a)
If we consider i pointing to west-east direction and j pointing to south-north direction, since the airspeed is 600 miles per hour headed to north, the jet velocity relative to the air is given by:
[tex]\vec{v_a}=600\hat{j}[/tex]Since the jet stream is is 110 miles per hour in the northeasterly direction, we have:
[tex]\begin{gathered} \vec{v_w}=110\cdot(\cos 45\degree\hat{i}+\sin 45\degree\hat{j}) \\ \vec{v_w}=55\sqrt[]{2}\hat{i}+55\sqrt[]{2}\hat{j} \end{gathered}[/tex]b)
The velocity of the jet relative to the ground is given by:
[tex]\begin{gathered} \vec{v_a}+\vec{v_w}=_{}55\sqrt[]{2}\hat{i}+(600+55\sqrt[]{2})\hat{j} \\ \vec{v_a}+\vec{v_w}=55\sqrt[]{2}\hat{i}+5(120+11\sqrt[]{2})\hat{j} \end{gathered}[/tex]c)
Then, the actual speed of the jet relative to the ground is given by:
[tex]\sqrt[]{\lbrack(55\sqrt[]{2})^2+(600+55\sqrt[]{2})^2}=\sqrt[]{372100+66000\sqrt[]{2}}=10\sqrt[]{3721+660\sqrt[]{2}}[/tex]Finally, the direction of the jet relative to the ground is given by:
[tex]\tan ^{-1}\frac{600+55\sqrt[]{2}}{55\sqrt[]{2}}=\tan ^{-1}(\frac{60}{11}\sqrt[]{2}+1)[/tex]