Vector Field / Line Integral (Part 3)

$\text{p1}=\{0,0,0\}$

$\text{p2}=\{1,2,3\}$

$\text{center}=\frac{\text{p1}+\text{p2}}{2} = \left\{\frac{1}{2},1,\frac{3}{2}\right\}$

$\text{radius}=\text{EuclideanDistance}[\text{p1},\text{center}] =\sqrt{\frac{7}{2}}$

The unit vectors that used: $\left( \begin{array}{ccc} -\frac{1}{\sqrt{14}} & -\sqrt{\frac{2}{7}} & -\frac{3}{\sqrt{14}} \\ \sqrt{\frac{13}{14}} & -\sqrt{\frac{2}{91}} & -\frac{3}{\sqrt{182}} \\ 0 & -\frac{3}{\sqrt{13}} & \frac{2}{\sqrt{13}} \\ \end{array} \right)$

Scale a vector $\vec{v}$ back to original coordinates: $\text{radius} \vec{v}+\text{center}]$

Compute a coordinate from \tau), the path integration variable, and $\theta$ , the angle the path is displaced. $\left\{\frac{1}{2} \left(\sqrt{13} \cos (\theta ) \cos (\tau )+\sin (\tau )\right),\\-\frac{\cos (\theta ) \cos (\tau )}{\sqrt{13}}-3 \sqrt{\frac{7}{26}} \sin (\theta ) \cos (\tau )+\sin (\tau ),\frac{\cos (\tau ) \left(2 \sqrt{14} \sin (\theta )-3 \cos (\theta )\right)}{2 \sqrt{13}}+\frac{3 \sin (\tau )}{2}\right\}$

The integral: $\int_0^{\pi } \frac{52 \sqrt{14} \sin (2 \theta ) (49 \sin (\tau )+61 \sin (2 \tau )) \sin ^2\left(\frac{\tau }{2}\right)-104 \sin ^3\left(\frac{\tau }{2}\right) \cos \left(\frac{\tau }{2}\right) (\cos (2 \theta ) (\cos (\tau )-37)+403 \cos (\tau )-299)+\\ 13 \sqrt{13} \cos (\theta ) \sin ^2\left(\frac{\tau }{2}\right) (38 \cos (\tau )-453 \cos (2 \tau )+411)+13 \sqrt{182} \sin (\theta ) \sin ^2\left(\frac{\tau }{2}\right) (-86 \cos (\tau )+81 \cos (2 \tau )+25)-\\ \sqrt{13} \sin ^2(\tau ) \cos (\tau ) \left(943 \sqrt{14} \sin (3 \theta )+919 \cos (3 \theta )\right)}{5408} \, d\tau =\\ \frac{1}{624} \left(50 \sqrt{14} \sin (2 \theta )+51 \pi \sqrt{182} \sin (\theta )+224 \cos (2 \theta )+294 \pi \sqrt{13} \cos (\theta )+2600\right)$

Maximize the result over $\theta$ and evaluate the maximization result numerically. The maximum is $10.9596177774$ . The angle from $\vec{u_2}$ is $0.524769623149$ radians.