Question 4.1: If the position vector of the International Space Station in......

We employ Algorithm 4.1.

Step 1:

r={\sqrt{\left(-5368\right)^{2}+\left(-1784\right)^{2}+3691^{2}}}=6754\,\mathrm{km}

Step 2:

I={\frac{-5368}{6754}}=-0.7947\;\;\;\;m={\frac{-1784}{6754}}=-0.2642\;\;\;\;n={\frac{3691}{6754}}=0.5462

Step 3:

\delta={\sin}^{-1}0.5464=33.12°

Step 4:

Since the direction cosine m is negative ,

\alpha=360^{\circ}-\cos^{-1}\left({\frac{I}{\cos\delta}}\right)=360^{\circ}-\cos^{-1}\left({\frac{-0.7947}{\cos33.12^{\circ}}}\right)=360^{\circ}-161.6^{\circ}=198.4^{\circ}