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What is the difference between the dot product and the scalar projection?

Is there any distinction between these products: scalar, dot, inner?Proof of equivalance beginning with the geometrical definition of the scalar (dot) product and the algebraic definition of the dot product.What separates the dot product from the scalar projection?What is the difference between and projection and a reflection, in vector transformation?Why is the dot product defined as a scalar and the cross product a normal vector?relationship between scalar product and tensor productWhy are the two dot product definitions equal?Taking the Dot product of two normalized vectors (unit vectors) vs the dot product of two non-normalized vectorsDifference between dot product and cross productWhy is the dot product of two vectors a scalar value?

4

4

$begingroup$

I don't understand the difference between the dot product of two vectors and the scalar projection of a vector onto another one.

To me it looks like they are both (geometrically) the length of the vector projection. I am wrong since their formulas are different, so can anyone explain why?

vectors projection

share|cite|improve this question

edited Mar 10 at 16:42

nbro

2,43363374

asked Dec 12 '14 at 4:46

John TrentnorJohn Trentnor

2313

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean by "the scalar projection"? Evidently, you don't use that phrase to mean the same thing as "the vector projection", so what do you mean by that term?
  $endgroup$
  – Gerry Myerson
  Dec 12 '14 at 5:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @GerryMyerson: I apologize if this is not a universal term. In my grade 12 class, we defined the scalar projection as $operatornamecomp_vecavecb:=dfracvecacdot vecb$
  $endgroup$
  – John Trentnor
  Dec 12 '14 at 5:27
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  We also learned that the vector projection is $operatornameproj_vecavecb:=operatornamecomp_vecavecbcdot dfracveca$. That is, the vector projection is the scalar projection multiplied by a unit vector in the direction of $a$.
  $endgroup$
  – John Trentnor
  Dec 12 '14 at 5:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then the dot product doesn't give the length of the vector projection, since the former is $acdot b$ while the latter is $acdot b/|a|$. I don't see what there is to explain.
  $endgroup$
  – Gerry Myerson
  Dec 12 '14 at 5:34
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well, yes, the two formulas are identical if $|a|=1$.
  $endgroup$
  – Gerry Myerson
  Dec 12 '14 at 5:41
  

  
  
  
  

 | 
show 3 more comments

4

4

$begingroup$

I don't understand the difference between the dot product of two vectors and the scalar projection of a vector onto another one.

To me it looks like they are both (geometrically) the length of the vector projection. I am wrong since their formulas are different, so can anyone explain why?

vectors projection

share|cite|improve this question

edited Mar 10 at 16:42

nbro

2,43363374

asked Dec 12 '14 at 4:46

John TrentnorJohn Trentnor

2313

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean by "the scalar projection"? Evidently, you don't use that phrase to mean the same thing as "the vector projection", so what do you mean by that term?
  $endgroup$
  – Gerry Myerson
  Dec 12 '14 at 5:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @GerryMyerson: I apologize if this is not a universal term. In my grade 12 class, we defined the scalar projection as $operatornamecomp_vecavecb:=dfracvecacdot vecb$
  $endgroup$
  – John Trentnor
  Dec 12 '14 at 5:27
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  We also learned that the vector projection is $operatornameproj_vecavecb:=operatornamecomp_vecavecbcdot dfracveca$. That is, the vector projection is the scalar projection multiplied by a unit vector in the direction of $a$.
  $endgroup$
  – John Trentnor
  Dec 12 '14 at 5:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then the dot product doesn't give the length of the vector projection, since the former is $acdot b$ while the latter is $acdot b/|a|$. I don't see what there is to explain.
  $endgroup$
  – Gerry Myerson
  Dec 12 '14 at 5:34
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well, yes, the two formulas are identical if $|a|=1$.
  $endgroup$
  – Gerry Myerson
  Dec 12 '14 at 5:41
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

I don't understand the difference between the dot product of two vectors and the scalar projection of a vector onto another one.

To me it looks like they are both (geometrically) the length of the vector projection. I am wrong since their formulas are different, so can anyone explain why?

vectors projection

share|cite|improve this question

edited Mar 10 at 16:42

nbro

2,43363374

asked Dec 12 '14 at 4:46

John TrentnorJohn Trentnor

2313

$endgroup$

share|cite|improve this question

edited Mar 10 at 16:42

nbro

2,43363374

asked Dec 12 '14 at 4:46

John TrentnorJohn Trentnor

2313

share|cite|improve this question

edited Mar 10 at 16:42

nbro

2,43363374

asked Dec 12 '14 at 4:46

John TrentnorJohn Trentnor

2313

edited Mar 10 at 16:42

nbro

2,43363374

edited Mar 10 at 16:42

nbro

2,43363374

asked Dec 12 '14 at 4:46

John TrentnorJohn Trentnor

2313

asked Dec 12 '14 at 4:46

John TrentnorJohn Trentnor

2313

2 Answers
2

active

oldest

votes

1

$begingroup$

The output of a dot product is a real number. The output of a projection is a vector. If you look at the formulas, the scalar projection does not depend on the length of the vector you are projecting onto.

According to Wikipeda, the scalar projection does not depend on the length of the vector being projected on. If you double the length of the second vector in the dot product, the dot product doubles.

share|cite|improve this answer

edited Dec 12 '14 at 5:23

answered Dec 12 '14 at 5:00

Ross MillikanRoss Millikan

299k24200374

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It seems to me from the wording of the question that OP is making some distinction between "scalar projection" and "vector projection", though I'm not sure what exactly the first one is supposed to be.
  $endgroup$
  – Gerry Myerson
  Dec 12 '14 at 5:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GerryMyerson: I found a Wikipedia page on scalar projection It is a vector in the direction of the second argument with the magnitude of the dot product between the first argument and the unit vector in the direction of the second argument. I have no clue why this would be of interest.
  $endgroup$
  – Ross Millikan
  Dec 12 '14 at 5:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. OP has just provided the definition his class is using, and it's a scalar, not a vector.
  $endgroup$
  – Gerry Myerson
  Dec 12 '14 at 5:30
  

  
  
  
  

add a comment | 

2

$begingroup$

Projection

The component of a vector a that is in the same direction as of vector b (Hence projection is a vector) Length of the projection does not depend on the length(magnitude) of b. See the image below

Projection has two parts:

(i) The direction where you're projecting onto. That's the unit vector in direction of b, which is computed by dividing b by the length of b. That is $$fracb$$

(ii) The component of a
in the direction of b. That is, the "shadow" or image of a when you project it onto b. This is computed by $$fraca⋅b$$ . because a⋅b=||a|| ||b|| cos(θ). Hence

||a||cos(θ)= $fraca⋅b$

and that gives you (as in the triangle figure), the length of a's projection on the direction of b

Putting it together, the projection of a onto b is a vector of length $$fraca⋅b$$

in the direction of $fracb$, i.e.
$$fraca.b fracb $$

Dot Product

It's simply the projection of one vector onto the other multiplied by the magnitude of other vector. The dot product tells you what amount of one vector goes in the direction of another (Thus its a scalar ) and hence do not have any direction .

a.b= ||a|| ||b|| cos(θ). Alternatively if a=(x1,y1) and b=(x2,y2) (Position vectors) the dot product is x1.x2+y1.y2 .

share|cite|improve this answer

edited Mar 10 at 16:47

nbro

2,43363374

answered Jan 11 '17 at 16:05

Vijay P RVijay P R

1214

$endgroup$

add a comment | 

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1

$begingroup$

The output of a dot product is a real number. The output of a projection is a vector. If you look at the formulas, the scalar projection does not depend on the length of the vector you are projecting onto.

According to Wikipeda, the scalar projection does not depend on the length of the vector being projected on. If you double the length of the second vector in the dot product, the dot product doubles.

share|cite|improve this answer

edited Dec 12 '14 at 5:23

answered Dec 12 '14 at 5:00

Ross MillikanRoss Millikan

299k24200374

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It seems to me from the wording of the question that OP is making some distinction between "scalar projection" and "vector projection", though I'm not sure what exactly the first one is supposed to be.
  $endgroup$
  – Gerry Myerson
  Dec 12 '14 at 5:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GerryMyerson: I found a Wikipedia page on scalar projection It is a vector in the direction of the second argument with the magnitude of the dot product between the first argument and the unit vector in the direction of the second argument. I have no clue why this would be of interest.
  $endgroup$
  – Ross Millikan
  Dec 12 '14 at 5:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. OP has just provided the definition his class is using, and it's a scalar, not a vector.
  $endgroup$
  – Gerry Myerson
  Dec 12 '14 at 5:30
  

  
  
  
  

add a comment | 

1

$begingroup$

The output of a dot product is a real number. The output of a projection is a vector. If you look at the formulas, the scalar projection does not depend on the length of the vector you are projecting onto.

According to Wikipeda, the scalar projection does not depend on the length of the vector being projected on. If you double the length of the second vector in the dot product, the dot product doubles.

share|cite|improve this answer

edited Dec 12 '14 at 5:23

answered Dec 12 '14 at 5:00

Ross MillikanRoss Millikan

299k24200374

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It seems to me from the wording of the question that OP is making some distinction between "scalar projection" and "vector projection", though I'm not sure what exactly the first one is supposed to be.
  $endgroup$
  – Gerry Myerson
  Dec 12 '14 at 5:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GerryMyerson: I found a Wikipedia page on scalar projection It is a vector in the direction of the second argument with the magnitude of the dot product between the first argument and the unit vector in the direction of the second argument. I have no clue why this would be of interest.
  $endgroup$
  – Ross Millikan
  Dec 12 '14 at 5:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. OP has just provided the definition his class is using, and it's a scalar, not a vector.
  $endgroup$
  – Gerry Myerson
  Dec 12 '14 at 5:30
  

  
  
  
  

add a comment | 

$begingroup$

The output of a dot product is a real number. The output of a projection is a vector. If you look at the formulas, the scalar projection does not depend on the length of the vector you are projecting onto.

According to Wikipeda, the scalar projection does not depend on the length of the vector being projected on. If you double the length of the second vector in the dot product, the dot product doubles.

share|cite|improve this answer

edited Dec 12 '14 at 5:23

answered Dec 12 '14 at 5:00

Ross MillikanRoss Millikan

299k24200374

$endgroup$

share|cite|improve this answer

edited Dec 12 '14 at 5:23

answered Dec 12 '14 at 5:00

Ross MillikanRoss Millikan

299k24200374

answered Dec 12 '14 at 5:00

Ross MillikanRoss Millikan

299k24200374

answered Dec 12 '14 at 5:00

Ross MillikanRoss Millikan

299k24200374

2

$begingroup$

Projection

The component of a vector a that is in the same direction as of vector b (Hence projection is a vector) Length of the projection does not depend on the length(magnitude) of b. See the image below

Projection has two parts:

(i) The direction where you're projecting onto. That's the unit vector in direction of b, which is computed by dividing b by the length of b. That is $$fracb$$

(ii) The component of a
in the direction of b. That is, the "shadow" or image of a when you project it onto b. This is computed by $$fraca⋅b$$ . because a⋅b=||a|| ||b|| cos(θ). Hence

||a||cos(θ)= $fraca⋅b$

and that gives you (as in the triangle figure), the length of a's projection on the direction of b

Putting it together, the projection of a onto b is a vector of length $$fraca⋅b$$

in the direction of $fracb$, i.e.
$$fraca.b fracb $$

Dot Product

It's simply the projection of one vector onto the other multiplied by the magnitude of other vector. The dot product tells you what amount of one vector goes in the direction of another (Thus its a scalar ) and hence do not have any direction .

a.b= ||a|| ||b|| cos(θ). Alternatively if a=(x1,y1) and b=(x2,y2) (Position vectors) the dot product is x1.x2+y1.y2 .

share|cite|improve this answer

edited Mar 10 at 16:47

nbro

2,43363374

answered Jan 11 '17 at 16:05

Vijay P RVijay P R

1214

$endgroup$

add a comment | 

2

$begingroup$

Projection

The component of a vector a that is in the same direction as of vector b (Hence projection is a vector) Length of the projection does not depend on the length(magnitude) of b. See the image below

Projection has two parts:

(i) The direction where you're projecting onto. That's the unit vector in direction of b, which is computed by dividing b by the length of b. That is $$fracb$$

(ii) The component of a
in the direction of b. That is, the "shadow" or image of a when you project it onto b. This is computed by $$fraca⋅b$$ . because a⋅b=||a|| ||b|| cos(θ). Hence

||a||cos(θ)= $fraca⋅b$

and that gives you (as in the triangle figure), the length of a's projection on the direction of b

Putting it together, the projection of a onto b is a vector of length $$fraca⋅b$$

in the direction of $fracb$, i.e.
$$fraca.b fracb $$

Dot Product

It's simply the projection of one vector onto the other multiplied by the magnitude of other vector. The dot product tells you what amount of one vector goes in the direction of another (Thus its a scalar ) and hence do not have any direction .

a.b= ||a|| ||b|| cos(θ). Alternatively if a=(x1,y1) and b=(x2,y2) (Position vectors) the dot product is x1.x2+y1.y2 .

share|cite|improve this answer

edited Mar 10 at 16:47

nbro

2,43363374

answered Jan 11 '17 at 16:05

Vijay P RVijay P R

1214

$endgroup$

add a comment | 

$begingroup$

Projection

The component of a vector a that is in the same direction as of vector b (Hence projection is a vector) Length of the projection does not depend on the length(magnitude) of b. See the image below

Projection has two parts:

(i) The direction where you're projecting onto. That's the unit vector in direction of b, which is computed by dividing b by the length of b. That is $$fracb$$

(ii) The component of a
in the direction of b. That is, the "shadow" or image of a when you project it onto b. This is computed by $$fraca⋅b$$ . because a⋅b=||a|| ||b|| cos(θ). Hence

||a||cos(θ)= $fraca⋅b$

and that gives you (as in the triangle figure), the length of a's projection on the direction of b

Putting it together, the projection of a onto b is a vector of length $$fraca⋅b$$

in the direction of $fracb$, i.e.
$$fraca.b fracb $$

Dot Product

It's simply the projection of one vector onto the other multiplied by the magnitude of other vector. The dot product tells you what amount of one vector goes in the direction of another (Thus its a scalar ) and hence do not have any direction .

a.b= ||a|| ||b|| cos(θ). Alternatively if a=(x1,y1) and b=(x2,y2) (Position vectors) the dot product is x1.x2+y1.y2 .

share|cite|improve this answer

edited Mar 10 at 16:47

nbro

2,43363374

answered Jan 11 '17 at 16:05

Vijay P RVijay P R

1214

$endgroup$

share|cite|improve this answer

edited Mar 10 at 16:47

nbro

2,43363374

answered Jan 11 '17 at 16:05

Vijay P RVijay P R

1214

edited Mar 10 at 16:47

nbro

2,43363374

edited Mar 10 at 16:47

nbro

2,43363374

answered Jan 11 '17 at 16:05

Vijay P RVijay P R

1214

answered Jan 11 '17 at 16:05

Vijay P RVijay P R

1214