Why is the dot product of two vectors a scalar value? 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. 
 A: The scalar product describes how an amount of one vector goes in the direction of another.
For example, if you move a set of heavy books on an inclined angle, then there is a horizontal component and a vertical component to the vector describing the force applied.
The scalar product in this case would then describe the amount of force going in the direction of the displacement. 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: $$(1,2)\cdot (3,4) = 1 (3) + 2(4) = 11$$
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 $(u\cdot v) u$.
A: 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.

The dot product is $|A|\,|B|\cos\theta$, not the vector $A'$.
A: 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.
A: 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...
A: The dot product of two vectors can be considered to be nothing more than a special, simple case of matrix multiplication. This is not immediately obvious because $u\cdot v$ does not look like the multiplication of two matrices, even if we understand that $u=(u_1,u_2)$ and $v=(v_1,v_2)$.
But if we format $u$ and $v$ as a 1×2 and 2×1 matrix respectively (either is a perfectly valid representation of a two-element vector), then we have
$u\cdot v = (u_1,u_2)\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = (u_1v_1+u_2v_2)$.
This is a 1×1 matrix, which is exactly what a scalar is. Thus, a dot product of vectors produces a scalar.
