Proving that the dot product is distributive?

I know that one can prove that the dot product, as defined "algebraically", is distributive. However, to show the algebraic formula for the dot product, one needs to use the distributive property in the geometric definition. How would one show, geometrically, that for Euclidean vectors $\mathbf{a},\mathbf{b},\mathbf{c}$, $$\mathbf{a}\cdot\mathbf{b}+\mathbf{a}\cdot\mathbf{c}=\mathbf{a}\cdot(\mathbf{b}+\mathbf{c})?$$

Solution 1:

In order to prove that the geometric definition of the (2-dimensional) dot product is distributive, we use the following diagram:

$\hspace{4.5 cm}$

Note that (whenever $A$ is non-zero) $$ \|B_A\| = \frac{B \cdot A}{\|A\|}\\ \|C_A\| = \frac{C \cdot A}{\|A\|}\\ \|B_A + C_A\| = \frac{(B + C) \cdot A}{\|A\|} $$ It is clear from the diagram that $$ \frac{(B + C) \cdot A}{\|A\|} = \|B_A + C_A\| = \|B_A\| + \|C_A\| = \frac{B \cdot A}{\|A\|}+ \frac{C \cdot A}{\|A\|} $$ the distributivity of the dot-product follows.${}$

Solution 2:

This proof is for the general case that considers non-coplanar vectors:

It suffices to prove that the sum of the individual projections of vectors b and c in the direction of vector a is equal to the projection of the vector sum b+c in the direction of a.

As shown in the figure below, the non-coplanar vectors under consideration can be brought to the following arrangement within a large enough cylinder "S" that runs parallel to the vector a. I have colored the vectors differently just to indicate that they need not lie on the same plane.

Observe that the projection of vector b in the direction of a is exactly the the distance (call it XY) between the two blue "cross-circles" X (that contains the tail of b) and Y (that contains the head of b). This can be seen clearly when the vector b is translated in space so that its tail sits on the axis of the cylinder. Here "cross-circles" means the circles parallel to the base of the cylinder and perpendicular to its axis.

Similarly, YZ is the projection of c in the direction of a because cross-circles Y and Z contain the tail and head of c respectively. Also, XZ is the projection of vector sum b+c in the direction of a.

It is evident that XY + YZ = XZ, which was what we wanted to show.