Establish the properties of the line \(L\) defined by \((1.41)\) for the general motion of a rigid body. That is, show that if \(P\) is a point of \(L\), then \(\boldsymbol{v}_{P}\) is proportional to \(\omega\); and show conversely that if \(v_{P}\) is proportional to \(\omega\), then \(P\) lies on \(L\). (Assume that \(\omega \neq 0\).)

The line L is defined by (1.41) and represents the axis of rotation. For an arbitrary point \(P\) on the line \(L\), the velocity vector can be described as \(\boldsymbol{v}_{P} = \boldsymbol{\omega} \times \boldsymbol{r}_{P}\), where \(\boldsymbol{r}_{P}\) is the position vector of the point \(P\) and \(\boldsymbol{\omega}\) is the angular velocity vector.

If point \(P\) is on line \(L\), then the line \(L\) is parallel to the position vector \(\boldsymbol{r}_{P}\). Since they are parallel, the cross product of \(\boldsymbol{\omega}\) and \(\boldsymbol{r}_P\) will also be parallel to \(\boldsymbol{\omega}\), which means \(\boldsymbol{v}_{P} = k\boldsymbol{\omega}\) for some constant \(k\). This shows that if \(P\) is a point of \(L\), then \(\boldsymbol{v}_{P}\) is proportional to \(\boldsymbol{\omega}\).

Now we need to show the converse. If \(\boldsymbol{v}_P\) is proportional to \(\boldsymbol{\omega}\), then there is a constant \(k\) such that \(\boldsymbol{v}_P = k\boldsymbol{\omega}\). Rearranging this equation, we obtain \(\boldsymbol{v}_{P} - k\boldsymbol{\omega} = 0\). Taking the cross product of this equation with the angular velocity vector \(\boldsymbol{\omega}\), we get \((\boldsymbol{v}_{P} - k\boldsymbol{\omega}) \times \boldsymbol{\omega} = 0\). Expanding the cross product, we have \(\boldsymbol{v}_{P} \times \boldsymbol{\omega} - k\boldsymbol{\omega} \times \boldsymbol{\omega} = 0\). Since the cross product of a vector with itself is always zero, the equation simplifies to \(\boldsymbol{v}_{P} \times \boldsymbol{\omega} = 0\). Recall the velocity vector formula, \(\boldsymbol{v}_{P} = \boldsymbol{\omega} \times \boldsymbol{r}_{P}\). Substituting this into the equation, we get \((\boldsymbol{\omega} \times \boldsymbol{r}_{P}) \times \boldsymbol{\omega} = 0\). Using the vector triple product identity, \(\boldsymbol{A} \times (\boldsymbol{B} \times \boldsymbol{C}) = (\boldsymbol{A} \cdot \boldsymbol{C})\boldsymbol{B} - (\boldsymbol{A} \cdot \boldsymbol{B})\boldsymbol{C}\), we can rewrite the above equation as \((\boldsymbol{\omega} \cdot \boldsymbol{\omega})\boldsymbol{r}_{P} - (\boldsymbol{\omega} \cdot \boldsymbol{r}_{P})\boldsymbol{\omega} = 0\). Or equivalently, \(\boldsymbol{r}_{P} = \frac{\boldsymbol{\omega} \cdot \boldsymbol{r}_{P}}{\boldsymbol{\omega} \cdot \boldsymbol{\omega}}\boldsymbol{\omega}\). This equation implies that the position vector \(\boldsymbol{r}_{P}\) is parallel to the angular velocity vector \(\boldsymbol{\omega}\). Therefore, point \(P\) must lie on the line of action \(L\). Thus, we have shown that if \(\boldsymbol{v}_{P}\) is proportional to \(\boldsymbol{\omega}\), then \(P\) lies on \(L\).