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Angle between vectors $vec a + vec b$ and $vec c$?
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If there are three vector $vec a$, $vec b$ and $vec c$ provided, what is the angle between $vec a + vec b$ and $vec c$? I understand how to calculate for angle between $vec a$ and $vec b$, $vec a$ and $vec c$ and $vec b$ and $vec c$ but what does the angle between $vec a + vec b$ and $vec c$ mean?
vectors inner-product-space
edited Aug 6 at 5:49
Jyrki Lahtonen
105k12161355
asked Aug 5 at 10:29
bison72
314
add a comment |Â
up vote
2
down vote
favorite
If there are three vector $vec a$, $vec b$ and $vec c$ provided, what is the angle between $vec a + vec b$ and $vec c$? I understand how to calculate for angle between $vec a$ and $vec b$, $vec a$ and $vec c$ and $vec b$ and $vec c$ but what does the angle between $vec a + vec b$ and $vec c$ mean?
vectors inner-product-space
edited Aug 6 at 5:49
Jyrki Lahtonen
105k12161355
asked Aug 5 at 10:29
bison72
314
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
If there are three vector $vec a$, $vec b$ and $vec c$ provided, what is the angle between $vec a + vec b$ and $vec c$? I understand how to calculate for angle between $vec a$ and $vec b$, $vec a$ and $vec c$ and $vec b$ and $vec c$ but what does the angle between $vec a + vec b$ and $vec c$ mean?
vectors inner-product-space
edited Aug 6 at 5:49
Jyrki Lahtonen
105k12161355
asked Aug 5 at 10:29
bison72
314
If there are three vector $vec a$, $vec b$ and $vec c$ provided, what is the angle between $vec a + vec b$ and $vec c$? I understand how to calculate for angle between $vec a$ and $vec b$, $vec a$ and $vec c$ and $vec b$ and $vec c$ but what does the angle between $vec a + vec b$ and $vec c$ mean?
vectors inner-product-space
edited Aug 6 at 5:49
Jyrki Lahtonen
105k12161355
asked Aug 5 at 10:29
bison72
314
edited Aug 6 at 5:49
Jyrki Lahtonen
105k12161355
edited Aug 6 at 5:49
Jyrki Lahtonen
105k12161355
edited Aug 6 at 5:49
Jyrki Lahtonen
105k12161355
105k12161355
asked Aug 5 at 10:29
bison72
314
asked Aug 5 at 10:29
bison72
314
asked Aug 5 at 10:29
bison72
314
314
add a comment |Â
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
7
down vote
You compute the sum of $a+b$, that is a vector, if it helps, you can give a new name, $d$.
Now you can compute the angle between $d$ and $c$ using the formula that you are familiar with.
answered Aug 5 at 10:34
Siong Thye Goh
78.1k134997
add a comment |Â
up vote
6
down vote
We have that $vec d=vec a+vec b$ then
$$cos (theta)=fracvec d cdot vec c$$
answered Aug 5 at 10:37
gimusi
65.5k73684
add a comment |Â
up vote
3
down vote
The important concept to keep in mind here is that when you add two vectors, $vec a$ and $vec b$, the result is just another vector. It has a magnitude, a direction (if its magnitude is positive), and components just like any other vector.
We may write the expression
$vec a + vec b$
in order to name this vector, but that does not make it into any new kind of object.
The formulas that apply to other vectors still apply.
To find the angle between $vec a + vec b$ and $vec c,$
take the formula you would use for the angle between any two vectors, put $vec a + vec b$
in the places where the first vector occurs in the formula,
and put $vec c$ in the places where the second vector occurs.
Look at other answers for details of what happens after you do this.
answered Aug 5 at 12:31
David K
48.3k340107
1
If vectors a=[1,1,0], b=[3,2,1], and c=[1,0,2], does the mean the angle between a +b and c will be the angle between [4,3,1] and [1,0,2]?
– bison72
Aug 5 at 14:59
1
@bison72 yes, this is how vector addition works.
– PaÃ…Âlo Ebermann
Aug 5 at 20:47
add a comment |Â
up vote
2
down vote
Using the definition of the scalar product we get
$$
(a+b)cdot c =
lVert a+b rVert lVert c rVert cos angle(a+b,c)
$$
where $lVert v rVert = sqrtvcdot v$.
For non-zero $a+b$ and $c$ we can solve for the angle
$$
angle(a+b,c) = arccos frac(a+b)cdot clVert a+b rVert lVert c rVert
$$
answered Aug 5 at 11:28
mvw
30.6k22251
add a comment |Â
up vote
2
down vote
Just $$cos(widehatveca+vecb,vecc)=frac(veca+vecb)vecc.$$
Here, $|veca+vecb|=sqrt^2+$ and $(veca+vecb)vecc=vecavecc+vecbvecc.$
answered Aug 5 at 10:31
Michael Rozenberg
88.2k1579180
Why someone down voted?
– Michael Rozenberg
Aug 5 at 15:37
Might be because someone thought it's not obvious that e.g. $ab$ is the scalar product of vectors and not simple product of their norms (because you've dropped the arrows used by the OP). But that's just my guess.
– Ruslan
Aug 5 at 21:35
@Ruslan But there is a context. Thank you for your trying! I'll fix it.
– Michael Rozenberg
Aug 6 at 4:11
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
You compute the sum of $a+b$, that is a vector, if it helps, you can give a new name, $d$.
Now you can compute the angle between $d$ and $c$ using the formula that you are familiar with.
answered Aug 5 at 10:34
Siong Thye Goh
78.1k134997
add a comment |Â
up vote
7
down vote
You compute the sum of $a+b$, that is a vector, if it helps, you can give a new name, $d$.
Now you can compute the angle between $d$ and $c$ using the formula that you are familiar with.
answered Aug 5 at 10:34
Siong Thye Goh
78.1k134997
add a comment |Â
up vote
7
down vote
up vote
7
down vote
You compute the sum of $a+b$, that is a vector, if it helps, you can give a new name, $d$.
Now you can compute the angle between $d$ and $c$ using the formula that you are familiar with.
answered Aug 5 at 10:34
Siong Thye Goh
78.1k134997
You compute the sum of $a+b$, that is a vector, if it helps, you can give a new name, $d$.
Now you can compute the angle between $d$ and $c$ using the formula that you are familiar with.
answered Aug 5 at 10:34
Siong Thye Goh
78.1k134997
answered Aug 5 at 10:34
Siong Thye Goh
78.1k134997
answered Aug 5 at 10:34
Siong Thye Goh
78.1k134997
answered Aug 5 at 10:34
Siong Thye Goh
78.1k134997
78.1k134997
add a comment |Â
add a comment |Â
up vote
6
down vote
We have that $vec d=vec a+vec b$ then
$$cos (theta)=fracvec d cdot vec c$$
answered Aug 5 at 10:37
gimusi
65.5k73684
add a comment |Â
up vote
6
down vote
We have that $vec d=vec a+vec b$ then
$$cos (theta)=fracvec d cdot vec c$$
answered Aug 5 at 10:37
gimusi
65.5k73684
add a comment |Â
up vote
6
down vote
up vote
6
down vote
We have that $vec d=vec a+vec b$ then
$$cos (theta)=fracvec d cdot vec c$$
answered Aug 5 at 10:37
gimusi
65.5k73684
We have that $vec d=vec a+vec b$ then
$$cos (theta)=fracvec d cdot vec c$$
answered Aug 5 at 10:37
gimusi
65.5k73684
answered Aug 5 at 10:37
gimusi
65.5k73684
answered Aug 5 at 10:37
gimusi
65.5k73684
answered Aug 5 at 10:37
gimusi
65.5k73684
65.5k73684
add a comment |Â
add a comment |Â
up vote
3
down vote
The important concept to keep in mind here is that when you add two vectors, $vec a$ and $vec b$, the result is just another vector. It has a magnitude, a direction (if its magnitude is positive), and components just like any other vector.
We may write the expression
$vec a + vec b$
in order to name this vector, but that does not make it into any new kind of object.
The formulas that apply to other vectors still apply.
To find the angle between $vec a + vec b$ and $vec c,$
take the formula you would use for the angle between any two vectors, put $vec a + vec b$
in the places where the first vector occurs in the formula,
and put $vec c$ in the places where the second vector occurs.
Look at other answers for details of what happens after you do this.
answered Aug 5 at 12:31
David K
48.3k340107
1
If vectors a=[1,1,0], b=[3,2,1], and c=[1,0,2], does the mean the angle between a +b and c will be the angle between [4,3,1] and [1,0,2]?
– bison72
Aug 5 at 14:59
1
@bison72 yes, this is how vector addition works.
– PaÃ…Âlo Ebermann
Aug 5 at 20:47
add a comment |Â
up vote
3
down vote
The important concept to keep in mind here is that when you add two vectors, $vec a$ and $vec b$, the result is just another vector. It has a magnitude, a direction (if its magnitude is positive), and components just like any other vector.
We may write the expression
$vec a + vec b$
in order to name this vector, but that does not make it into any new kind of object.
The formulas that apply to other vectors still apply.
To find the angle between $vec a + vec b$ and $vec c,$
take the formula you would use for the angle between any two vectors, put $vec a + vec b$
in the places where the first vector occurs in the formula,
and put $vec c$ in the places where the second vector occurs.
Look at other answers for details of what happens after you do this.
answered Aug 5 at 12:31
David K
48.3k340107
1
If vectors a=[1,1,0], b=[3,2,1], and c=[1,0,2], does the mean the angle between a +b and c will be the angle between [4,3,1] and [1,0,2]?
– bison72
Aug 5 at 14:59
1
@bison72 yes, this is how vector addition works.
– PaÃ…Âlo Ebermann
Aug 5 at 20:47
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The important concept to keep in mind here is that when you add two vectors, $vec a$ and $vec b$, the result is just another vector. It has a magnitude, a direction (if its magnitude is positive), and components just like any other vector.
We may write the expression
$vec a + vec b$
in order to name this vector, but that does not make it into any new kind of object.
The formulas that apply to other vectors still apply.
To find the angle between $vec a + vec b$ and $vec c,$
take the formula you would use for the angle between any two vectors, put $vec a + vec b$
in the places where the first vector occurs in the formula,
and put $vec c$ in the places where the second vector occurs.
Look at other answers for details of what happens after you do this.
answered Aug 5 at 12:31
David K
48.3k340107
The important concept to keep in mind here is that when you add two vectors, $vec a$ and $vec b$, the result is just another vector. It has a magnitude, a direction (if its magnitude is positive), and components just like any other vector.
We may write the expression
$vec a + vec b$
in order to name this vector, but that does not make it into any new kind of object.
The formulas that apply to other vectors still apply.
To find the angle between $vec a + vec b$ and $vec c,$
take the formula you would use for the angle between any two vectors, put $vec a + vec b$
in the places where the first vector occurs in the formula,
and put $vec c$ in the places where the second vector occurs.
Look at other answers for details of what happens after you do this.
answered Aug 5 at 12:31
David K
48.3k340107
answered Aug 5 at 12:31
David K
48.3k340107
answered Aug 5 at 12:31
David K
48.3k340107
answered Aug 5 at 12:31
David K
48.3k340107
48.3k340107
1
If vectors a=[1,1,0], b=[3,2,1], and c=[1,0,2], does the mean the angle between a +b and c will be the angle between [4,3,1] and [1,0,2]?
– bison72
Aug 5 at 14:59
1
@bison72 yes, this is how vector addition works.
– PaÃ…Âlo Ebermann
Aug 5 at 20:47
add a comment |Â
1
If vectors a=[1,1,0], b=[3,2,1], and c=[1,0,2], does the mean the angle between a +b and c will be the angle between [4,3,1] and [1,0,2]?
– bison72
Aug 5 at 14:59
1
@bison72 yes, this is how vector addition works.
– PaÃ…Âlo Ebermann
Aug 5 at 20:47
1
1
If vectors a=[1,1,0], b=[3,2,1], and c=[1,0,2], does the mean the angle between a +b and c will be the angle between [4,3,1] and [1,0,2]?
– bison72
Aug 5 at 14:59
If vectors a=[1,1,0], b=[3,2,1], and c=[1,0,2], does the mean the angle between a +b and c will be the angle between [4,3,1] and [1,0,2]?
– bison72
Aug 5 at 14:59
1
1
@bison72 yes, this is how vector addition works.
– PaÃ…Âlo Ebermann
Aug 5 at 20:47
@bison72 yes, this is how vector addition works.
– PaÃ…Âlo Ebermann
Aug 5 at 20:47
add a comment |Â
up vote
2
down vote
Using the definition of the scalar product we get
$$
(a+b)cdot c =
lVert a+b rVert lVert c rVert cos angle(a+b,c)
$$
where $lVert v rVert = sqrtvcdot v$.
For non-zero $a+b$ and $c$ we can solve for the angle
$$
angle(a+b,c) = arccos frac(a+b)cdot clVert a+b rVert lVert c rVert
$$
answered Aug 5 at 11:28
mvw
30.6k22251
add a comment |Â
up vote
2
down vote
Using the definition of the scalar product we get
$$
(a+b)cdot c =
lVert a+b rVert lVert c rVert cos angle(a+b,c)
$$
where $lVert v rVert = sqrtvcdot v$.
For non-zero $a+b$ and $c$ we can solve for the angle
$$
angle(a+b,c) = arccos frac(a+b)cdot clVert a+b rVert lVert c rVert
$$
answered Aug 5 at 11:28
mvw
30.6k22251
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Using the definition of the scalar product we get
$$
(a+b)cdot c =
lVert a+b rVert lVert c rVert cos angle(a+b,c)
$$
where $lVert v rVert = sqrtvcdot v$.
For non-zero $a+b$ and $c$ we can solve for the angle
$$
angle(a+b,c) = arccos frac(a+b)cdot clVert a+b rVert lVert c rVert
$$
answered Aug 5 at 11:28
mvw
30.6k22251
Using the definition of the scalar product we get
$$
(a+b)cdot c =
lVert a+b rVert lVert c rVert cos angle(a+b,c)
$$
where $lVert v rVert = sqrtvcdot v$.
For non-zero $a+b$ and $c$ we can solve for the angle
$$
angle(a+b,c) = arccos frac(a+b)cdot clVert a+b rVert lVert c rVert
$$
answered Aug 5 at 11:28
mvw
30.6k22251
edited Aug 5 at 11:33
answered Aug 5 at 11:28
mvw
30.6k22251
answered Aug 5 at 11:28
mvw
30.6k22251
answered Aug 5 at 11:28
mvw
30.6k22251
30.6k22251
add a comment |Â
add a comment |Â
up vote
2
down vote
Just $$cos(widehatveca+vecb,vecc)=frac(veca+vecb)vecc.$$
Here, $|veca+vecb|=sqrt^2+$ and $(veca+vecb)vecc=vecavecc+vecbvecc.$
answered Aug 5 at 10:31
Michael Rozenberg
88.2k1579180
Why someone down voted?
– Michael Rozenberg
Aug 5 at 15:37
Might be because someone thought it's not obvious that e.g. $ab$ is the scalar product of vectors and not simple product of their norms (because you've dropped the arrows used by the OP). But that's just my guess.
– Ruslan
Aug 5 at 21:35
@Ruslan But there is a context. Thank you for your trying! I'll fix it.
– Michael Rozenberg
Aug 6 at 4:11
add a comment |Â
up vote
2
down vote
Just $$cos(widehatveca+vecb,vecc)=frac(veca+vecb)vecc.$$
Here, $|veca+vecb|=sqrt^2+$ and $(veca+vecb)vecc=vecavecc+vecbvecc.$
answered Aug 5 at 10:31
Michael Rozenberg
88.2k1579180
Why someone down voted?
– Michael Rozenberg
Aug 5 at 15:37
Might be because someone thought it's not obvious that e.g. $ab$ is the scalar product of vectors and not simple product of their norms (because you've dropped the arrows used by the OP). But that's just my guess.
– Ruslan
Aug 5 at 21:35
@Ruslan But there is a context. Thank you for your trying! I'll fix it.
– Michael Rozenberg
Aug 6 at 4:11
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Just $$cos(widehatveca+vecb,vecc)=frac(veca+vecb)vecc.$$
Here, $|veca+vecb|=sqrt^2+$ and $(veca+vecb)vecc=vecavecc+vecbvecc.$
answered Aug 5 at 10:31
Michael Rozenberg
88.2k1579180
Just $$cos(widehatveca+vecb,vecc)=frac(veca+vecb)vecc.$$
Here, $|veca+vecb|=sqrt^2+$ and $(veca+vecb)vecc=vecavecc+vecbvecc.$
answered Aug 5 at 10:31
Michael Rozenberg
88.2k1579180
edited Aug 6 at 4:17
answered Aug 5 at 10:31
Michael Rozenberg
88.2k1579180
answered Aug 5 at 10:31
Michael Rozenberg
88.2k1579180
answered Aug 5 at 10:31
Michael Rozenberg
88.2k1579180
88.2k1579180
Why someone down voted?
– Michael Rozenberg
Aug 5 at 15:37
Might be because someone thought it's not obvious that e.g. $ab$ is the scalar product of vectors and not simple product of their norms (because you've dropped the arrows used by the OP). But that's just my guess.
– Ruslan
Aug 5 at 21:35
@Ruslan But there is a context. Thank you for your trying! I'll fix it.
– Michael Rozenberg
Aug 6 at 4:11
add a comment |Â
Why someone down voted?
– Michael Rozenberg
Aug 5 at 15:37
Might be because someone thought it's not obvious that e.g. $ab$ is the scalar product of vectors and not simple product of their norms (because you've dropped the arrows used by the OP). But that's just my guess.
– Ruslan
Aug 5 at 21:35
@Ruslan But there is a context. Thank you for your trying! I'll fix it.
– Michael Rozenberg
Aug 6 at 4:11
Why someone down voted?
– Michael Rozenberg
Aug 5 at 15:37
Why someone down voted?
– Michael Rozenberg
Aug 5 at 15:37
Might be because someone thought it's not obvious that e.g. $ab$ is the scalar product of vectors and not simple product of their norms (because you've dropped the arrows used by the OP). But that's just my guess.
– Ruslan
Aug 5 at 21:35
Might be because someone thought it's not obvious that e.g. $ab$ is the scalar product of vectors and not simple product of their norms (because you've dropped the arrows used by the OP). But that's just my guess.
– Ruslan
Aug 5 at 21:35
@Ruslan But there is a context. Thank you for your trying! I'll fix it.
– Michael Rozenberg
Aug 6 at 4:11
@Ruslan But there is a context. Thank you for your trying! I'll fix it.
– Michael Rozenberg
Aug 6 at 4:11
add a comment |Â
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