross product here? If so, it's still not quite correct. You want the distance from C to the line going through AB?$endgroup$
– Dylan
Dec 13 '18 at 16:53
$begingroup$
Are you looking for the distance to the line through $A$ and $B$ or to the line segement $overline{AB}$? If the latter, then you also have to deal with the case that the perpendicular from $C$ doesn’t intersect the segment at all.
$endgroup$
– amd
Dec 13 '18 at 18:02
$begingroup$
Are you looking for the distance to the line through $A$ and $B$ or to the line segement $overline{AB}$? If the latter, then you also have to deal with the case that the perpendicular from $C$ doesn’t intersect the segment at all.
$endgroup$
– amd
Dec 13 '18 at 18:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $theta$ be the angle between the vectors $overrightarrow{AC}$ and $overrightarrow{AB}$. Then, the distance, i.e. the length of the perpendicular drawn from $C$ to $AB$ is
$$d = vertoverrightarrow{AC}vertsintheta$$
You also know the cross product
$$ vertoverrightarrow{AC} times overrightarrow{AB}vert = vertoverrightarrow{AC}vertvertoverrightarrow{AB}vertsintheta $$
therefore
$$ d = frac{vert overrightarrow{AC} times vec{AB} vert}{vertvec{AB}vert} $$
EDIT: To find the perpendicular vector, consider the projection vector of $overrightarrow{AC}$ on the line $AB$. Let's call this $overrightarrow{D}$. Then its length is given by
$$ vertoverrightarrow{AD}vert = vertoverrightarrow{AC}vertcostheta = frac{overrightarrow{AC}cdot overrightarrow{AB}}{vert overrightarrow{AB} vert} $$
where the dot product identity was used. Since $overrightarrow{AD}$ is in the same direction as $overrightarrow{AB}$, you can find its coordinates by "scaling" $overrightarrow{AB}$, i.e.
$$ overrightarrow{AD} = vert overrightarrow{AD} vert frac{overrightarrow{AB}}{vertoverrightarrow{AB}vert} = frac{overrightarrow{AC}cdot overrightarrow{AB}}{vert overrightarrow{AB} vert^2} overrightarrow{AB}$$
The perpendicular vector, $overrightarrow{DC}$, is computed by the difference
$$ overrightarrow{DC} = overrightarrow{AC} - overrightarrow{AD} $$
The 3 components of $overrightarrow{DC}$ are the $x$, $y$, $z$ distances you're looking for.
answered Dec 13 '18 at 16:59
DylanDylan
12.4k31026
$endgroup$
$begingroup$
Thank you very much! Just to be sure, for x distance, I can take x values of the points, for y distance, I take y and for z distance, I take z values of points?
$endgroup$
– Marin Kovač
Dec 13 '18 at 19:26
$begingroup$
@MarinKovač It doesn't work like that. You need all 3 coordinates to compute the cross product. The result is just the total distance. What you're asking is finding the perpendicular vector from C to AB, which is entirely different.
$endgroup$
– Dylan
Dec 14 '18 at 6:39
$begingroup$
I added some more information on the perpendicular vector. Hopefully you know how to compute the cross and dot products.
$endgroup$
– Dylan
Dec 14 '18 at 6:52
$begingroup$
Thanks, that helped. :D
$endgroup$
– Marin Kovač
Dec 14 '18 at 13:31
add a comment |
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1 Answer
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$begingroup$
Let $theta$ be the angle between the vectors $overrightarrow{AC}$ and $overrightarrow{AB}$. Then, the distance, i.e. the length of the perpendicular drawn from $C$ to $AB$ is
$$d = vertoverrightarrow{AC}vertsintheta$$
You also know the cross product
$$ vertoverrightarrow{AC} times overrightarrow{AB}vert = vertoverrightarrow{AC}vertvertoverrightarrow{AB}vertsintheta $$
therefore
$$ d = frac{vert overrightarrow{AC} times vec{AB} vert}{vertvec{AB}vert} $$
EDIT: To find the perpendicular vector, consider the projection vector of $overrightarrow{AC}$ on the line $AB$. Let's call this $overrightarrow{D}$. Then its length is given by
$$ vertoverrightarrow{AD}vert = vertoverrightarrow{AC}vertcostheta = frac{overrightarrow{AC}cdot overrightarrow{AB}}{vert overrightarrow{AB} vert} $$
where the dot product identity was used. Since $overrightarrow{AD}$ is in the same direction as $overrightarrow{AB}$, you can find its coordinates by "scaling" $overrightarrow{AB}$, i.e.
$$ overrightarrow{AD} = vert overrightarrow{AD} vert frac{overrightarrow{AB}}{vertoverrightarrow{AB}vert} = frac{overrightarrow{AC}cdot overrightarrow{AB}}{vert overrightarrow{AB} vert^2} overrightarrow{AB}$$
The perpendicular vector, $overrightarrow{DC}$, is computed by the difference
$$ overrightarrow{DC} = overrightarrow{AC} - overrightarrow{AD} $$
The 3 components of $overrightarrow{DC}$ are the $x$, $y$, $z$ distances you're looking for.
answered Dec 13 '18 at 16:59
DylanDylan
12.4k31026
$endgroup$
$begingroup$
Thank you very much! Just to be sure, for x distance, I can take x values of the points, for y distance, I take y and for z distance, I take z values of points?
$endgroup$
– Marin Kovač
Dec 13 '18 at 19:26
$begingroup$
@MarinKovač It doesn't work like that. You need all 3 coordinates to compute the cross product. The result is just the total distance. What you're asking is finding the perpendicular vector from C to AB, which is entirely different.
$endgroup$
– Dylan
Dec 14 '18 at 6:39
$begingroup$
I added some more information on the perpendicular vector. Hopefully you know how to compute the cross and dot products.
$endgroup$
– Dylan
Dec 14 '18 at 6:52
$begingroup$
Thanks, that helped. :D
$endgroup$
– Marin Kovač
Dec 14 '18 at 13:31
add a comment |
$begingroup$
Let $theta$ be the angle between the vectors $overrightarrow{AC}$ and $overrightarrow{AB}$. Then, the distance, i.e. the length of the perpendicular drawn from $C$ to $AB$ is
$$d = vertoverrightarrow{AC}vertsintheta$$
You also know the cross product
$$ vertoverrightarrow{AC} times overrightarrow{AB}vert = vertoverrightarrow{AC}vertvertoverrightarrow{AB}vertsintheta $$
therefore
$$ d = frac{vert overrightarrow{AC} times vec{AB} vert}{vertvec{AB}vert} $$
EDIT: To find the perpendicular vector, consider the projection vector of $overrightarrow{AC}$ on the line $AB$. Let's call this $overrightarrow{D}$. Then its length is given by
$$ vertoverrightarrow{AD}vert = vertoverrightarrow{AC}vertcostheta = frac{overrightarrow{AC}cdot overrightarrow{AB}}{vert overrightarrow{AB} vert} $$
where the dot product identity was used. Since $overrightarrow{AD}$ is in the same direction as $overrightarrow{AB}$, you can find its coordinates by "scaling" $overrightarrow{AB}$, i.e.
$$ overrightarrow{AD} = vert overrightarrow{AD} vert frac{overrightarrow{AB}}{vertoverrightarrow{AB}vert} = frac{overrightarrow{AC}cdot overrightarrow{AB}}{vert overrightarrow{AB} vert^2} overrightarrow{AB}$$
The perpendicular vector, $overrightarrow{DC}$, is computed by the difference
$$ overrightarrow{DC} = overrightarrow{AC} - overrightarrow{AD} $$
The 3 components of $overrightarrow{DC}$ are the $x$, $y$, $z$ distances you're looking for.
answered Dec 13 '18 at 16:59
DylanDylan
12.4k31026
$endgroup$
$begingroup$
Thank you very much! Just to be sure, for x distance, I can take x values of the points, for y distance, I take y and for z distance, I take z values of points?
$endgroup$
– Marin Kovač
Dec 13 '18 at 19:26
$begingroup$
@MarinKovač It doesn't work like that. You need all 3 coordinates to compute the cross product. The result is just the total distance. What you're asking is finding the perpendicular vector from C to AB, which is entirely different.
$endgroup$
– Dylan
Dec 14 '18 at 6:39
$begingroup$
I added some more information on the perpendicular vector. Hopefully you know how to compute the cross and dot products.
$endgroup$
– Dylan
Dec 14 '18 at 6:52
$begingroup$
Thanks, that helped. :D
$endgroup$
– Marin Kovač
Dec 14 '18 at 13:31
add a comment |
$begingroup$
Let $theta$ be the angle between the vectors $overrightarrow{AC}$ and $overrightarrow{AB}$. Then, the distance, i.e. the length of the perpendicular drawn from $C$ to $AB$ is
$$d = vertoverrightarrow{AC}vertsintheta$$
You also know the cross product
$$ vertoverrightarrow{AC} times overrightarrow{AB}vert = vertoverrightarrow{AC}vertvertoverrightarrow{AB}vertsintheta $$
therefore
$$ d = frac{vert overrightarrow{AC} times vec{AB} vert}{vertvec{AB}vert} $$
EDIT: To find the perpendicular vector, consider the projection vector of $overrightarrow{AC}$ on the line $AB$. Let's call this $overrightarrow{D}$. Then its length is given by
$$ vertoverrightarrow{AD}vert = vertoverrightarrow{AC}vertcostheta = frac{overrightarrow{AC}cdot overrightarrow{AB}}{vert overrightarrow{AB} vert} $$
where the dot product identity was used. Since $overrightarrow{AD}$ is in the same direction as $overrightarrow{AB}$, you can find its coordinates by "scaling" $overrightarrow{AB}$, i.e.
$$ overrightarrow{AD} = vert overrightarrow{AD} vert frac{overrightarrow{AB}}{vertoverrightarrow{AB}vert} = frac{overrightarrow{AC}cdot overrightarrow{AB}}{vert overrightarrow{AB} vert^2} overrightarrow{AB}$$
The perpendicular vector, $overrightarrow{DC}$, is computed by the difference
$$ overrightarrow{DC} = overrightarrow{AC} - overrightarrow{AD} $$
The 3 components of $overrightarrow{DC}$ are the $x$, $y$, $z$ distances you're looking for.
answered Dec 13 '18 at 16:59
DylanDylan
12.4k31026
$endgroup$
Let $theta$ be the angle between the vectors $overrightarrow{AC}$ and $overrightarrow{AB}$. Then, the distance, i.e. the length of the perpendicular drawn from $C$ to $AB$ is
$$d = vertoverrightarrow{AC}vertsintheta$$
You also know the cross product
$$ vertoverrightarrow{AC} times overrightarrow{AB}vert = vertoverrightarrow{AC}vertvertoverrightarrow{AB}vertsintheta $$
therefore
$$ d = frac{vert overrightarrow{AC} times vec{AB} vert}{vertvec{AB}vert} $$
EDIT: To find the perpendicular vector, consider the projection vector of $overrightarrow{AC}$ on the line $AB$. Let's call this $overrightarrow{D}$. Then its length is given by
$$ vertoverrightarrow{AD}vert = vertoverrightarrow{AC}vertcostheta = frac{overrightarrow{AC}cdot overrightarrow{AB}}{vert overrightarrow{AB} vert} $$
where the dot product identity was used. Since $overrightarrow{AD}$ is in the same direction as $overrightarrow{AB}$, you can find its coordinates by "scaling" $overrightarrow{AB}$, i.e.
$$ overrightarrow{AD} = vert overrightarrow{AD} vert frac{overrightarrow{AB}}{vertoverrightarrow{AB}vert} = frac{overrightarrow{AC}cdot overrightarrow{AB}}{vert overrightarrow{AB} vert^2} overrightarrow{AB}$$
The perpendicular vector, $overrightarrow{DC}$, is computed by the difference
$$ overrightarrow{DC} = overrightarrow{AC} - overrightarrow{AD} $$
The 3 components of $overrightarrow{DC}$ are the $x$, $y$, $z$ distances you're looking for.
answered Dec 13 '18 at 16:59
DylanDylan
12.4k31026
edited Dec 14 '18 at 6:50
answered Dec 13 '18 at 16:59
DylanDylan
12.4k31026
answered Dec 13 '18 at 16:59
DylanDylan
12.4k31026
answered Dec 13 '18 at 16:59
DylanDylan
12.4k31026
12.4k31026
$begingroup$
Thank you very much! Just to be sure, for x distance, I can take x values of the points, for y distance, I take y and for z distance, I take z values of points?
$endgroup$
– Marin Kovač
Dec 13 '18 at 19:26
$begingroup$
@MarinKovač It doesn't work like that. You need all 3 coordinates to compute the cross product. The result is just the total distance. What you're asking is finding the perpendicular vector from C to AB, which is entirely different.
$endgroup$
– Dylan
Dec 14 '18 at 6:39
$begingroup$
I added some more information on the perpendicular vector. Hopefully you know how to compute the cross and dot products.
$endgroup$
– Dylan
Dec 14 '18 at 6:52
$begingroup$
Thanks, that helped. :D
$endgroup$
– Marin Kovač
Dec 14 '18 at 13:31
add a comment |
$begingroup$
Thank you very much! Just to be sure, for x distance, I can take x values of the points, for y distance, I take y and for z distance, I take z values of points?
$endgroup$
– Marin Kovač
Dec 13 '18 at 19:26
$begingroup$
@MarinKovač It doesn't work like that. You need all 3 coordinates to compute the cross product. The result is just the total distance. What you're asking is finding the perpendicular vector from C to AB, which is entirely different.
$endgroup$
– Dylan
Dec 14 '18 at 6:39
$begingroup$
I added some more information on the perpendicular vector. Hopefully you know how to compute the cross and dot products.
$endgroup$
– Dylan
Dec 14 '18 at 6:52
$begingroup$
Thanks, that helped. :D
$endgroup$
– Marin Kovač
Dec 14 '18 at 13:31
$begingroup$
Thank you very much! Just to be sure, for x distance, I can take x values of the points, for y distance, I take y and for z distance, I take z values of points?
$endgroup$
– Marin Kovač
Dec 13 '18 at 19:26
$begingroup$
Thank you very much! Just to be sure, for x distance, I can take x values of the points, for y distance, I take y and for z distance, I take z values of points?
$endgroup$
– Marin Kovač
Dec 13 '18 at 19:26
$begingroup$
@MarinKovač It doesn't work like that. You need all 3 coordinates to compute the cross product. The result is just the total distance. What you're asking is finding the perpendicular vector from C to AB, which is entirely different.
$endgroup$
– Dylan
Dec 14 '18 at 6:39
$begingroup$
@MarinKovač It doesn't work like that. You need all 3 coordinates to compute the cross product. The result is just the total distance. What you're asking is finding the perpendicular vector from C to AB, which is entirely different.
$endgroup$
– Dylan
Dec 14 '18 at 6:39
$begingroup$
I added some more information on the perpendicular vector. Hopefully you know how to compute the cross and dot products.
$endgroup$
– Dylan
Dec 14 '18 at 6:52
$begingroup$
I added some more information on the perpendicular vector. Hopefully you know how to compute the cross and dot products.
$endgroup$
– Dylan
Dec 14 '18 at 6:52
$begingroup$
Thanks, that helped. :D
$endgroup$
– Marin Kovač
Dec 14 '18 at 13:31
$begingroup$
Thanks, that helped. :D
$endgroup$
– Marin Kovač
Dec 14 '18 at 13:31
add a comment |
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$begingroup$
Does
*refer to the cross product here? If so, it's still not quite correct. You want the distance from C to the line going through AB?$endgroup$
– Dylan
Dec 13 '18 at 16:53
$begingroup$
Are you looking for the distance to the line through $A$ and $B$ or to the line segement $overline{AB}$? If the latter, then you also have to deal with the case that the perpendicular from $C$ doesn’t intersect the segment at all.
$endgroup$
– amd
Dec 13 '18 at 18:02