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

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edited 8 mins ago

Jyrki Lahtonen

104k12160355

asked 19 hours ago

bison72

152

add a comment | 

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

share|cite|improve this question

edited 8 mins ago

Jyrki Lahtonen

104k12160355

asked 19 hours ago

bison72

152

add a comment | 

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

share|cite|improve this question

edited 8 mins ago

Jyrki Lahtonen

104k12160355

asked 19 hours ago

bison72

152

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

share|cite|improve this question

edited 8 mins ago

Jyrki Lahtonen

104k12160355

asked 19 hours ago

bison72

152

share|cite|improve this question

share|cite|improve this question

edited 8 mins ago

Jyrki Lahtonen

104k12160355

edited 8 mins ago

Jyrki Lahtonen

104k12160355

edited 8 mins ago

Jyrki Lahtonen

104k12160355

104k12160355

asked 19 hours ago

bison72

152

asked 19 hours ago

bison72

152

asked 19 hours ago

bison72

152

152

add a comment | 

add a comment | 

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.

share|cite|improve this answer

answered 19 hours ago

Siong Thye Goh

76.3k124592

add a comment | 

up vote
7
down vote

We have that $vec d=vec a+vec b$ then

$$cos (theta)=fracvec d cdot vec c$$

share|cite|improve this answer

answered 19 hours ago

gimusi

63.4k73379

add a comment | 

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.

share|cite|improve this answer

answered 17 hours ago

David K

48k339106

• 
  
  
  

  
  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
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @bison72 yes, this is how vector addition works.
  – PaÅ­lo Ebermann
  9 hours ago
  

  
  
  
  

add a comment | 

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
$$
angle(a+b,c) = arccos frac(a+b)cdot clVert a+b rVert lVert c rVert
$$

share|cite|improve this answer

edited 18 hours ago

answered 18 hours ago

mvw

30.1k22150

add a comment | 

up vote
2
down vote

Just $$cos(widehatveca+vecb,vecc)=frac(veca+vecb)vecc.$$

Here, $|veca+vecb|=sqrt$ and $(veca+vecb)vecc=vecavecc+vecbvecc.$

share|cite|improve this answer

edited 1 hour ago

answered 19 hours ago

Michael Rozenberg

86.9k1575178

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why someone down voted?
  – Michael Rozenberg
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ruslan But there is a context. Thank you for your trying! I'll fix it.
  – Michael Rozenberg
  1 hour ago
  
  

  
  
  
  

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

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.

share|cite|improve this answer

answered 19 hours ago

Siong Thye Goh

76.3k124592

add a comment | 

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.

share|cite|improve this answer

answered 19 hours ago

Siong Thye Goh

76.3k124592

add a comment | 

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.

share|cite|improve this answer

answered 19 hours ago

Siong Thye Goh

76.3k124592

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.

share|cite|improve this answer

answered 19 hours ago

Siong Thye Goh

76.3k124592

share|cite|improve this answer

share|cite|improve this answer

answered 19 hours ago

Siong Thye Goh

76.3k124592

answered 19 hours ago

Siong Thye Goh

76.3k124592

answered 19 hours ago

Siong Thye Goh

76.3k124592

76.3k124592

add a comment | 

add a comment | 

up vote
7
down vote

We have that $vec d=vec a+vec b$ then

$$cos (theta)=fracvec d cdot vec c$$

share|cite|improve this answer

answered 19 hours ago

gimusi

63.4k73379

add a comment | 

up vote
7
down vote

We have that $vec d=vec a+vec b$ then

$$cos (theta)=fracvec d cdot vec c$$

share|cite|improve this answer

answered 19 hours ago

gimusi

63.4k73379

add a comment | 

up vote
7
down vote

up vote
7
down vote

We have that $vec d=vec a+vec b$ then

$$cos (theta)=fracvec d cdot vec c$$

share|cite|improve this answer

answered 19 hours ago

gimusi

63.4k73379

We have that $vec d=vec a+vec b$ then

$$cos (theta)=fracvec d cdot vec c$$

share|cite|improve this answer

answered 19 hours ago

gimusi

63.4k73379

share|cite|improve this answer

share|cite|improve this answer

answered 19 hours ago

gimusi

63.4k73379

answered 19 hours ago

gimusi

63.4k73379

answered 19 hours ago

gimusi

63.4k73379

63.4k73379

add a comment | 

add a comment | 

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.

share|cite|improve this answer

answered 17 hours ago

David K

48k339106

• 
  
  
  

  
  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
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @bison72 yes, this is how vector addition works.
  – PaÅ­lo Ebermann
  9 hours ago
  

  
  
  
  

add a comment | 

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.

share|cite|improve this answer

answered 17 hours ago

David K

48k339106

• 
  
  
  

  
  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
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @bison72 yes, this is how vector addition works.
  – PaÅ­lo Ebermann
  9 hours ago
  

  
  
  
  

add a comment | 

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.

share|cite|improve this answer

answered 17 hours ago

David K

48k339106

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.

share|cite|improve this answer

answered 17 hours ago

David K

48k339106

share|cite|improve this answer

share|cite|improve this answer

answered 17 hours ago

David K

48k339106

answered 17 hours ago

David K

48k339106

answered 17 hours ago

David K

48k339106

48k339106

• 
  
  
  

  
  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
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @bison72 yes, this is how vector addition works.
  – PaÅ­lo Ebermann
  9 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  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
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @bison72 yes, this is how vector addition works.
  – PaÅ­lo Ebermann
  9 hours ago
  

  
  
  
  

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
14 hours ago

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
14 hours ago

1

1

@bison72 yes, this is how vector addition works.
– PaÅ­lo Ebermann
9 hours ago

@bison72 yes, this is how vector addition works.
– PaÅ­lo Ebermann
9 hours ago

add a comment | 

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
$$
angle(a+b,c) = arccos frac(a+b)cdot clVert a+b rVert lVert c rVert
$$

share|cite|improve this answer

edited 18 hours ago

answered 18 hours ago

mvw

30.1k22150

add a comment | 

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
$$
angle(a+b,c) = arccos frac(a+b)cdot clVert a+b rVert lVert c rVert
$$

share|cite|improve this answer

edited 18 hours ago

answered 18 hours ago

mvw

30.1k22150

add a comment | 

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
$$
angle(a+b,c) = arccos frac(a+b)cdot clVert a+b rVert lVert c rVert
$$

share|cite|improve this answer

edited 18 hours ago

answered 18 hours ago

mvw

30.1k22150

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
$$
angle(a+b,c) = arccos frac(a+b)cdot clVert a+b rVert lVert c rVert
$$

share|cite|improve this answer

edited 18 hours ago

answered 18 hours ago

mvw

30.1k22150

share|cite|improve this answer

share|cite|improve this answer

edited 18 hours ago

edited 18 hours ago

edited 18 hours ago

answered 18 hours ago

mvw

30.1k22150

answered 18 hours ago

mvw

30.1k22150

answered 18 hours ago

mvw

30.1k22150

30.1k22150

add a comment | 

add a comment | 

up vote
2
down vote

Just $$cos(widehatveca+vecb,vecc)=frac(veca+vecb)vecc.$$

Here, $|veca+vecb|=sqrt$ and $(veca+vecb)vecc=vecavecc+vecbvecc.$

share|cite|improve this answer

edited 1 hour ago

answered 19 hours ago

Michael Rozenberg

86.9k1575178

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why someone down voted?
  – Michael Rozenberg
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ruslan But there is a context. Thank you for your trying! I'll fix it.
  – Michael Rozenberg
  1 hour ago
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

Just $$cos(widehatveca+vecb,vecc)=frac(veca+vecb)vecc.$$

Here, $|veca+vecb|=sqrt$ and $(veca+vecb)vecc=vecavecc+vecbvecc.$

share|cite|improve this answer

edited 1 hour ago

answered 19 hours ago

Michael Rozenberg

86.9k1575178

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why someone down voted?
  – Michael Rozenberg
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ruslan But there is a context. Thank you for your trying! I'll fix it.
  – Michael Rozenberg
  1 hour ago
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

Just $$cos(widehatveca+vecb,vecc)=frac(veca+vecb)vecc.$$

Here, $|veca+vecb|=sqrt$ and $(veca+vecb)vecc=vecavecc+vecbvecc.$

share|cite|improve this answer

edited 1 hour ago

answered 19 hours ago

Michael Rozenberg

86.9k1575178

Just $$cos(widehatveca+vecb,vecc)=frac(veca+vecb)vecc.$$

Here, $|veca+vecb|=sqrt$ and $(veca+vecb)vecc=vecavecc+vecbvecc.$

share|cite|improve this answer

edited 1 hour ago

answered 19 hours ago

Michael Rozenberg

86.9k1575178

share|cite|improve this answer

share|cite|improve this answer

edited 1 hour ago

edited 1 hour ago

edited 1 hour ago

answered 19 hours ago

Michael Rozenberg

86.9k1575178

answered 19 hours ago

Michael Rozenberg

86.9k1575178

answered 19 hours ago

Michael Rozenberg

86.9k1575178

86.9k1575178

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why someone down voted?
  – Michael Rozenberg
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ruslan But there is a context. Thank you for your trying! I'll fix it.
  – Michael Rozenberg
  1 hour ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why someone down voted?
  – Michael Rozenberg
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ruslan But there is a context. Thank you for your trying! I'll fix it.
  – Michael Rozenberg
  1 hour ago
  
  

  
  
  
  

Why someone down voted?
– Michael Rozenberg
14 hours ago

Why someone down voted?
– Michael Rozenberg
14 hours ago

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
8 hours ago

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
8 hours ago

@Ruslan But there is a context. Thank you for your trying! I'll fix it.
– Michael Rozenberg
1 hour ago

@Ruslan But there is a context. Thank you for your trying! I'll fix it.
– Michael Rozenberg
1 hour ago

add a comment | 

 

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