Ellipse lying on a circular cylinder

The given equation is a parametric representation of an ellipse with a center of $(0, 0, 0)$ .

The distance between each point on the ellipse from its center $(0, 0, 0)$ is $d = \sqrt{(4 \cos t + 5 \sin t)^2 + (3 \cos t + 10 \sin t)^2 + (-3 \cos t + \frac{50}{3} \sin t)^2} = \sqrt{\frac{3319}{9}\sin^2 t + 34}$ . Since the maximum of $\sin^2 t$ is $1$ at at $t = 90°$ , the semi-major axis is $a = \sqrt{\frac{3319}{9} \cdot 1 + 34} = \frac{5}{3}\sqrt{145}$ with a vector of $\overrightarrow{OA} = (4, 3, -3) \cos 90° + (5, 10, \frac{50}{3}) \sin 90° = (5, 10, \frac{50}{3})$ , and since the minimum of $\sin^2 t$ is $0$ at $t = 0°$ , the semi-minor axis is $b = \sqrt{\frac{3319}{9} \cdot 0 + 34} = \sqrt{34}$ with a vector of $\overrightarrow{OB} = (4, 3, -3) \cos 0° + (5, 10, \frac{50}{3}) \sin 0° = (4, 3, -3)$ .

The vector perpendicular to both $\overrightarrow{OA}$ and $\overrightarrow{OB}$ is $\overrightarrow{OC} = \overrightarrow{OA} \times \overrightarrow{OB} = (48, -49, 15)$ .

An ellipse lying on the surface of a circular cylinder whose axis is along the unit vector $\mathbf{a} = (a_x, a_y, a_z)$ will be one where $\mathbf{a} \perp \overrightarrow{OB}$ so that:

$4a_x + 3a_y - 3a_z = 0$

and $\cos \theta = \cfrac{b}{a} = \cfrac{\overrightarrow{OC} \cdot \mathbf{a}}{|\overrightarrow{OC}| \cdot |\mathbf{a}|}$ or $\cfrac{\sqrt{34}}{\frac{5}{3}\sqrt{145}} = \cfrac{48a_x - 49a_y + 15a_z}{\sqrt{48^2 + 49^2 + 15^2} \cdot 1}$ , which rearranges to:

$240a_x - 245a_y + 75a_z = 102$

Also, since $\mathbf{a}$ is a unit vector:

$a_x^2 + a_y^2 + a_z^2 = 1$

These three equations solve to $\mathbf{a} = (\frac{1}{725}(144 - 3\sqrt{3319}), \frac{1}{725}(-147 - 6\sqrt{3319}), \frac{1}{725}(45 - 10\sqrt{3319}))$ or $\mathbf{a} = (\frac{1}{725}(144 + 3\sqrt{3319}), \frac{1}{725}(-147 + 6\sqrt{3319}), \frac{1}{725}(45 + 10\sqrt{3319}))$ , the second of which has a higher $|a_z|$ value.

Therefore, the sum of $|a_x| + |a_y| + |a_z| = |\frac{1}{725}(144 + 3\sqrt{3319})| + |\frac{1}{725}(-147 + 6\sqrt{3319})| + |\frac{1}{725}(45 + 10\sqrt{3319})| = \frac{1}{725}(42 + 19\sqrt{3319}) \approx \boxed{1.56773}$ .