Distance an arbitrary point is found along a given vector
Say I have a vector in 2D space defined by two points $(x_1, y_1)$ and $(x_2, y_2)$: $$vec{v}=(x_2 - x_1, y_2 - y_1)$$ I would like to find how far along that vector an arbitrary point $(x_3, y_3)$ is. This very woolly language$^*$, so I've attempted to create a diagram showing the sitation.
In this diagram, the quantity I'm interested in $a$, which I can calculate using Pythagoras' theorem if I know $b$ and $c$. I know $c$, which is the length of vector $(x_3 - x_1,y_3 - y_1)$, given by, $$c = sqrt{(x_3 - x_1)^2 + (y_3 - y_1)^2}$$ So, now I need to calculate $b$: the length of a vector – that I'll call $vec{u}$ – that is perpendicular to $vec{v}$ and passes through point $(x_3, y_3)$. For $vec{u}$ and $vec{v}$ to be perpendicular the dot product must be zero. That is,
$$vec{v}cdot vec{u}=0$$
$$(x_4-x_3)(x_2-x_1)+(y_4-y_3)(y_2-y_1)=0$$
This is where I begin to falter: one equation with two unknowns, $y_4$ & $x_4$. I expect there is some obvious constraint on $vec{u}$ that I should be using to eliminate an unknown, but my sleep-deprived mind is offering no help. Can someone point out what I've missed?
$$$$
$^*$I really want to use the word project to describe how my arbitrary point $(x_3, y_3)$ is placed along that vector. Is this the correct terminology?
geometry
asked Nov 28 at 17:19
Lyngbakr
1084
add a comment |
Say I have a vector in 2D space defined by two points $(x_1, y_1)$ and $(x_2, y_2)$: $$vec{v}=(x_2 - x_1, y_2 - y_1)$$ I would like to find how far along that vector an arbitrary point $(x_3, y_3)$ is. This very woolly language$^*$, so I've attempted to create a diagram showing the sitation.
In this diagram, the quantity I'm interested in $a$, which I can calculate using Pythagoras' theorem if I know $b$ and $c$. I know $c$, which is the length of vector $(x_3 - x_1,y_3 - y_1)$, given by, $$c = sqrt{(x_3 - x_1)^2 + (y_3 - y_1)^2}$$ So, now I need to calculate $b$: the length of a vector – that I'll call $vec{u}$ – that is perpendicular to $vec{v}$ and passes through point $(x_3, y_3)$. For $vec{u}$ and $vec{v}$ to be perpendicular the dot product must be zero. That is,
$$vec{v}cdot vec{u}=0$$
$$(x_4-x_3)(x_2-x_1)+(y_4-y_3)(y_2-y_1)=0$$
This is where I begin to falter: one equation with two unknowns, $y_4$ & $x_4$. I expect there is some obvious constraint on $vec{u}$ that I should be using to eliminate an unknown, but my sleep-deprived mind is offering no help. Can someone point out what I've missed?
$$$$
$^*$I really want to use the word project to describe how my arbitrary point $(x_3, y_3)$ is placed along that vector. Is this the correct terminology?
geometry
asked Nov 28 at 17:19
Lyngbakr
1084
add a comment |
Say I have a vector in 2D space defined by two points $(x_1, y_1)$ and $(x_2, y_2)$: $$vec{v}=(x_2 - x_1, y_2 - y_1)$$ I would like to find how far along that vector an arbitrary point $(x_3, y_3)$ is. This very woolly language$^*$, so I've attempted to create a diagram showing the sitation.
In this diagram, the quantity I'm interested in $a$, which I can calculate using Pythagoras' theorem if I know $b$ and $c$. I know $c$, which is the length of vector $(x_3 - x_1,y_3 - y_1)$, given by, $$c = sqrt{(x_3 - x_1)^2 + (y_3 - y_1)^2}$$ So, now I need to calculate $b$: the length of a vector – that I'll call $vec{u}$ – that is perpendicular to $vec{v}$ and passes through point $(x_3, y_3)$. For $vec{u}$ and $vec{v}$ to be perpendicular the dot product must be zero. That is,
$$vec{v}cdot vec{u}=0$$
$$(x_4-x_3)(x_2-x_1)+(y_4-y_3)(y_2-y_1)=0$$
This is where I begin to falter: one equation with two unknowns, $y_4$ & $x_4$. I expect there is some obvious constraint on $vec{u}$ that I should be using to eliminate an unknown, but my sleep-deprived mind is offering no help. Can someone point out what I've missed?
$$$$
$^*$I really want to use the word project to describe how my arbitrary point $(x_3, y_3)$ is placed along that vector. Is this the correct terminology?
geometry
asked Nov 28 at 17:19
Lyngbakr
1084
Say I have a vector in 2D space defined by two points $(x_1, y_1)$ and $(x_2, y_2)$: $$vec{v}=(x_2 - x_1, y_2 - y_1)$$ I would like to find how far along that vector an arbitrary point $(x_3, y_3)$ is. This very woolly language$^*$, so I've attempted to create a diagram showing the sitation.
In this diagram, the quantity I'm interested in $a$, which I can calculate using Pythagoras' theorem if I know $b$ and $c$. I know $c$, which is the length of vector $(x_3 - x_1,y_3 - y_1)$, given by, $$c = sqrt{(x_3 - x_1)^2 + (y_3 - y_1)^2}$$ So, now I need to calculate $b$: the length of a vector – that I'll call $vec{u}$ – that is perpendicular to $vec{v}$ and passes through point $(x_3, y_3)$. For $vec{u}$ and $vec{v}$ to be perpendicular the dot product must be zero. That is,
$$vec{v}cdot vec{u}=0$$
$$(x_4-x_3)(x_2-x_1)+(y_4-y_3)(y_2-y_1)=0$$
This is where I begin to falter: one equation with two unknowns, $y_4$ & $x_4$. I expect there is some obvious constraint on $vec{u}$ that I should be using to eliminate an unknown, but my sleep-deprived mind is offering no help. Can someone point out what I've missed?
$$$$
$^*$I really want to use the word project to describe how my arbitrary point $(x_3, y_3)$ is placed along that vector. Is this the correct terminology?
geometry
geometry
asked Nov 28 at 17:19
Lyngbakr
1084
asked Nov 28 at 17:19
Lyngbakr
1084
edited Nov 28 at 17:26
asked Nov 28 at 17:19
Lyngbakr
1084
asked Nov 28 at 17:19
Lyngbakr
1084
asked Nov 28 at 17:19
Lyngbakr
1084
1084
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2 Answers
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You are right in calling this a projection. If $(x_1,y_1)$ is the origin, then you can project ${bf u} = (x_3,y_3)$ onto $bf v$ thus:
$${rm proj}_{bf v}{bf u} = frac{bf u cdot v}{bf v cdot v}{bf v}.$$
If $(x_1,y_1)$ is not the origin, then just shift the frame of reference to make it so.
answered Nov 28 at 17:45
Théophile
19.4k12946
1
See also the Wikipedia article on vector projection.
– Théophile
Nov 28 at 17:46
Perfect! Thanks.
– Lyngbakr
Nov 28 at 18:03
add a comment |
Consider the vectors $v_1 = (x_2-x_1,y_2-y_1)$ and $v_2 = (x_3-x_1,y_3-y_1)$. You can compute the cosine of the angle between these two vectors as follows:
$cos theta = frac{<v_1,v_2>}{||v_1||||v_2||}$,
where $<v_1,v_2>$ is the dot-product and $||.||$ is the norm. Once you have done this computation, it is easy to see that $a = c cos theta$ and $b = c sin theta$.
The quantity $a$ is the projection you speak of and it is related to the dot-product as described above.
answered Nov 28 at 17:48
Aditya Dua
80418
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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You are right in calling this a projection. If $(x_1,y_1)$ is the origin, then you can project ${bf u} = (x_3,y_3)$ onto $bf v$ thus:
$${rm proj}_{bf v}{bf u} = frac{bf u cdot v}{bf v cdot v}{bf v}.$$
If $(x_1,y_1)$ is not the origin, then just shift the frame of reference to make it so.
answered Nov 28 at 17:45
Théophile
19.4k12946
1
See also the Wikipedia article on vector projection.
– Théophile
Nov 28 at 17:46
Perfect! Thanks.
– Lyngbakr
Nov 28 at 18:03
add a comment |
You are right in calling this a projection. If $(x_1,y_1)$ is the origin, then you can project ${bf u} = (x_3,y_3)$ onto $bf v$ thus:
$${rm proj}_{bf v}{bf u} = frac{bf u cdot v}{bf v cdot v}{bf v}.$$
If $(x_1,y_1)$ is not the origin, then just shift the frame of reference to make it so.
answered Nov 28 at 17:45
Théophile
19.4k12946
1
See also the Wikipedia article on vector projection.
– Théophile
Nov 28 at 17:46
Perfect! Thanks.
– Lyngbakr
Nov 28 at 18:03
add a comment |
You are right in calling this a projection. If $(x_1,y_1)$ is the origin, then you can project ${bf u} = (x_3,y_3)$ onto $bf v$ thus:
$${rm proj}_{bf v}{bf u} = frac{bf u cdot v}{bf v cdot v}{bf v}.$$
If $(x_1,y_1)$ is not the origin, then just shift the frame of reference to make it so.
answered Nov 28 at 17:45
Théophile
19.4k12946
You are right in calling this a projection. If $(x_1,y_1)$ is the origin, then you can project ${bf u} = (x_3,y_3)$ onto $bf v$ thus:
$${rm proj}_{bf v}{bf u} = frac{bf u cdot v}{bf v cdot v}{bf v}.$$
If $(x_1,y_1)$ is not the origin, then just shift the frame of reference to make it so.
answered Nov 28 at 17:45
Théophile
19.4k12946
answered Nov 28 at 17:45
Théophile
19.4k12946
answered Nov 28 at 17:45
Théophile
19.4k12946
answered Nov 28 at 17:45
Théophile
19.4k12946
19.4k12946
1
See also the Wikipedia article on vector projection.
– Théophile
Nov 28 at 17:46
Perfect! Thanks.
– Lyngbakr
Nov 28 at 18:03
add a comment |
1
See also the Wikipedia article on vector projection.
– Théophile
Nov 28 at 17:46
Perfect! Thanks.
– Lyngbakr
Nov 28 at 18:03
1
1
See also the Wikipedia article on vector projection.
– Théophile
Nov 28 at 17:46
See also the Wikipedia article on vector projection.
– Théophile
Nov 28 at 17:46
Perfect! Thanks.
– Lyngbakr
Nov 28 at 18:03
Perfect! Thanks.
– Lyngbakr
Nov 28 at 18:03
add a comment |
Consider the vectors $v_1 = (x_2-x_1,y_2-y_1)$ and $v_2 = (x_3-x_1,y_3-y_1)$. You can compute the cosine of the angle between these two vectors as follows:
$cos theta = frac{<v_1,v_2>}{||v_1||||v_2||}$,
where $<v_1,v_2>$ is the dot-product and $||.||$ is the norm. Once you have done this computation, it is easy to see that $a = c cos theta$ and $b = c sin theta$.
The quantity $a$ is the projection you speak of and it is related to the dot-product as described above.
answered Nov 28 at 17:48
Aditya Dua
80418
add a comment |
Consider the vectors $v_1 = (x_2-x_1,y_2-y_1)$ and $v_2 = (x_3-x_1,y_3-y_1)$. You can compute the cosine of the angle between these two vectors as follows:
$cos theta = frac{<v_1,v_2>}{||v_1||||v_2||}$,
where $<v_1,v_2>$ is the dot-product and $||.||$ is the norm. Once you have done this computation, it is easy to see that $a = c cos theta$ and $b = c sin theta$.
The quantity $a$ is the projection you speak of and it is related to the dot-product as described above.
answered Nov 28 at 17:48
Aditya Dua
80418
add a comment |
Consider the vectors $v_1 = (x_2-x_1,y_2-y_1)$ and $v_2 = (x_3-x_1,y_3-y_1)$. You can compute the cosine of the angle between these two vectors as follows:
$cos theta = frac{<v_1,v_2>}{||v_1||||v_2||}$,
where $<v_1,v_2>$ is the dot-product and $||.||$ is the norm. Once you have done this computation, it is easy to see that $a = c cos theta$ and $b = c sin theta$.
The quantity $a$ is the projection you speak of and it is related to the dot-product as described above.
answered Nov 28 at 17:48
Aditya Dua
80418
Consider the vectors $v_1 = (x_2-x_1,y_2-y_1)$ and $v_2 = (x_3-x_1,y_3-y_1)$. You can compute the cosine of the angle between these two vectors as follows:
$cos theta = frac{<v_1,v_2>}{||v_1||||v_2||}$,
where $<v_1,v_2>$ is the dot-product and $||.||$ is the norm. Once you have done this computation, it is easy to see that $a = c cos theta$ and $b = c sin theta$.
The quantity $a$ is the projection you speak of and it is related to the dot-product as described above.
answered Nov 28 at 17:48
Aditya Dua
80418
answered Nov 28 at 17:48
Aditya Dua
80418
answered Nov 28 at 17:48
Aditya Dua
80418
answered Nov 28 at 17:48
Aditya Dua
80418
80418
add a comment |
add a comment |
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