I'm having some trouble seeing why dot products are said to give scalar values. As a far as I can see, it just gives another vector that is projected onto one of the 2 original vectors. How, then, is the result a scalar quantity. Can someone please explain this to me? Thank you.


No, it doesn't give another vector. It gives the product of the length of one vector by the length of the projection of the other. This is a scalar.

You may have been misled by some figure.

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The dot product is $|A|\,|B|\cos\theta$, not the vector $A'$.

  • $\begingroup$ So it gives the length of A's horizontal length multiplied to the magnitude of B? $\endgroup$ – Ethan Chan Sep 26 '18 at 9:25
  • $\begingroup$ @EthanChan: horizontal is inappropriate here. $\endgroup$ – Yves Daoust Sep 26 '18 at 9:26
  • $\begingroup$ What word would be better for describing Acos(theta) in that case? $\endgroup$ – Ethan Chan Sep 26 '18 at 9:26
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    $\begingroup$ @EthanChan: as in my answer. $\endgroup$ – Yves Daoust Sep 26 '18 at 9:27

$$(1,2)\cdot (3,4) = 1 (3) + 2(4) = 11$$

is a scalar.

I think you are confusing dot product with projection.

Suppose $u$ is a unit vector, we can project $v$ onto $u$ and its length would be $|u\cdot v|$ while the projection would be $(u\cdot v) u$.

  • $\begingroup$ But in all the diagrams I see, the dot product is projected onto one of the vectors. Why is this? (eg: encrypted-tbn0.gstatic.com/…). $\endgroup$ – Ethan Chan Sep 26 '18 at 9:20
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    $\begingroup$ @EthanChan $A\cdot B$ is the length of the projection of $B$ onto $A$. The length of a vector is a scalar. $\endgroup$ – 5xum Sep 26 '18 at 9:22
  • $\begingroup$ @5xum So it treats the second vector as a scalar, and the projects it onto it, to show the length of A⋅B? $\endgroup$ – Ethan Chan Sep 26 '18 at 9:23
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    $\begingroup$ It is the length. half of the magnitude of the cross product , $\frac12|A \times B|$ gives the area of the triangle. $\endgroup$ – Siong Thye Goh Sep 26 '18 at 13:01
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    $\begingroup$ @costrom It's the length. You can remember that because it's largest when the two vectors is colinear (when the area of the triangle is smallest) $\endgroup$ – 5xum Sep 26 '18 at 13:38

The scalar product describes how an amount of one vector goes in the direction of another.

If you moved a set of heavy books on an inclined angle, then there is a horizontal component and a vertical component to the vector descrining the force applied.

The scalar product in this case would then describes the amount of force going in the direction of the displacemen. The work done here, is defined to be the force exerted multiplied by displacement of the books, the force here is defined to be the force in the direction of the displacement.


A dot product, by definition, is a mapping that takes two vectors and returns a scalar.

For example, the standard dot product on $\mathbb R^n$ takes two vectors, $x=(x_1,\dots, x_n)$ and $y=(y_1,\dots, y_n)$, and returns their dot product, $$\langle x,y\rangle = \sum_{i=1}^n x_iy_i$$ which is a real number, and thus, a scalar.


I think it would be easier to start from the basics until we reach to the actual definition of the dot product. Here's my view:

You know the the geometrical definition of multiplication--I'm referring to scalars. The thing I'm referring to is Descartes' geometrical definition of it (just look it up on google images and you'll see what it looks like).

So, if you take the same intuition and apply it on coordinates. Analogously, apply Descartes' definition of multiplication to the two vectors. You may start to notice that the multiplication between the vector looks familiar. It's at this point you should understand that you are now multiplying two lengths together without any regard to the actual coordinates themselves.

This is where projection comes in.

When you project one vector onto another, you have to take into account the unit vector that looms behind descartes' definition, i.e., the reference--really, just think of it as a conversion tool between different units.

Once you get to this point, the projection definition of the dot product is exactly like the cosine definition of it. Once you realize that, the standard component interpretation comes out of the cosine definition.

Also, may not be a tangent, but keep in mind the angle between the vectors is considered only.

So, in a way, the dot product is like the multiplication we already use, but multiplication in a different setting...

So, all in all, from Descartes' intuition of multiplication, you can apply the same logic to coordinates and get a result that seems like it ignores the fact that multiplication is being done on coordinates.

Also, to mention why you get a scalar...I think the dot product is too much like Descartes' description, hence why you get a scalar...


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