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An excerpt from my Physics Notes. The topic is solving for angular momentum in rigid bodies.

$$\vec{r} \times (\vec{\omega}\times \vec{r})=(\vec{r} \cdot \vec{r})\vec{\omega}-(\vec{r} \cdot \vec{\omega})\vec{r}$$ $$\vec{\omega}=(\omega_x,\omega_y,\omega_z)=\begin{bmatrix} \omega_x\\\omega_y\\ \omega_z\end{bmatrix}$$ $$(\vec{r} \cdot \vec{r})\vec{\omega}-(\vec{r} \cdot \vec{\omega})\bf{\vec{r}}=\begin{bmatrix} (x^2+y^2+z^2)\omega_x\\(x^2+y^2+z^2)\omega_y\\(x^2+y^2+z^2)\omega_z\end{bmatrix}-\begin{bmatrix}(x\omega_x+y\omega_y+z\omega_z) \bf{x}\\(x\omega_x+y\omega_y+z\omega_z)\bf{y}\\(x\omega_x+y\omega_y+z\omega_z)\bf{z} \end{bmatrix}$$

My question is concerning the very last $\bf{\vec{r}}$ (bolded in the work above). When it's turned into a matrix, why is it treated as$\begin{bmatrix} x\\y\\ z\end{bmatrix}$ instead of $\begin{bmatrix} (x+y+z)\\(x+y+z)\\(x+y+z)\end{bmatrix}$ like the $\vec{r}$ just before it, that was dotted with $\vec{\omega}$?

Any insight is appreciated.

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  • $\begingroup$ Check the definition of the scalar product. Your $\vec{r}$ has always been just $[x,y,z]$, not $[x+y+z, x+y+z, x+y+z]$... $\endgroup$
    – MsTais
    Commented Feb 5, 2017 at 2:52

2 Answers 2

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The scalar product is defined:

\begin{equation} \vec{r}\cdot \vec{\omega} = [x,y,z] \left[\begin{matrix} \omega_x\\ \omega_y\\ \omega_z \end{matrix}\right] \end{equation} This construction means that you multiply each member of the row by each corresponding member of the column. That's how you get: \begin{equation} \vec{r}\cdot \vec{\omega} = x\omega_x+y\omega_y+z\omega_z \end{equation} which is a scalar, not a vector any more (that's why the operation is called scalar product). If now you multiply this product by $\vec{r}$ one more time you get:

\begin{equation} (\vec{r}\cdot \vec{\omega})\vec{r} = (x\omega_x+y\omega_y+z\omega_z) \left[\begin{matrix} \omega_x\\ \omega_y\\ \omega_z \end{matrix}\right] \end{equation} This gives you what you have obtained at the end=) Goodluck! Refresh vector calculus.

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  • $\begingroup$ Thank you! I've never seen dot product written in matrices form and it is very insightful. $\endgroup$
    – Jess L
    Commented Feb 5, 2017 at 3:51
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Because the components of $\vec r$ are $ \left(\begin{array}{c}x\\ y \\ z\end{array}\right), $ not $ \left(\begin{array}{c}x+y+z\\ x+y+z\\ x+y+z\end{array}\right)\, . $ At no point do you use the latter incorrect expression.

$\vec r\cdot \vec \omega$ is the scalar $x\omega_x+y\omega_y +z\omega_z$ and it's a common factor to all the components of $\vec r$. Likewise $\vec r\cdot \vec r=x^2+y^2+z^2$ is a common factor in front of $\vec \omega$.

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