Direct understanding of vector projection

Question

I'm aware that we can project an arbitrary vector $v$ onto a unit vector $\hat{\mathbf{k}}$ by $(\mathbf{v} \cdot \hat{\mathbf{k}})\hat{\mathbf{k}}$

But why is this true?

I would imagine that such a clean result has a correspondingly clean derivation. But I can't see it.

The Wikipedia page on vector projection http://en.wikipedia.org/wiki/Vector_projection makes use of the dot product rule in order to prove it. A cancellation occurs which removes $cos(\theta)$.

To illustrate, if we work in 2D, we take a vector $\mathbf{k}$ on the unit circle, and then rotate it 90° counterclockwise to get $\mathbf{k}'$.

Now given an arbitrary unit vector $\mathbf{v}$, we can write $\mathbf{v}$ in terms of $\mathbf{k}$ and $\mathbf{k'}$: $\mathbf{v} = \cos(\theta) \mathbf{k} + \sin(\theta) \mathbf{k}'$ where $\theta$ is the angle between $\mathbf{v}$ and $\mathbf{k}$.

So the projection in the $\mathbf{k}$ direction is $\cos(\theta)$

And if we are allowed the dot product rule, then for unit $\mathbf{k}$ and $\mathbf{v}$, we have $\mathbf{k} \cdot \mathbf{v} = \cos(\theta)$

So the projection in the $\mathbf{k}$ direction is $k \cdot v$

But then this motivates the question: Why is the dot product rule true?

The Wikipedia page on the dot product rule surprisingly defines $\cos(\theta)$ as $\mathbf{k \cdot v}$

This doesn't make any sense to me. $\cos(\theta)$ is defined far more clearly as projection of circular motion, i.e. defined geometrically.

It must be possible to set up some geometrical construction, whereby the dot-product falls out as a consequence.

But how to go about this?

EDIT: Link: How to understand dot product is the angle's cosine?

Answer 1

I finally got a derivation from scratch:

This addressed a hole in my understanding -- I had never enquired as to what motivates the definition of the dot operator.

As far as I can see, we just go ahead and define it, then show that it behaves nicely with algebraic manipulations. Then we show, using Pythagoras, that u, v perpendicular gives: $\mathbf u \cdot \mathbf v = 0$.

Finally we use all previous results to get the projection result.

I'm not typing it up at the moment as RSI is paining me; if somebody would care to do it then I am grateful! (what would be a good way of making the diagrams?)

Answer 2

This does have a clean derivation, as long as you accept that orthogonality implies zero dot product. Certainly the (orthogonal) projection of $\mathbf v$ onto $\mathbf k$ should be a multiple $\lambda\mathbf k$ of $\mathbf k$. This projection should have the property that $\mathbf v - \lambda\mathbf k$ is orthogonal to $\mathbf k$ (draw a picture!), so $$ (\mathbf v - \lambda\mathbf k) \cdot \mathbf k = 0. $$ Expand this out and solve for $\lambda$ to find that $\lambda = \mathbf v \cdot \mathbf k$.