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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?

\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: