I Dot product definition: deriving component form

  • Thread starter ibkev
  • Start date
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43
##
\newcommand{\ihat}{\hat{\boldsymbol{\imath}}}
\newcommand{\jhat}{\hat{\boldsymbol{\jmath}}}
\newcommand{\khat}{\hat{\boldsymbol{k}}}
##
Several times now I've seen the following technique for deriving the component form of the dot product. It always felt clean and simple until last night when I noticed for the first time that I don't understand why it's ok to do the step that I've annotated below. Can someone clarify this please?

Starting from:
$$
\begin{aligned}
\vec{u} &= u_x \ihat + u_y \jhat + u_z \khat \\
\vec{v} &= v_x \ihat + v_y \jhat + v_z \khat
\end{aligned}
$$
we can write out the dot product as:
$$
\begin{aligned}
\vec{u} \cdot \vec{v}
=&\begin{pmatrix} u_x \ihat + u_y \jhat + u_z \khat \end{pmatrix} \cdot
\begin{pmatrix} v_x \ihat + v_y \jhat + v_z \khat \end{pmatrix} \\
=& \text{why can we do this next part?} \\
=& \text{it's just straight multiplication of each term in the algebraic expression rather than an overall dot product} \\
=&u_x v_x \ihat \cdot \ihat + u_x v_y \ihat \cdot \jhat + u_x v_z \ihat \cdot \khat +\\
&u_y v_x \jhat \cdot \ihat + u_y v_y \jhat \cdot \jhat + u_y v_z \jhat \cdot \khat +\\
&u_z v_x \khat \cdot \ihat + u_z v_y \khat \cdot \jhat + u_z v_z \khat \cdot \khat
\end{aligned}
$$
Substituting these (which came from ## \cos{\theta} ##):
$$
\begin{aligned}
\ihat \cdot \ihat = \jhat \cdot \jhat = \khat \cdot \khat = 1 \\
\ihat \cdot \jhat = \ihat \cdot \khat = \jhat \cdot \khat = 0
\end{aligned}
$$
we find most of the terms cancel and we are left with:
$$ \vec{u} \cdot \vec{v} = u_x v_x + u_y v_y + u_z v_z $$

Edit: as I've typed this out, I've realized that I could insert the following, that adds a tiny bit of extra clarity:

$$
\begin{aligned}
\vec{u} \cdot \vec{v}
=&\begin{pmatrix} u_x \ihat + u_y \jhat + u_z \khat \end{pmatrix} \cdot
\begin{pmatrix} v_x \ihat + v_y \jhat + v_z \khat \end{pmatrix} \\
=&u_x \ihat \cdot v_x \ihat + u_x \ihat \cdot v_y \jhat + u_x \ihat \cdot v_z \khat + \\
&u_y \jhat \cdot v_x \ihat + u_y \jhat \cdot v_y \jhat + u_y \jhat \cdot v_z \khat + \\
&u_z \khat \cdot v_x \ihat + u_z \khat \cdot v_y \jhat + u_z \khat \cdot v_z \khat
\end{aligned}
$$

but I'm still not clear on what property of dot products allows me to go from this first line to the second?

As usual, simply explaining the question seems to lead to the answer. My problem goes away if I think of the initial vectors as:

$$
\begin{aligned}
\vec{u} \cdot \vec{v}
=&\begin{pmatrix} \vec{a} + \vec{b} + \vec{c} \end{pmatrix} \cdot
\begin{pmatrix} \vec{d} +\vec{e} + \vec{f} \end{pmatrix} \\
=&\vec{a} \cdot \vec{d} + \vec{a} \cdot \vec{e} + \vec{a} \cdot \vec{f} + \\
&\vec{b} \cdot \vec{d} + \vec{b} \cdot \vec{e} + \vec{b} \cdot \vec{f} + \\
&\vec{c} \cdot \vec{d} + \vec{c} \cdot \vec{e} + \vec{c} \cdot \vec{f}
\end{aligned}
$$

So I believe this is the distributive law in action?
 
Last edited:

fresh_42

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Several times now I've seen the following technique for deriving the component form of the dot product.
Derive from what? Something has to be defined from the start in order to change a representation.

What you call a derivation is simply the bilinear continuation of the "fact" that the inner product defines orthonormality on your coordinate vectors. However, where does this "fact" come from? How can you define orthonormality without already having an inner product?
 
119
43
Derive from what? Something has to be defined from the start in order to change a representation.

What you call a derivation is simply the bilinear continuation of the "fact" that the inner product defines orthonormality on your coordinate vectors. However, where does this "fact" come from? How can you define orthonormality without already having an inner product?
I've been taking the following as a definition for the dot product:
$$ \vec{a} \cdot \vec{b} = \lvert a \rvert \lvert b \rvert \cos {\theta} $$
To feel comfortable with where that came from I was thinking of the dot product from the perspective of determining the angle between two vectors. The angle should be independent of whatever lengths the two vectors may have so:
$$ \hat{a} \cdot \hat{b} = \cos{\theta} $$
which seemed geometrically like an intuitive starting point. Then I plugged in the i/j/k unit vectors to show i.i=1, i.j=0, etc.
And substituting ## \hat{a} = \frac{\vec{a}}{\lvert a \rvert} ## I can get back to the original definition:
$$ \vec{a} \cdot \vec{b} = \lvert a \rvert \lvert b \rvert \cos {\theta} $$

p.s. I edited my original post at the bottom. Do you agree that its the distributive law that I was using?
 
Last edited:

fresh_42

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Let me sort this out:
You defined the dot product by the definition of the cosine: ##\cos \sphericalangle(\vec{a},\vec{b}) =: \dfrac{\text{ adjacent }}{\text{ hypotenuse }} =: \dfrac{\vec{a}}{|\vec{a}|} \cdot \dfrac{\vec{b}}{|\vec{b}|}##. Now you are asking, why this definition of a product is bilinear: ##\vec{a}\cdot (\vec{b}+\vec{c})=\vec{a}\vec{b}+\vec{a}\vec{c}##.

I assume this comes down to the intercept theorem(s), but I do not have the deduction in mind.
 
119
43
Let me sort this out:
You defined the dot product by the definition of the cosine: ##\cos \sphericalangle(\vec{a},\vec{b}) =: \dfrac{\text{ adjacent }}{\text{ hypotenuse }} =: \dfrac{\vec{a}}{|\vec{a}|} \cdot \dfrac{\vec{b}}{|\vec{b}|}##. Now you are asking, why this definition of a product is bilinear: ##\vec{a}\cdot (\vec{b}+\vec{c})=\vec{a}\vec{b}+\vec{a}\vec{c}##.

I assume this comes down to the intercept theorem(s), but I do not have the deduction in mind.
Thanks for replying - I've got this all sorted out now :)
 

Ray Vickson

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##
\newcommand{\ihat}{\hat{\boldsymbol{\imath}}}
\newcommand{\jhat}{\hat{\boldsymbol{\jmath}}}
\newcommand{\khat}{\hat{\boldsymbol{k}}}
##
Several times now I've seen the following technique for deriving the component form of the dot product. It always felt clean and simple until last night when I noticed for the first time that I don't understand why it's ok to do the step that I've annotated below. Can someone clarify this please?

Starting from:
$$
\begin{aligned}
\vec{u} &= u_x \ihat + u_y \jhat + u_z \khat \\
\vec{v} &= v_x \ihat + v_y \jhat + v_z \khat
\end{aligned}
$$
we can write out the dot product as:
$$
\begin{aligned}
\vec{u} \cdot \vec{v}
=&\begin{pmatrix} u_x \ihat + u_y \jhat + u_z \khat \end{pmatrix} \cdot
\begin{pmatrix} v_x \ihat + v_y \jhat + v_z \khat \end{pmatrix} \\
=& \text{why can we do this next part?} \\
=& \text{it's just straight multiplication of each term in the algebraic expression rather than an overall dot product} \\
=&u_x v_x \ihat \cdot \ihat + u_x v_y \ihat \cdot \jhat + u_x v_z \ihat \cdot \khat +\\
&u_y v_x \jhat \cdot \ihat + u_y v_y \jhat \cdot \jhat + u_y v_z \jhat \cdot \khat +\\
&u_z v_x \khat \cdot \ihat + u_z v_y \khat \cdot \jhat + u_z v_z \khat \cdot \khat
\end{aligned}
$$
Substituting these (which came from ## \cos{\theta} ##):
$$
\begin{aligned}
\ihat \cdot \ihat = \jhat \cdot \jhat = \khat \cdot \khat = 1 \\
\ihat \cdot \jhat = \ihat \cdot \khat = \jhat \cdot \khat = 0
\end{aligned}
$$
we find most of the terms cancel and we are left with:
$$ \vec{u} \cdot \vec{v} = u_x v_x + u_y v_y + u_z v_z $$

Edit: as I've typed this out, I've realized that I could insert the following, that adds a tiny bit of extra clarity:

$$
\begin{aligned}
\vec{u} \cdot \vec{v}
=&\begin{pmatrix} u_x \ihat + u_y \jhat + u_z \khat \end{pmatrix} \cdot
\begin{pmatrix} v_x \ihat + v_y \jhat + v_z \khat \end{pmatrix} \\
=&u_x \ihat \cdot v_x \ihat + u_x \ihat \cdot v_y \jhat + u_x \ihat \cdot v_z \khat + \\
&u_y \jhat \cdot v_x \ihat + u_y \jhat \cdot v_y \jhat + u_y \jhat \cdot v_z \khat + \\
&u_z \khat \cdot v_x \ihat + u_z \khat \cdot v_y \jhat + u_z \khat \cdot v_z \khat
\end{aligned}
$$

but I'm still not clear on what property of dot products allows me to go from this first line to the second?

As usual, simply explaining the question seems to lead to the answer. My problem goes away if I think of the initial vectors as:

$$
\begin{aligned}
\vec{u} \cdot \vec{v}
=&\begin{pmatrix} \vec{a} + \vec{b} + \vec{c} \end{pmatrix} \cdot
\begin{pmatrix} \vec{d} +\vec{e} + \vec{f} \end{pmatrix} \\
=&\vec{a} \cdot \vec{d} + \vec{a} \cdot \vec{e} + \vec{a} \cdot \vec{f} + \\
&\vec{b} \cdot \vec{d} + \vec{b} \cdot \vec{e} + \vec{b} \cdot \vec{f} + \\
&\vec{c} \cdot \vec{d} + \vec{c} \cdot \vec{e} + \vec{c} \cdot \vec{f}
\end{aligned}
$$

So I believe this is the distributive law in action?
You need to make assumptions about "##\cdot##", such as the following:
1. For a scalar ##c##, ##(c \vec{a}) \cdot \vec{b} = c (\vec{a} \cdot \vec{b})##
2. Commutativity: ##\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}.##
3. Distributvity: ##\vec{a} \cdot ( \vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}.##
4. Special values: ##\vec{u} \cdot \vec{u} = 1## for ##\vec{u} = \vec{i}, \vec{j}, \vec{k}##, and ##\vec{u} \cdot \vec{v} =0## for ##\vec{u}, \vec{v} \in \{ \vec{i}, \vec{j}, \vec{k} \}## with ##\vec{u} \neq \vec{v}.##

Unless you make some such assumptions (or else define "##\cdot##" in some other way) there is no way you could derive what you want, as we can easily enough cook up examples where some of 1--4 fail and the resulting product is not anything useful.
 

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