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

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    $\begingroup$ My answer here is not explicit but describes exactly the proof needed.math.stackexchange.com/q/731939/31475 $\endgroup$
    – Emily
    Jan 18, 2015 at 15:02
  • $\begingroup$ What definition of the dot product would you like to start with? Are we specifically talking about 2 or 3 dimensional space? $\endgroup$ Jan 18, 2015 at 15:04
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    $\begingroup$ The definition that says that $\mathbf{a}\cdot\mathbf{b}=\|\mathbf{a}\|\|\mathbf{b}\|\cos\alpha$, where $\alpha$ is the angle between $\mathbf{a}$ and $\mathbf{b}$. $\endgroup$
    – user197402
    Jan 18, 2015 at 15:05
  • $\begingroup$ "to show the algebraic formula for the dot product, one needs to use the distributive property in the geometric definition." - why? $\endgroup$
    – GFauxPas
    Jan 18, 2015 at 15:06
  • $\begingroup$ Well, at least wikipedia uses it: en.wikipedia.org/wiki/… $\endgroup$
    – user197402
    Jan 18, 2015 at 15:07

2 Answers 2


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

$\hspace{4.5 cm}$(dot product diagram)

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.${}$

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    $\begingroup$ How do we know this is true $ \|B_A + C_A\| = \|B_A\| + \|C_A\| $ . It seems obvious from picture , but how can i be sure ? $\endgroup$
    – Milan
    Sep 4, 2019 at 16:41
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    $\begingroup$ @Milan In a more rigorous proof, one would separately consider the case where the components $B_A, C_A$ point in opposite directions. In this context however, it suffices to note that the length of a combination of two segments placed next to each other along the same line is the sum of the lengths of the segments. $\endgroup$ Sep 5, 2019 at 7:02
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    $\begingroup$ How does the distributivity of the dot-product follow from the 2-dimensional case? A,B and C are 3 vectors, they don't need to lie in one 2-dimensional plane. $\endgroup$
    – SomeName
    Nov 23, 2019 at 10:03
  • $\begingroup$ @SomeName I was only talking about the dot product in the 2-dimensional case since the asker didn’t specify what exactly they wanted; I do not claim that the properties in the 3D case follow. See the other answer for the 3D case. $\endgroup$ Nov 23, 2019 at 20:00
  • $\begingroup$ @SomeName, for geometric vectors $b,c$ in higher dimensions, resolve each of them (and their sum) into orthogonal components with respect to the direction $a.$ Then by definition the components normal to the direction of $a$ will vanish. So you only deal with components parallel to the direction of $a,$ which reduces the case to the one considered above. $\endgroup$
    – Allawonder
    Jan 16, 2022 at 10:30

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.

(fig : vectors within cylinder)

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.

  • $\begingroup$ Im impressed by the usage of the simple yet so easy to be missed fact that the projection of a vector B onto another vector A remains unaffected under cylindrical rotation of B around vector A thumbs up! and of course thanks! $\endgroup$
    – Aditya
    Apr 3, 2021 at 20:26

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