Is a unit vector really unitless and dimensionless?

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That's the question I wanted to ask. According to my textbooks, a unit vector has no units and no dimensions, it is only used to specify direction, it only shows the orientation of a corresponding vector in space. I think it's true, or that's what it looks like. It makes sense because a unit vector is defined as 'a vector' divided by its magnitude. Since we have the same numerical value in numerator and the denominator, a unit vector has a magnitude of 1 unit. Likewise, we have the same unit in both numerator and the denominator, that makes a unit vector 'unitless', and hence dimensionless. That's why I think a unit vector has no dimensions. (Please correct me if I'm wrong)

But...a big but.. Another question naturally comes to our mind. Why if I say, "a force of 1 N due east" ?? Or "a displacement of 1m, 30° NOE" ?

Both force and displacement are vector quantities, and both have a magnitude of 1 unit in the above two examples.
My question is, can we call these two "unit vectors" ?? That's what I'm struggling to understand. There's no reason why we can't call these two unit vectors. Because both have a magnitude of 1 unit, and both are vectors. But..... Both have units, and hence both are not dimensionless.

Please explain it to me, thanks

vectors

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asked 30 mins ago

π times e

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up vote
1
down vote

favorite

That's the question I wanted to ask. According to my textbooks, a unit vector has no units and no dimensions, it is only used to specify direction, it only shows the orientation of a corresponding vector in space. I think it's true, or that's what it looks like. It makes sense because a unit vector is defined as 'a vector' divided by its magnitude. Since we have the same numerical value in numerator and the denominator, a unit vector has a magnitude of 1 unit. Likewise, we have the same unit in both numerator and the denominator, that makes a unit vector 'unitless', and hence dimensionless. That's why I think a unit vector has no dimensions. (Please correct me if I'm wrong)

But...a big but.. Another question naturally comes to our mind. Why if I say, "a force of 1 N due east" ?? Or "a displacement of 1m, 30° NOE" ?

Both force and displacement are vector quantities, and both have a magnitude of 1 unit in the above two examples.
My question is, can we call these two "unit vectors" ?? That's what I'm struggling to understand. There's no reason why we can't call these two unit vectors. Because both have a magnitude of 1 unit, and both are vectors. But..... Both have units, and hence both are not dimensionless.

Please explain it to me, thanks

vectors

share|cite|improve this question

asked 30 mins ago

π times e

273

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

That's the question I wanted to ask. According to my textbooks, a unit vector has no units and no dimensions, it is only used to specify direction, it only shows the orientation of a corresponding vector in space. I think it's true, or that's what it looks like. It makes sense because a unit vector is defined as 'a vector' divided by its magnitude. Since we have the same numerical value in numerator and the denominator, a unit vector has a magnitude of 1 unit. Likewise, we have the same unit in both numerator and the denominator, that makes a unit vector 'unitless', and hence dimensionless. That's why I think a unit vector has no dimensions. (Please correct me if I'm wrong)

But...a big but.. Another question naturally comes to our mind. Why if I say, "a force of 1 N due east" ?? Or "a displacement of 1m, 30° NOE" ?

Both force and displacement are vector quantities, and both have a magnitude of 1 unit in the above two examples.
My question is, can we call these two "unit vectors" ?? That's what I'm struggling to understand. There's no reason why we can't call these two unit vectors. Because both have a magnitude of 1 unit, and both are vectors. But..... Both have units, and hence both are not dimensionless.

Please explain it to me, thanks

vectors

share|cite|improve this question

asked 30 mins ago

π times e

273

That's the question I wanted to ask. According to my textbooks, a unit vector has no units and no dimensions, it is only used to specify direction, it only shows the orientation of a corresponding vector in space. I think it's true, or that's what it looks like. It makes sense because a unit vector is defined as 'a vector' divided by its magnitude. Since we have the same numerical value in numerator and the denominator, a unit vector has a magnitude of 1 unit. Likewise, we have the same unit in both numerator and the denominator, that makes a unit vector 'unitless', and hence dimensionless. That's why I think a unit vector has no dimensions. (Please correct me if I'm wrong)

But...a big but.. Another question naturally comes to our mind. Why if I say, "a force of 1 N due east" ?? Or "a displacement of 1m, 30° NOE" ?

Both force and displacement are vector quantities, and both have a magnitude of 1 unit in the above two examples.
My question is, can we call these two "unit vectors" ?? That's what I'm struggling to understand. There's no reason why we can't call these two unit vectors. Because both have a magnitude of 1 unit, and both are vectors. But..... Both have units, and hence both are not dimensionless.

Please explain it to me, thanks

vectors

vectors

share|cite|improve this question

asked 30 mins ago

π times e

273

share|cite|improve this question

asked 30 mins ago

π times e

273

share|cite|improve this question

share|cite|improve this question

asked 30 mins ago

π times e

273

asked 30 mins ago

π times e

273

asked 30 mins ago

π times e

273

273

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
2
down vote

Why if I say, "a force of 1 N due east" ?? Or "a displacement of 1m, 30° NOE" ?
Both force and displacement are vector quantities, and both have a magnitude of 1 unit in the above two examples. My question is, can we call these two "unit vectors" ??

No, those are not unit vectors. Let $textbfF=(1 textN)hattextbfx$. (Some people notate $hattextbfx$ as $hattextbfi$.) Then the unit vector in the direction of $textbfF$ is

$fractextbfF = hattextbfx$,

which is not the same as $textbfF$. It has different units, and it is not true that $|textbfF|=|hattextbfx|=1$, since things with incompatible units can never be equal.

share|cite|improve this answer

answered 24 mins ago

Ben Crowell

45.3k3147275

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot for the quick and precise explanation. I have a question, you said the unit vector ("x cap", don't know how to use mathjax yet, sorry) has different units, than the units of F. But aren't unit vectors supposed to be unitless? What are the units of that 'unit vector', btw?
  – Ï€ times e
  15 mins ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

Something to realize is that your vector of length $1 rm N$ only has "unit" length because your chose to measure your force in Newtons. If you chose some other unit like pounds then you would not have $1$ pound of force.

On the other hand, your actual unit vectors are indeed unitless, so they always have a (unitless) magnitude of $1$. This is because unit vectors are defined as the ratio between two things with the same units.

share|cite|improve this answer

answered 18 mins ago

Aaron Stevens

4,3321624

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up vote
0
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If $vecv$ is a vector with a physical unit, then its unit vector is defined as:
$$hatv=fracvecv$$
Where:
$$||vecv||=sqrtsum_i v_i^2$$
where every components $v_i$ has the physical unit. This clearly means that the unit vector is dimensionless.

share|cite|improve this answer

answered 16 mins ago

Kevin De Notariis

1717

New contributor

Kevin De Notariis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

Why if I say, "a force of 1 N due east" ?? Or "a displacement of 1m, 30° NOE" ?
Both force and displacement are vector quantities, and both have a magnitude of 1 unit in the above two examples. My question is, can we call these two "unit vectors" ??

No, those are not unit vectors. Let $textbfF=(1 textN)hattextbfx$. (Some people notate $hattextbfx$ as $hattextbfi$.) Then the unit vector in the direction of $textbfF$ is

$fractextbfF = hattextbfx$,

which is not the same as $textbfF$. It has different units, and it is not true that $|textbfF|=|hattextbfx|=1$, since things with incompatible units can never be equal.

share|cite|improve this answer

answered 24 mins ago

Ben Crowell

45.3k3147275

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot for the quick and precise explanation. I have a question, you said the unit vector ("x cap", don't know how to use mathjax yet, sorry) has different units, than the units of F. But aren't unit vectors supposed to be unitless? What are the units of that 'unit vector', btw?
  – Ï€ times e
  15 mins ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

Why if I say, "a force of 1 N due east" ?? Or "a displacement of 1m, 30° NOE" ?
Both force and displacement are vector quantities, and both have a magnitude of 1 unit in the above two examples. My question is, can we call these two "unit vectors" ??

No, those are not unit vectors. Let $textbfF=(1 textN)hattextbfx$. (Some people notate $hattextbfx$ as $hattextbfi$.) Then the unit vector in the direction of $textbfF$ is

$fractextbfF = hattextbfx$,

which is not the same as $textbfF$. It has different units, and it is not true that $|textbfF|=|hattextbfx|=1$, since things with incompatible units can never be equal.

share|cite|improve this answer

answered 24 mins ago

Ben Crowell

45.3k3147275

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot for the quick and precise explanation. I have a question, you said the unit vector ("x cap", don't know how to use mathjax yet, sorry) has different units, than the units of F. But aren't unit vectors supposed to be unitless? What are the units of that 'unit vector', btw?
  – Ï€ times e
  15 mins ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

Why if I say, "a force of 1 N due east" ?? Or "a displacement of 1m, 30° NOE" ?
Both force and displacement are vector quantities, and both have a magnitude of 1 unit in the above two examples. My question is, can we call these two "unit vectors" ??

No, those are not unit vectors. Let $textbfF=(1 textN)hattextbfx$. (Some people notate $hattextbfx$ as $hattextbfi$.) Then the unit vector in the direction of $textbfF$ is

$fractextbfF = hattextbfx$,

which is not the same as $textbfF$. It has different units, and it is not true that $|textbfF|=|hattextbfx|=1$, since things with incompatible units can never be equal.

share|cite|improve this answer

answered 24 mins ago

Ben Crowell

45.3k3147275

Why if I say, "a force of 1 N due east" ?? Or "a displacement of 1m, 30° NOE" ?
Both force and displacement are vector quantities, and both have a magnitude of 1 unit in the above two examples. My question is, can we call these two "unit vectors" ??

No, those are not unit vectors. Let $textbfF=(1 textN)hattextbfx$. (Some people notate $hattextbfx$ as $hattextbfi$.) Then the unit vector in the direction of $textbfF$ is

$fractextbfF = hattextbfx$,

which is not the same as $textbfF$. It has different units, and it is not true that $|textbfF|=|hattextbfx|=1$, since things with incompatible units can never be equal.

share|cite|improve this answer

answered 24 mins ago

Ben Crowell

45.3k3147275

share|cite|improve this answer

share|cite|improve this answer

answered 24 mins ago

Ben Crowell

45.3k3147275

answered 24 mins ago

Ben Crowell

45.3k3147275

answered 24 mins ago

Ben Crowell

45.3k3147275

45.3k3147275

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot for the quick and precise explanation. I have a question, you said the unit vector ("x cap", don't know how to use mathjax yet, sorry) has different units, than the units of F. But aren't unit vectors supposed to be unitless? What are the units of that 'unit vector', btw?
  – Ï€ times e
  15 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot for the quick and precise explanation. I have a question, you said the unit vector ("x cap", don't know how to use mathjax yet, sorry) has different units, than the units of F. But aren't unit vectors supposed to be unitless? What are the units of that 'unit vector', btw?
  – Ï€ times e
  15 mins ago
  

  
  
  
  

Thanks a lot for the quick and precise explanation. I have a question, you said the unit vector ("x cap", don't know how to use mathjax yet, sorry) has different units, than the units of F. But aren't unit vectors supposed to be unitless? What are the units of that 'unit vector', btw?
– Ï€ times e
15 mins ago

Thanks a lot for the quick and precise explanation. I have a question, you said the unit vector ("x cap", don't know how to use mathjax yet, sorry) has different units, than the units of F. But aren't unit vectors supposed to be unitless? What are the units of that 'unit vector', btw?
– Ï€ times e
15 mins ago

add a comment | 

up vote
0
down vote

Something to realize is that your vector of length $1 rm N$ only has "unit" length because your chose to measure your force in Newtons. If you chose some other unit like pounds then you would not have $1$ pound of force.

On the other hand, your actual unit vectors are indeed unitless, so they always have a (unitless) magnitude of $1$. This is because unit vectors are defined as the ratio between two things with the same units.

share|cite|improve this answer

answered 18 mins ago

Aaron Stevens

4,3321624

add a comment | 

up vote
0
down vote

Something to realize is that your vector of length $1 rm N$ only has "unit" length because your chose to measure your force in Newtons. If you chose some other unit like pounds then you would not have $1$ pound of force.

On the other hand, your actual unit vectors are indeed unitless, so they always have a (unitless) magnitude of $1$. This is because unit vectors are defined as the ratio between two things with the same units.

share|cite|improve this answer

answered 18 mins ago

Aaron Stevens

4,3321624

add a comment | 

up vote
0
down vote

up vote
0
down vote

Something to realize is that your vector of length $1 rm N$ only has "unit" length because your chose to measure your force in Newtons. If you chose some other unit like pounds then you would not have $1$ pound of force.

On the other hand, your actual unit vectors are indeed unitless, so they always have a (unitless) magnitude of $1$. This is because unit vectors are defined as the ratio between two things with the same units.

share|cite|improve this answer

answered 18 mins ago

Aaron Stevens

4,3321624

Something to realize is that your vector of length $1 rm N$ only has "unit" length because your chose to measure your force in Newtons. If you chose some other unit like pounds then you would not have $1$ pound of force.

On the other hand, your actual unit vectors are indeed unitless, so they always have a (unitless) magnitude of $1$. This is because unit vectors are defined as the ratio between two things with the same units.

share|cite|improve this answer

answered 18 mins ago

Aaron Stevens

4,3321624

share|cite|improve this answer

share|cite|improve this answer

answered 18 mins ago

Aaron Stevens

4,3321624

answered 18 mins ago

Aaron Stevens

4,3321624

answered 18 mins ago

Aaron Stevens

4,3321624

4,3321624

add a comment | 

add a comment | 

up vote
0
down vote

If $vecv$ is a vector with a physical unit, then its unit vector is defined as:
$$hatv=fracvecv$$
Where:
$$||vecv||=sqrtsum_i v_i^2$$
where every components $v_i$ has the physical unit. This clearly means that the unit vector is dimensionless.

share|cite|improve this answer

answered 16 mins ago

Kevin De Notariis

1717

New contributor

Kevin De Notariis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
0
down vote

If $vecv$ is a vector with a physical unit, then its unit vector is defined as:
$$hatv=fracvecv$$
Where:
$$||vecv||=sqrtsum_i v_i^2$$
where every components $v_i$ has the physical unit. This clearly means that the unit vector is dimensionless.

share|cite|improve this answer

answered 16 mins ago

Kevin De Notariis

1717

New contributor

Kevin De Notariis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
0
down vote

up vote
0
down vote

If $vecv$ is a vector with a physical unit, then its unit vector is defined as:
$$hatv=fracvecv$$
Where:
$$||vecv||=sqrtsum_i v_i^2$$
where every components $v_i$ has the physical unit. This clearly means that the unit vector is dimensionless.

share|cite|improve this answer

answered 16 mins ago

Kevin De Notariis

1717

New contributor

Kevin De Notariis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

If $vecv$ is a vector with a physical unit, then its unit vector is defined as:
$$hatv=fracvecv$$
Where:
$$||vecv||=sqrtsum_i v_i^2$$
where every components $v_i$ has the physical unit. This clearly means that the unit vector is dimensionless.

share|cite|improve this answer

answered 16 mins ago

Kevin De Notariis

1717

New contributor

Kevin De Notariis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this answer

share|cite|improve this answer

answered 16 mins ago

Kevin De Notariis

1717

New contributor

Kevin De Notariis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 16 mins ago

Kevin De Notariis

1717

answered 16 mins ago

Kevin De Notariis

1717

1717

New contributor

Kevin De Notariis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

Kevin De Notariis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Kevin De Notariis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

add a comment | 

 

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