Projection of vector into an axis, along a direction
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I recently learned how I can project a vector $overline{a}$ onto another one $overline{b}$. I was wondering how I could achieve the same effect, but project the vector $overline{a}$ along a direction. For example, in this 
, we can see that the direction of the projection is the axis $overline{b}$ normal. This I know how to do. What I really wanted to do, is that the projection was made in the direction of the vector $overline{a}$ normal.
While searching, I found something about Thales Theorem, but I have no clue of how I could use it to my advantage. If someone could enlighten me, I would be much appreciated.
Thank you.
EDIT: I want to get the vector $overline{a3}$:
linear-algebra
asked Dec 6 '18 at 1:07
jorgenorojorgenoro
33
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add a comment |
$begingroup$
I recently learned how I can project a vector $overline{a}$ onto another one $overline{b}$. I was wondering how I could achieve the same effect, but project the vector $overline{a}$ along a direction. For example, in this , we can see that the direction of the projection is the axis $overline{b}$ normal. This I know how to do. What I really wanted to do, is that the projection was made in the direction of the vector $overline{a}$ normal.
While searching, I found something about Thales Theorem, but I have no clue of how I could use it to my advantage. If someone could enlighten me, I would be much appreciated.
Thank you.
EDIT: I want to get the vector $overline{a3}$:
linear-algebra
asked Dec 6 '18 at 1:07
jorgenorojorgenoro
33
$endgroup$
$begingroup$
If you know how to project a vector onto another vector, and you want to project a vector along a direction, why not just take a vector in that direction, and then project onto that vector?
$endgroup$
– Gerry Myerson
Dec 6 '18 at 1:27
add a comment |
$begingroup$
I recently learned how I can project a vector $overline{a}$ onto another one $overline{b}$. I was wondering how I could achieve the same effect, but project the vector $overline{a}$ along a direction. For example, in this , we can see that the direction of the projection is the axis $overline{b}$ normal. This I know how to do. What I really wanted to do, is that the projection was made in the direction of the vector $overline{a}$ normal.
While searching, I found something about Thales Theorem, but I have no clue of how I could use it to my advantage. If someone could enlighten me, I would be much appreciated.
Thank you.
EDIT: I want to get the vector $overline{a3}$:
linear-algebra
asked Dec 6 '18 at 1:07
jorgenorojorgenoro
33
$endgroup$
I recently learned how I can project a vector $overline{a}$ onto another one $overline{b}$. I was wondering how I could achieve the same effect, but project the vector $overline{a}$ along a direction. For example, in this , we can see that the direction of the projection is the axis $overline{b}$ normal. This I know how to do. What I really wanted to do, is that the projection was made in the direction of the vector $overline{a}$ normal.
While searching, I found something about Thales Theorem, but I have no clue of how I could use it to my advantage. If someone could enlighten me, I would be much appreciated.
Thank you.
EDIT: I want to get the vector $overline{a3}$:
linear-algebra
linear-algebra
asked Dec 6 '18 at 1:07
jorgenorojorgenoro
33
asked Dec 6 '18 at 1:07
jorgenorojorgenoro
33
edited Dec 6 '18 at 11:02
jorgenoro
asked Dec 6 '18 at 1:07
jorgenorojorgenoro
33
asked Dec 6 '18 at 1:07
jorgenorojorgenoro
33
asked Dec 6 '18 at 1:07
jorgenorojorgenoro
33
33
$begingroup$
If you know how to project a vector onto another vector, and you want to project a vector along a direction, why not just take a vector in that direction, and then project onto that vector?
$endgroup$
– Gerry Myerson
Dec 6 '18 at 1:27
add a comment |
$begingroup$
If you know how to project a vector onto another vector, and you want to project a vector along a direction, why not just take a vector in that direction, and then project onto that vector?
$endgroup$
– Gerry Myerson
Dec 6 '18 at 1:27
$begingroup$
If you know how to project a vector onto another vector, and you want to project a vector along a direction, why not just take a vector in that direction, and then project onto that vector?
$endgroup$
– Gerry Myerson
Dec 6 '18 at 1:27
$begingroup$
If you know how to project a vector onto another vector, and you want to project a vector along a direction, why not just take a vector in that direction, and then project onto that vector?
$endgroup$
– Gerry Myerson
Dec 6 '18 at 1:27
add a comment |
5 Answers
5
active
oldest
votes
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For the first projection, the one you know, you want the dot product $langle a_1, a_2rangle$ to be $0$, where $a_1 = lambda b$ and $a_2 = a-lambda b$. This yields $lambda = frac{langle b,arangle}{langle b,brangle}$ as you know.
In the second case, you want $langle c,arangle=0$, where $c=lambda b-a$. This yields $lambda = frac{langle a, arangle}{langle b,arangle}$ and hence
$$
a_3 = lambda b = frac{langle a,arangle}{langle b,arangle} , b.
$$
answered Dec 6 '18 at 11:28
ChristophChristoph
11.9k1642
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I look into this and try to figure this out. It will take me some time. I'll leave a comment if I am able to do this.
$endgroup$
– jorgenoro
Dec 6 '18 at 12:01
add a comment |
$begingroup$
So the projection of a vector $overline{a}$ onto a vector $overline{b}$ is $frac{langle overline{a},overline{b}rangleoverline{b}}{|overline{b}|^2}$.
answered Dec 6 '18 at 1:37
nafhgoodnafhgood
1,797422
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OP says he/she already knows that, right?
$endgroup$
– Gerry Myerson
Dec 6 '18 at 5:58
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Yes, this I know. I edited the post to show exactly what I want.
$endgroup$
– jorgenoro
Dec 6 '18 at 11:03
add a comment |
$begingroup$
Note that $a_1 + a_2 = a$ so once you have the projection $a_1$ then $a_2=a-a_1$ and you can project onto that vector in the usual way.
answered Dec 6 '18 at 1:58
CyclotomicFieldCyclotomicField
2,3481314
$endgroup$
$begingroup$
How can this help me to get the vector $overline{c}$ (depicted in the edited post) ?
$endgroup$
– jorgenoro
Dec 6 '18 at 11:05
add a comment |
$begingroup$
Projection of $bar a$ onto $bar b$ is $frac{<bar a, bar b> bar b}{|a|_2} $ and if you want to project onto hyper plane perpendicular to $bar b$ then it is $bar a - frac{<bar a, bar b> bar b}{|a|_2}$
answered Dec 6 '18 at 1:59
zerozero
708
$endgroup$
$begingroup$
But is that what OP wants to do? That's not at all clear to me.
$endgroup$
– Gerry Myerson
Dec 6 '18 at 5:56
$begingroup$
I don't understand this answer either. It seems that you are explaining to me how I can achieve the projection shown on the image above. What I want to do, is to take $overline{a}$, project onto $overline{b}$, along the direction of a vector $overline{c}$, where $overline{c}$ is normal to $overline{a}$.
$endgroup$
– jorgenoro
Dec 6 '18 at 10:54
$begingroup$
Updated post makes things clear thanks
$endgroup$
– zero
Dec 7 '18 at 22:29
add a comment |
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So, vectors $a$ and $b$ are given, and you want a vector $c$ such that $c$ is orthogonal to $a$ and such that $a+c$ is parallel to $b$. As equations, that's $acdot c=0$ and $a+c=kb$ for some real number $k$. So we get $acdot(a+c)=acdot kb$, which is $acdot a+acdot c=kacdot b$, which is $acdot a=kacdot b$, which is $k={acdot aover acdot b}$, so finally $$c=kb-a={acdot aover acdot b}b-a$$
answered Dec 6 '18 at 11:25
Gerry MyersonGerry Myerson
146k8147299
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In a comment you write that you want $c$, which I've given, but in the question you say you want $a_3$, which is $a+c$, which is $kb$, which is ${acdot aover acdot b}b$.
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– Gerry Myerson
Dec 6 '18 at 11:30
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You are right, that comment is wrong. It shouldn't be $overline{c}$ but $overline{a3}$. I will chew on this and try to figure this out and the above answer by Cristoph as well. Thank you.
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– jorgenoro
Dec 6 '18 at 11:59
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For the first projection, the one you know, you want the dot product $langle a_1, a_2rangle$ to be $0$, where $a_1 = lambda b$ and $a_2 = a-lambda b$. This yields $lambda = frac{langle b,arangle}{langle b,brangle}$ as you know.
In the second case, you want $langle c,arangle=0$, where $c=lambda b-a$. This yields $lambda = frac{langle a, arangle}{langle b,arangle}$ and hence
$$
a_3 = lambda b = frac{langle a,arangle}{langle b,arangle} , b.
$$
answered Dec 6 '18 at 11:28
ChristophChristoph
11.9k1642
$endgroup$
$begingroup$
I look into this and try to figure this out. It will take me some time. I'll leave a comment if I am able to do this.
$endgroup$
– jorgenoro
Dec 6 '18 at 12:01
add a comment |
$begingroup$
For the first projection, the one you know, you want the dot product $langle a_1, a_2rangle$ to be $0$, where $a_1 = lambda b$ and $a_2 = a-lambda b$. This yields $lambda = frac{langle b,arangle}{langle b,brangle}$ as you know.
In the second case, you want $langle c,arangle=0$, where $c=lambda b-a$. This yields $lambda = frac{langle a, arangle}{langle b,arangle}$ and hence
$$
a_3 = lambda b = frac{langle a,arangle}{langle b,arangle} , b.
$$
answered Dec 6 '18 at 11:28
ChristophChristoph
11.9k1642
$endgroup$
$begingroup$
I look into this and try to figure this out. It will take me some time. I'll leave a comment if I am able to do this.
$endgroup$
– jorgenoro
Dec 6 '18 at 12:01
add a comment |
$begingroup$
For the first projection, the one you know, you want the dot product $langle a_1, a_2rangle$ to be $0$, where $a_1 = lambda b$ and $a_2 = a-lambda b$. This yields $lambda = frac{langle b,arangle}{langle b,brangle}$ as you know.
In the second case, you want $langle c,arangle=0$, where $c=lambda b-a$. This yields $lambda = frac{langle a, arangle}{langle b,arangle}$ and hence
$$
a_3 = lambda b = frac{langle a,arangle}{langle b,arangle} , b.
$$
answered Dec 6 '18 at 11:28
ChristophChristoph
11.9k1642
$endgroup$
For the first projection, the one you know, you want the dot product $langle a_1, a_2rangle$ to be $0$, where $a_1 = lambda b$ and $a_2 = a-lambda b$. This yields $lambda = frac{langle b,arangle}{langle b,brangle}$ as you know.
In the second case, you want $langle c,arangle=0$, where $c=lambda b-a$. This yields $lambda = frac{langle a, arangle}{langle b,arangle}$ and hence
$$
a_3 = lambda b = frac{langle a,arangle}{langle b,arangle} , b.
$$
answered Dec 6 '18 at 11:28
ChristophChristoph
11.9k1642
answered Dec 6 '18 at 11:28
ChristophChristoph
11.9k1642
answered Dec 6 '18 at 11:28
ChristophChristoph
11.9k1642
answered Dec 6 '18 at 11:28
ChristophChristoph
11.9k1642
11.9k1642
$begingroup$
I look into this and try to figure this out. It will take me some time. I'll leave a comment if I am able to do this.
$endgroup$
– jorgenoro
Dec 6 '18 at 12:01
add a comment |
$begingroup$
I look into this and try to figure this out. It will take me some time. I'll leave a comment if I am able to do this.
$endgroup$
– jorgenoro
Dec 6 '18 at 12:01
$begingroup$
I look into this and try to figure this out. It will take me some time. I'll leave a comment if I am able to do this.
$endgroup$
– jorgenoro
Dec 6 '18 at 12:01
$begingroup$
I look into this and try to figure this out. It will take me some time. I'll leave a comment if I am able to do this.
$endgroup$
– jorgenoro
Dec 6 '18 at 12:01
add a comment |
$begingroup$
So the projection of a vector $overline{a}$ onto a vector $overline{b}$ is $frac{langle overline{a},overline{b}rangleoverline{b}}{|overline{b}|^2}$.
answered Dec 6 '18 at 1:37
nafhgoodnafhgood
1,797422
$endgroup$
$begingroup$
OP says he/she already knows that, right?
$endgroup$
– Gerry Myerson
Dec 6 '18 at 5:58
$begingroup$
Yes, this I know. I edited the post to show exactly what I want.
$endgroup$
– jorgenoro
Dec 6 '18 at 11:03
add a comment |
$begingroup$
So the projection of a vector $overline{a}$ onto a vector $overline{b}$ is $frac{langle overline{a},overline{b}rangleoverline{b}}{|overline{b}|^2}$.
answered Dec 6 '18 at 1:37
nafhgoodnafhgood
1,797422
$endgroup$
$begingroup$
OP says he/she already knows that, right?
$endgroup$
– Gerry Myerson
Dec 6 '18 at 5:58
$begingroup$
Yes, this I know. I edited the post to show exactly what I want.
$endgroup$
– jorgenoro
Dec 6 '18 at 11:03
add a comment |
$begingroup$
So the projection of a vector $overline{a}$ onto a vector $overline{b}$ is $frac{langle overline{a},overline{b}rangleoverline{b}}{|overline{b}|^2}$.
answered Dec 6 '18 at 1:37
nafhgoodnafhgood
1,797422
$endgroup$
So the projection of a vector $overline{a}$ onto a vector $overline{b}$ is $frac{langle overline{a},overline{b}rangleoverline{b}}{|overline{b}|^2}$.
answered Dec 6 '18 at 1:37
nafhgoodnafhgood
1,797422
answered Dec 6 '18 at 1:37
nafhgoodnafhgood
1,797422
answered Dec 6 '18 at 1:37
nafhgoodnafhgood
1,797422
answered Dec 6 '18 at 1:37
nafhgoodnafhgood
1,797422
1,797422
$begingroup$
OP says he/she already knows that, right?
$endgroup$
– Gerry Myerson
Dec 6 '18 at 5:58
$begingroup$
Yes, this I know. I edited the post to show exactly what I want.
$endgroup$
– jorgenoro
Dec 6 '18 at 11:03
add a comment |
$begingroup$
OP says he/she already knows that, right?
$endgroup$
– Gerry Myerson
Dec 6 '18 at 5:58
$begingroup$
Yes, this I know. I edited the post to show exactly what I want.
$endgroup$
– jorgenoro
Dec 6 '18 at 11:03
$begingroup$
OP says he/she already knows that, right?
$endgroup$
– Gerry Myerson
Dec 6 '18 at 5:58
$begingroup$
OP says he/she already knows that, right?
$endgroup$
– Gerry Myerson
Dec 6 '18 at 5:58
$begingroup$
Yes, this I know. I edited the post to show exactly what I want.
$endgroup$
– jorgenoro
Dec 6 '18 at 11:03
$begingroup$
Yes, this I know. I edited the post to show exactly what I want.
$endgroup$
– jorgenoro
Dec 6 '18 at 11:03
add a comment |
$begingroup$
Note that $a_1 + a_2 = a$ so once you have the projection $a_1$ then $a_2=a-a_1$ and you can project onto that vector in the usual way.
answered Dec 6 '18 at 1:58
CyclotomicFieldCyclotomicField
2,3481314
$endgroup$
$begingroup$
How can this help me to get the vector $overline{c}$ (depicted in the edited post) ?
$endgroup$
– jorgenoro
Dec 6 '18 at 11:05
add a comment |
$begingroup$
Note that $a_1 + a_2 = a$ so once you have the projection $a_1$ then $a_2=a-a_1$ and you can project onto that vector in the usual way.
answered Dec 6 '18 at 1:58
CyclotomicFieldCyclotomicField
2,3481314
$endgroup$
$begingroup$
How can this help me to get the vector $overline{c}$ (depicted in the edited post) ?
$endgroup$
– jorgenoro
Dec 6 '18 at 11:05
add a comment |
$begingroup$
Note that $a_1 + a_2 = a$ so once you have the projection $a_1$ then $a_2=a-a_1$ and you can project onto that vector in the usual way.
answered Dec 6 '18 at 1:58
CyclotomicFieldCyclotomicField
2,3481314
$endgroup$
Note that $a_1 + a_2 = a$ so once you have the projection $a_1$ then $a_2=a-a_1$ and you can project onto that vector in the usual way.
answered Dec 6 '18 at 1:58
CyclotomicFieldCyclotomicField
2,3481314
answered Dec 6 '18 at 1:58
CyclotomicFieldCyclotomicField
2,3481314
answered Dec 6 '18 at 1:58
CyclotomicFieldCyclotomicField
2,3481314
answered Dec 6 '18 at 1:58
CyclotomicFieldCyclotomicField
2,3481314
2,3481314
$begingroup$
How can this help me to get the vector $overline{c}$ (depicted in the edited post) ?
$endgroup$
– jorgenoro
Dec 6 '18 at 11:05
add a comment |
$begingroup$
How can this help me to get the vector $overline{c}$ (depicted in the edited post) ?
$endgroup$
– jorgenoro
Dec 6 '18 at 11:05
$begingroup$
How can this help me to get the vector $overline{c}$ (depicted in the edited post) ?
$endgroup$
– jorgenoro
Dec 6 '18 at 11:05
$begingroup$
How can this help me to get the vector $overline{c}$ (depicted in the edited post) ?
$endgroup$
– jorgenoro
Dec 6 '18 at 11:05
add a comment |
$begingroup$
Projection of $bar a$ onto $bar b$ is $frac{<bar a, bar b> bar b}{|a|_2} $ and if you want to project onto hyper plane perpendicular to $bar b$ then it is $bar a - frac{<bar a, bar b> bar b}{|a|_2}$
answered Dec 6 '18 at 1:59
zerozero
708
$endgroup$
$begingroup$
But is that what OP wants to do? That's not at all clear to me.
$endgroup$
– Gerry Myerson
Dec 6 '18 at 5:56
$begingroup$
I don't understand this answer either. It seems that you are explaining to me how I can achieve the projection shown on the image above. What I want to do, is to take $overline{a}$, project onto $overline{b}$, along the direction of a vector $overline{c}$, where $overline{c}$ is normal to $overline{a}$.
$endgroup$
– jorgenoro
Dec 6 '18 at 10:54
$begingroup$
Updated post makes things clear thanks
$endgroup$
– zero
Dec 7 '18 at 22:29
add a comment |
$begingroup$
Projection of $bar a$ onto $bar b$ is $frac{<bar a, bar b> bar b}{|a|_2} $ and if you want to project onto hyper plane perpendicular to $bar b$ then it is $bar a - frac{<bar a, bar b> bar b}{|a|_2}$
answered Dec 6 '18 at 1:59
zerozero
708
$endgroup$
$begingroup$
But is that what OP wants to do? That's not at all clear to me.
$endgroup$
– Gerry Myerson
Dec 6 '18 at 5:56
$begingroup$
I don't understand this answer either. It seems that you are explaining to me how I can achieve the projection shown on the image above. What I want to do, is to take $overline{a}$, project onto $overline{b}$, along the direction of a vector $overline{c}$, where $overline{c}$ is normal to $overline{a}$.
$endgroup$
– jorgenoro
Dec 6 '18 at 10:54
$begingroup$
Updated post makes things clear thanks
$endgroup$
– zero
Dec 7 '18 at 22:29
add a comment |
$begingroup$
Projection of $bar a$ onto $bar b$ is $frac{<bar a, bar b> bar b}{|a|_2} $ and if you want to project onto hyper plane perpendicular to $bar b$ then it is $bar a - frac{<bar a, bar b> bar b}{|a|_2}$
answered Dec 6 '18 at 1:59
zerozero
708
$endgroup$
Projection of $bar a$ onto $bar b$ is $frac{<bar a, bar b> bar b}{|a|_2} $ and if you want to project onto hyper plane perpendicular to $bar b$ then it is $bar a - frac{<bar a, bar b> bar b}{|a|_2}$
answered Dec 6 '18 at 1:59
zerozero
708
answered Dec 6 '18 at 1:59
zerozero
708
answered Dec 6 '18 at 1:59
zerozero
708
answered Dec 6 '18 at 1:59
zerozero
708
708
$begingroup$
But is that what OP wants to do? That's not at all clear to me.
$endgroup$
– Gerry Myerson
Dec 6 '18 at 5:56
$begingroup$
I don't understand this answer either. It seems that you are explaining to me how I can achieve the projection shown on the image above. What I want to do, is to take $overline{a}$, project onto $overline{b}$, along the direction of a vector $overline{c}$, where $overline{c}$ is normal to $overline{a}$.
$endgroup$
– jorgenoro
Dec 6 '18 at 10:54
$begingroup$
Updated post makes things clear thanks
$endgroup$
– zero
Dec 7 '18 at 22:29
add a comment |
$begingroup$
But is that what OP wants to do? That's not at all clear to me.
$endgroup$
– Gerry Myerson
Dec 6 '18 at 5:56
$begingroup$
I don't understand this answer either. It seems that you are explaining to me how I can achieve the projection shown on the image above. What I want to do, is to take $overline{a}$, project onto $overline{b}$, along the direction of a vector $overline{c}$, where $overline{c}$ is normal to $overline{a}$.
$endgroup$
– jorgenoro
Dec 6 '18 at 10:54
$begingroup$
Updated post makes things clear thanks
$endgroup$
– zero
Dec 7 '18 at 22:29
$begingroup$
But is that what OP wants to do? That's not at all clear to me.
$endgroup$
– Gerry Myerson
Dec 6 '18 at 5:56
$begingroup$
But is that what OP wants to do? That's not at all clear to me.
$endgroup$
– Gerry Myerson
Dec 6 '18 at 5:56
$begingroup$
I don't understand this answer either. It seems that you are explaining to me how I can achieve the projection shown on the image above. What I want to do, is to take $overline{a}$, project onto $overline{b}$, along the direction of a vector $overline{c}$, where $overline{c}$ is normal to $overline{a}$.
$endgroup$
– jorgenoro
Dec 6 '18 at 10:54
$begingroup$
I don't understand this answer either. It seems that you are explaining to me how I can achieve the projection shown on the image above. What I want to do, is to take $overline{a}$, project onto $overline{b}$, along the direction of a vector $overline{c}$, where $overline{c}$ is normal to $overline{a}$.
$endgroup$
– jorgenoro
Dec 6 '18 at 10:54
$begingroup$
Updated post makes things clear thanks
$endgroup$
– zero
Dec 7 '18 at 22:29
$begingroup$
Updated post makes things clear thanks
$endgroup$
– zero
Dec 7 '18 at 22:29
add a comment |
$begingroup$
So, vectors $a$ and $b$ are given, and you want a vector $c$ such that $c$ is orthogonal to $a$ and such that $a+c$ is parallel to $b$. As equations, that's $acdot c=0$ and $a+c=kb$ for some real number $k$. So we get $acdot(a+c)=acdot kb$, which is $acdot a+acdot c=kacdot b$, which is $acdot a=kacdot b$, which is $k={acdot aover acdot b}$, so finally $$c=kb-a={acdot aover acdot b}b-a$$
answered Dec 6 '18 at 11:25
Gerry MyersonGerry Myerson
146k8147299
$endgroup$
$begingroup$
In a comment you write that you want $c$, which I've given, but in the question you say you want $a_3$, which is $a+c$, which is $kb$, which is ${acdot aover acdot b}b$.
$endgroup$
– Gerry Myerson
Dec 6 '18 at 11:30
$begingroup$
You are right, that comment is wrong. It shouldn't be $overline{c}$ but $overline{a3}$. I will chew on this and try to figure this out and the above answer by Cristoph as well. Thank you.
$endgroup$
– jorgenoro
Dec 6 '18 at 11:59
add a comment |
$begingroup$
So, vectors $a$ and $b$ are given, and you want a vector $c$ such that $c$ is orthogonal to $a$ and such that $a+c$ is parallel to $b$. As equations, that's $acdot c=0$ and $a+c=kb$ for some real number $k$. So we get $acdot(a+c)=acdot kb$, which is $acdot a+acdot c=kacdot b$, which is $acdot a=kacdot b$, which is $k={acdot aover acdot b}$, so finally $$c=kb-a={acdot aover acdot b}b-a$$
answered Dec 6 '18 at 11:25
Gerry MyersonGerry Myerson
146k8147299
$endgroup$
$begingroup$
In a comment you write that you want $c$, which I've given, but in the question you say you want $a_3$, which is $a+c$, which is $kb$, which is ${acdot aover acdot b}b$.
$endgroup$
– Gerry Myerson
Dec 6 '18 at 11:30
$begingroup$
You are right, that comment is wrong. It shouldn't be $overline{c}$ but $overline{a3}$. I will chew on this and try to figure this out and the above answer by Cristoph as well. Thank you.
$endgroup$
– jorgenoro
Dec 6 '18 at 11:59
add a comment |
$begingroup$
So, vectors $a$ and $b$ are given, and you want a vector $c$ such that $c$ is orthogonal to $a$ and such that $a+c$ is parallel to $b$. As equations, that's $acdot c=0$ and $a+c=kb$ for some real number $k$. So we get $acdot(a+c)=acdot kb$, which is $acdot a+acdot c=kacdot b$, which is $acdot a=kacdot b$, which is $k={acdot aover acdot b}$, so finally $$c=kb-a={acdot aover acdot b}b-a$$
answered Dec 6 '18 at 11:25
Gerry MyersonGerry Myerson
146k8147299
$endgroup$
So, vectors $a$ and $b$ are given, and you want a vector $c$ such that $c$ is orthogonal to $a$ and such that $a+c$ is parallel to $b$. As equations, that's $acdot c=0$ and $a+c=kb$ for some real number $k$. So we get $acdot(a+c)=acdot kb$, which is $acdot a+acdot c=kacdot b$, which is $acdot a=kacdot b$, which is $k={acdot aover acdot b}$, so finally $$c=kb-a={acdot aover acdot b}b-a$$
answered Dec 6 '18 at 11:25
Gerry MyersonGerry Myerson
146k8147299
answered Dec 6 '18 at 11:25
Gerry MyersonGerry Myerson
146k8147299
answered Dec 6 '18 at 11:25
Gerry MyersonGerry Myerson
146k8147299
answered Dec 6 '18 at 11:25
Gerry MyersonGerry Myerson
146k8147299
146k8147299
$begingroup$
In a comment you write that you want $c$, which I've given, but in the question you say you want $a_3$, which is $a+c$, which is $kb$, which is ${acdot aover acdot b}b$.
$endgroup$
– Gerry Myerson
Dec 6 '18 at 11:30
$begingroup$
You are right, that comment is wrong. It shouldn't be $overline{c}$ but $overline{a3}$. I will chew on this and try to figure this out and the above answer by Cristoph as well. Thank you.
$endgroup$
– jorgenoro
Dec 6 '18 at 11:59
add a comment |
$begingroup$
In a comment you write that you want $c$, which I've given, but in the question you say you want $a_3$, which is $a+c$, which is $kb$, which is ${acdot aover acdot b}b$.
$endgroup$
– Gerry Myerson
Dec 6 '18 at 11:30
$begingroup$
You are right, that comment is wrong. It shouldn't be $overline{c}$ but $overline{a3}$. I will chew on this and try to figure this out and the above answer by Cristoph as well. Thank you.
$endgroup$
– jorgenoro
Dec 6 '18 at 11:59
$begingroup$
In a comment you write that you want $c$, which I've given, but in the question you say you want $a_3$, which is $a+c$, which is $kb$, which is ${acdot aover acdot b}b$.
$endgroup$
– Gerry Myerson
Dec 6 '18 at 11:30
$begingroup$
In a comment you write that you want $c$, which I've given, but in the question you say you want $a_3$, which is $a+c$, which is $kb$, which is ${acdot aover acdot b}b$.
$endgroup$
– Gerry Myerson
Dec 6 '18 at 11:30
$begingroup$
You are right, that comment is wrong. It shouldn't be $overline{c}$ but $overline{a3}$. I will chew on this and try to figure this out and the above answer by Cristoph as well. Thank you.
$endgroup$
– jorgenoro
Dec 6 '18 at 11:59
$begingroup$
You are right, that comment is wrong. It shouldn't be $overline{c}$ but $overline{a3}$. I will chew on this and try to figure this out and the above answer by Cristoph as well. Thank you.
$endgroup$
– jorgenoro
Dec 6 '18 at 11:59
add a comment |
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$begingroup$
If you know how to project a vector onto another vector, and you want to project a vector along a direction, why not just take a vector in that direction, and then project onto that vector?
$endgroup$
– Gerry Myerson
Dec 6 '18 at 1:27