I came upon this proof of equivalence between the geometric and algebraic definitions of the dot product. I do not understand why this person multiplies the two vectors together, that's not the dot product. The dot product is the product of the component of one vector going in the same direction as the other, and the other one itself. Why would this person multiply the two actual vectors together to get the algebraic definition when this I not the dot product?

Here is the proof:

$$\vec{a} \cdot \vec{b} = \left|\vec{a} \right|\cdot \left| \vec{b} \right| \cos \theta$$

Decomposing the vector into its unit vectors (assuming 2D for simplicity):

$$\begin{eqnarray} a &=& a_x\hat{i} + a_y \hat{j}\\ b &=& b_x \hat{i} + b_y \hat{j} \end{eqnarray}$$

Multiplying the two vectors (loosely using the term 'multiplication') i.e., taking their dot product literally:

$$\vec{a} \cdot \vec{b} = a_x \hat{i} \cdot b_x \hat{i} + a_x \hat{i} \cdot b_y \hat{j} + a_y \hat{j} \cdot b_x \hat{i} + a_y \hat{j} \cdot b_y \hat{j} = a_x b_x + a_y b_y, $$

since $\hat{i} \cdot \hat{i} = 1$ and $\hat{i} \cdot \hat{j} = 0$ (i.e., angle between $\hat{i}$ and $\hat{j}$ is $90^\circ$) and that $a_x,a_y,b_x,b_y$ are scalars. Q.E.D.

  • $\begingroup$ I don't understand why this is a proof of geometric and algebraic equivalence. Algebraically, you argued $$\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y.$$ Geometrically, you argued that $$\vec{a} \cdot \vec{b} = \left| \vec{a} \right| \cdot \left| \vec{b} \right| \cos \theta = \sqrt{a_x^2 + a_y^2} \sqrt{b_x^2 + b_y^2} \cos \theta. $$ Why are they the same??? $\endgroup$
    – gt6989b
    Commented May 1, 2014 at 21:03
  • $\begingroup$ There should be an equivalence between the two. How do you make the connection? On one hand the dot product is defined to be a⃗ ⋅b⃗ =∣∣a⃗ ∣∣⋅∣∣b⃗ ∣∣cosθ and on the other it is a⃗ ⋅b⃗ =axi^⋅bxi^+axi^⋅byj^+ayj^⋅bxi^+ayj^⋅byj^=axbx+ayby. $\endgroup$ Commented May 1, 2014 at 21:08
  • $\begingroup$ I understand completely the geometric interpretation but how do you get this other way of calculating the dot product from the geometric interpretation? $\endgroup$ Commented May 1, 2014 at 21:09
  • $\begingroup$ Well, if $\vec{a} = a_x \hat{i} + a_y \hat{j}$, by definition, $$\left| \vec{a} \right| = \vec{a} \cdot \vec{a} = a_x^2 + a_y^2.$$ $\endgroup$
    – gt6989b
    Commented May 1, 2014 at 21:11
  • $\begingroup$ But how does∣∣a⃗ ∣∣⋅∣∣b⃗ ∣∣cosθ = axbx+ayby? That's really my question. $\endgroup$ Commented May 1, 2014 at 21:15

2 Answers 2


Consider the following, doing the $2$-D case, which can be generalized to $n$-D.

Vector $A$ with coordinates $(x_A,y_A)$

Vector $B$ with coordinates $(x_B,y_B)$

Vectors A and B

The dot product of those two vectors is : \begin{align*} A\cdot B &= AB\cos(\theta) \\ &= AB\cos(\alpha−\beta)\qquad(\mathrm{since}\,\theta=\alpha-\beta) \\ &= AB(\cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)) \\ &= AB\cos(\alpha)\cos(\beta) + AB\sin(\alpha)\sin(\beta) \\ &= A\cos(\alpha)B\cos(\beta) + A\sin(\alpha)B\sin(\beta) \\ &= x_Ax_B + y_Ay_B \end{align*}


The proof assumes that the dot product is linear, which is not trivial to prove without the standard algebraic definition.

The more straightforward proof would be as follows: Create a triangle with the two vectors $a$ and $b$ so that the third side is $a-b$. Define the dot product as $a\cdot b=a_1b_1+a_2b_2$. Then note that $$||x||^2=x_1^2+x_2^2=(x_1,x_2)\cdot (x_1,x_2)$$ so the magnitude squared of a vector equals the vector dotted with itself. Then by the law of cosines, letting $\theta$ denote the angle between $a$ and $b$ and recalling that $a-b$ is the side opposite $\theta$ we get $$||a-b||^2=||a||^2+||b||^2-2||a||\,||b||\cos\theta$$ Using the magnitude squared/dot product relationship above gives $$(a-b)\cdot (a-b)=a\cdot a+b\cdot b-2||a||\,||b||\cos\theta$$ Clearly the dot product is linear and symmetric by our algebraic definition, so the left side can be re-written as $$a\cdot a+b\cdot b-2a\cdot b=a\cdot a+b\cdot b-2||a||\,||b||\cos\theta$$ from which it follows that $$a\cdot b=||a||\,||b||\cos\theta$$

  • $\begingroup$ nice and elegant $\endgroup$
    – gt6989b
    Commented May 1, 2014 at 21:39
  • $\begingroup$ I know this but how would you go from the geometric to the algebraic? $\endgroup$ Commented May 1, 2014 at 21:46
  • $\begingroup$ Starting with the algebraic seems strange because it is fundamentally a geometric thing. $\endgroup$ Commented May 1, 2014 at 21:47
  • $\begingroup$ @KingSquirrel I don't know if I would say that... it's equally algebraic and geometric, especially in higher dimensions. As to your first question, as I mentioned you would have to prove that it is a linear operator, meaning that $a\cdot (k_1x+k_2y)=k_1(a\cdot x)+k_2(a\cdot y)$ and so on, and also show it commutes. I have no idea how that would be done from the 'geometric' definition. $\endgroup$
    – user142299
    Commented May 1, 2014 at 21:49

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