Need proof for a property in Analytic Geometry

Question

When I was taught Vector Calculus, we assumed the following properties of the dot product (without proof):

$$\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = \vec{\mathbf{b}} \cdot \vec{\mathbf{a}}$$

$$\vec{\mathbf{a}} \cdot (\vec{\mathbf{b}} + \vec{\mathbf{c}}) = \vec{\mathbf{a}} \cdot \vec{\mathbf{b}} + \vec{\mathbf{a}} \cdot \vec{\mathbf{c}}$$

And then derived the following component form of the Dot Product:

$$\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = a_1 b_1 + a_2 b_2 + a_3 b_3$$

In G. E. Hay's Vector and Tensor Analysis, this is done in exact reverse order. First, he derives $\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = a_1 b_1 + a_2 b_2 + a_3 b_3$ and then proves both the commutativity and the distributivity over vector addition properties of the Dot Product.

For deriving $\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = a_1 b_1 + a_2 b_2 + a_3 b_3$, he uses the formula that the cosine of the acute angle between two lines is the sum of the products of corresponding direction cosines of the two lines. He writes nearby: by a formula of analytic geometry.

Now, I was taught this formula in analytic geometry. But, the proof used the dot product in component form directly.

So, I want a proof for "cosine of the acute angle between two lines is the sum of the products of corresponding direction cosines of the two lines" that does not involve $\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = a_1 b_1 + a_2 b_2 + a_3 b_3$ because without such a proof the proof in G. E. Hay becomes circular reasoning.

Answer 1

Let a, b, and c be any three vectors. We can always resolve a and b into components in the direction of c whose unit vector is u. The diagram below shows a.u + b.u $= ($ a + b $)$ u.

If we multiply both sides by the scalar $\lambda$; where c $ = \lambda$ u, we have

a.c + b.c = (a + b)c.

Added:

In the revised picture, we have a . u = |a| . |u| $\cos \gamma$ = |a| $\cos \gamma$ = OH.

Similarly, b . u = … = |b| $\cos \alpha$ = AK = HM;

And, (a+b) . u = |a+b| . $\cos \beta$ = OM = OH + HM

∴ (a + b) . u =a . u + b . u