Answer:
The tower deviates [tex]64^\circ36'[/tex] from the vertical.
Step-by-step explanation:
Having 3 point of our new plane we can construct vectors on the plane by substacting 2 of them:
[tex]v_1=(0,3,2)-(0,0,0)\\v_2=(1,3,0)-(0,0,0)[/tex]
These vectors are on the plane, so a cross product between them will give us a vector perpendicular to the plane:
[tex](0,3,2)\times(1,3,0)=\left[\begin{array}{ccc}i&j&k\\0&3&2\\1&3&0\end{array}\right] =(-6,2,-3)[/tex]
Asuming that the aliens used our conventions, the original plane was perpendicular to the z axis, so that a perpendicular vector to that plane was
(0,0,1)
We know that a dot product between 2 vectors |V.W| = |V| |W| cos(α), where α is the angle between them. If we use the vector perpendicular to this plane, and the one perpendicular to the original plane, α will represent the deviation angle of our new plane.
[tex]\|(-6,2,-3)\|=7\\\|(0,0,1)\|=1[/tex]
[tex](-6,2,-3)\odot(0,0,1)= 7 cos(\alpha )\\\\ -3=7 cos(\alpha ) \\ \alpha=arc\ cos \frac{-3}{7} =115.37 ^\circ\\[/tex]
Since this angle is greater than 90 degrees it means that the vector we calculated as perpendicular to the plane points towards negative z (this can be seen by the -3 z component)
To fix this we can calculate a new perpendicular vector, or simply compare ir with the vector (0,0,-1). The latter is easier:
[tex](-6,2,-3)\odot(0,0,-1)= 7 cos(\alpha )\\\\ 3=7 cos(\alpha ) \\ \alpha=arc\ cos \frac{3}{7} =64.6^\circ =64^\circ36'[/tex]
