Why does the dot product represent the length of the projection of vector a onto vector b only in the case where b is of unit length? I'm trying to understand the dot product and projections at a deeper, intuitive level, and I'm struggling with this question. Why does the length of the projection, i.e. "shadow" of one vector on the other, have anything to do with the length of that other vector?
[Math] Why does dot product represent length of projection onto *unit* vector
projectionvector-spacesvectors
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 .
This is really simple, Dunja. First, normalize the vector $b$ by dividing each of its coordinates by the length of $b$ (which is $\sqrt 5$, here). By this, you shrink the vector to length one. Then put a minus in front of this normalized vector to make it opposite direction. Done.
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Best Answer
This really boils down to the long-standing question of "what is the best intuitive understanding of the dot product" which, admittedly, is no dobut a mystery to many as it very often comes up.
What I would say with regard to that question - and which thus also answers the one in your title - is that the dot product is a "mathematically nicefied" version of the projection operation: namely, the intuition is that $\mathbf{a} \cdot \hat{\mathbf{u}}$, for a unit vector $\hat{\mathbf{u}}$, is indeed that it is the component of $\mathbf{a}$ in the direction of $\hat{\mathbf{u}}$, and we could start the construction of the dot product from there.
The question, of course, arises as to what to do if $\mathbf{u}$ is not a unit vector. One option would be just to say that it doesn't matter: just give the projection (component) in the direction anyways. However, that leads to some problems. For one, let's consider commutativity. Ideally, we'd like it if our product were commutative:
$$\mathbf{a} \cdot \hat{\mathbf{u}} = \hat{\mathbf{u}} \cdot \mathbf{a}$$
however, if we are going down the line of description I just gave and just say it gives the projection to the second operand as a pure directional indicator, then we have an asymmetric, non-commutative operation.
Yet fortunately, it turns out that, very nicely, these problems go away if we decide that
$$\mathbf{a} \cdot \mathbf{u}$$
when $\mathbf{u}$ is not a unit vector, is to be the projection scaled by the length of the vector of projection, here, $\mathbf{u}$. Given that the projection already scales by the length of the vector being projected, effectively, this causes both vectors to be treated "on equal footing", so to speak, and the operation now commutes. Moreover, it becomes even bilinear, which is even better, and thus makes it a tensor (perhaps even the canonical example!), which allows all sorts of generalizations in more sophisticated mathematical usage, including the ability to define Riemannian manifolds and talk about curved, non-Euclidean spaces, and also the ability to provide infinite-dimensional spaces with useful topological structures permitting things like infinite linear combinations, as we get when we go down the inner product space sequence.
In short, the dot product represents projection onto a unit vector because at that point, the scaling coefficient we introduce to make it mathematically "nice" and "even-handed" equals $1$ when the vector of projection is such.