I am about to enter the class Engineering Physics II. Alas, much of my mastery of vector manipulation is predicated on something I don't understand that must be taken as an assumption for me to proceed; that's killing me. With that prelude, I can't seem to associate the sensible geometric definition of the dot product: $$AB\cos{\theta}$$ with the (seemingly) painfully arbitrary algebraic definition of the dot product, derivable from: $$(A_x\hat{i} + A_y\hat{j}+A_z\hat{z}) \bullet (B_x\hat{i} + B_y\hat{j}+B_z\hat{z})$$

I simply can't wrap my mind around how the geometry translates to this is any terms. (I can come to the conclusion that this statement simplifies to $A_xB_x+A_yB_y+A_zB_z=\vec{A}\bullet\vec{B}$, as $\hat{i}\hat{i}=0$, among others. But I CANNOT come to terms with this or any other more"top-level" statement.) I can neither prove this algebraically nor think, again, of a geometric case which can be constructed to demonstrate this. To prove that I tried, let me post some of my mathematical convulsions:

First, try to put the equations in more similar terms: $$\sqrt{A_x^2+A_y^2+A_z^2}\sqrt{B_x^2+B_y^2+B_z^2}\cos{\theta} = (A_x\hat{i} + A_y\hat{j}+A_z\hat{z}) \bullet (B_x\hat{i} + B_y\hat{j}+B_z\hat{z}) \space \mathrm{where} \space \theta =|\angle{A}-\angle{B}|$$ But I also states the right side of the equation in more explicit terms $$ (\sqrt{A_x^2+A_y^2+A_z^2}\hat{i}+ \sqrt{B_x^2+B_y^2+B_z^2}\hat{j} +\small{\small{\bullet\bullet\bullet}} \space \mathrm{ect}$$ For the sake of it being thus far fruitless, I have excluded the diagram.

Thank you for any help!

  • 1
    $\begingroup$ As of now, this post doesn't contain the algebraic definition of the dot product: ($A \cdot B := A_xB_x + A_yB_y + A_zB_z)$. $\endgroup$ – Thomas Belulovich Jan 5 '14 at 21:08
  • $\begingroup$ Hmmm... This is the equivalency that my textbook used to derive it. Thank you. I'll clarify this. $\endgroup$ – user1833028 Jan 5 '14 at 21:12

Consider this:

Both algebraic and geometric are commutative.

The algebraic dot product is linear: easily seen from the definition.

The geometric product is linear: scalar multiplication is easily checked. To check for summation, align $A,B$ on a coordinate system so that they are on the xy-plane with $B$ along the y-axis, then $|A|\cos\theta$ is just the $y$ coordinate of $A$, and then the rest follow.

The definition of both match each other at all pair of standard basis $e_{i}$: $e_{i}\cdot e_{j}=\delta_{i,j}$ in both case.

Hence they must match each other complete. Basically, it's just essentially a sort of uniqueness theorem: you want some sort of "dot product" that satisfy certain condition, and there is only one that come out. Any definition that match those condition must produce the same product.

And these conditions are very clearly motivated by physical issue: commutative is because you can rotate the world around and the physic should not change; linearity because it make sense that total energy spent by 2 people pushing something should be the sum of energy spent by each; orthogonal is because it make sense that no energy are spent if you don't move the object at all; and normalization is just choosing an arbitrary choice of unit.


To understand that it's the same thing you have to know about orthogonal projections. In particular, the orthogonal projection of $B$ onto the straight line spanned by $A$ is

$$ \pi_A(B) = \dfrac{A \bullet B}{A\bullet A} A \ . $$

Hence, if $\theta = \angle{AB}$ is an acute angle,

$$ \cos\theta = \dfrac{||\pi_A(B)||}{||B||} = \dfrac{A \bullet B}{A\bullet A}\dfrac{||A||}{||B||} = \dfrac{A\bullet B}{||A||\cdot||B||} \ . $$


The relationship between the geometric and algebraic viewpoints of the dot product is nicely expressed via the law of cosines.

The dot product $A \cdot B = A_xB_x + A_yB_y + A_zC_z$ has amazingly useful properties:

  1. $A \cdot A = A_x^2+A_y^2+A_z^2 = |A|^2$

  2. $A \cdot (B + C) = A \cdot B + A \cdot C$

  3. $(A+B) \cdot C = A \cdot C + B \cdot C.$

  4. $A \cdot B = B \cdot A$.

I will attempt to prove $A \cdot B = |A||B| \cos \theta$ using the above properties and the law of cosines.

A triangle $LMN$ in space can be made out of the three vectors $A = \vec{LM},B = \vec{LN},$ and $A-B = \vec{MN}$.

The law of cosines then tells us

$$|A-B|^2 = |A|^2 + |B|^2 - 2|A||B|\cos \theta$$

where $\theta$ is the angle formed by vectors $A$ and $B$.

Now let's use that property (1.) from above to write this in terms of dot products:

$$(A-B)\cdot (A-B) = A \cdot A + B \cdot B - 2 |A||B| \cos \theta.$$

Expand the left side out:

$$A \cdot A + B \cdot B - 2 A \cdot B = A \cdot A + B \cdot B - 2 |A||B| \cos \theta.$$


$$-2 A \cdot B = -2 |A||B| \cos \theta \implies A \cdot B = |A||B| \cos \theta.$$


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