Answer:
[tex]\theta=cos^{-1}(\frac{2}{3})[/tex]
Step-by-step explanation:
We are given that measurements of three points on the ground gave coordinates of (0,0,0),(1,2,0) and (0,2,1)
We have to find the angle by which the tower now deviate from the vertical
We find cross product of <1,2,0> and <0,2,1>
[tex]<1,2,0> \times <0,2,1>=\begin{vmatrix}i&j&k\\1&2&0\\0&2&1\end{vmatrix}[/tex]
[tex]<1,2,0> \times <0,2,1>=2\hat{i}-\hat{j}+2\hat{k}[/tex]
Now, we are finding the angle between [tex]<1,2,0> \times <0,2,1> [/tex]and vertical vector <0,0,1>
Angle between two vectors formula
[tex]cos\theta=\frac{a.b}{\mid a\mid\cdot\mid b\mid }[/tex]
Now, using this formula
[tex]cos\theta=\frac{2}{1\cdot 3}=\frac{2}{3}[/tex]
[tex]\theta=cos^{-1}(\frac{2}{3})[/tex]
Hence, the tower deviate from the vertical by the angle [tex]\theta=cos^{-1}(\frac{2}{3})[/tex]
