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.
[Math] Why is the dot product of two vectors a scalar value
intuitionlinear algebravectors
Related Solutions
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 $$\frac{b}{||b||}$$
(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 $$\frac{a⋅b}{||b||}$$ . because a⋅b=||a|| ||b|| cos(θ). Hence
||a||cos(θ)= $\frac{a⋅b}{||b||}$
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 $$\frac{a⋅b}{||b||}$$
in the direction of $\frac{b}{||b||}$, i.e. $$\frac{a.b}{||b||} \frac{b}{||b||} $$
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 .
Well, in a general way the dot product is an operation defined between two objects that satisfies some properties (check out this page on wikipedia for example: http://en.wikipedia.org/wiki/Dot_product), for a 3D vector it turns out that if you define the dot product as: $$ \mathbf{A} \cdot \mathbf{B} = a_xb_x + a_yb_y + a_zb_z $$ these properties are fullfilled. Note that $ a_xb_x + a_yb_y + a_zb_z $ is ascalar and not a vector anymore, because you are summing up products between scalar quantities (i.e. the components of your vectors).
Now consider the dot product between the vector $ \mathbf{A}= a_x \mathbf{i} + a_y \mathbf{j} + a_z \mathbf{k} $ and the vector $ \mathbf{i} $, using the formula above, you get: $$ \mathbf{A} \cdot \mathbf{i} = a_x = \mid \mathbf{A} \mid cos\theta $$ where $\theta$ is the angle between $ \mathbf{A} $ and $ \mathbf{i} $, i.e. $ \mathbf{A} \cdot \mathbf{i}$ gives you the projection of $ \mathbf{A} $ on $\mathbf{i} $ (try to sketch the vectors in 2D). So, in a general way, you have that $ \mathbf{A} \cdot \mathbf{B} $ gives you the projection of $\mathbf{A} $ on $\mathbf{B}$, i.e $ \mid \mathbf{A} \mid \mid \mathbf{B} \mid cos\theta $.
Best Answer
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 $A'$.