Exercise 4

According to Exercise 3 of Interlude 1 the magnitude of a vector is given by:

\[\left| \vec{A} \right| = \sqrt{ \vec{A} \cdot \vec{A}}\]

or written with components:

\[\left| \vec{A} \right| = \sqrt{a_x^2 +a_y^2 +a_z^2 }\]

The components of \(\vec{A}\) and \(\vec{B}\) are as follows:

\[\vec{A}=\left( \begin{matrix} 2\\-3\\1\end{matrix}\right)\]

\[\vec{B}=\left( \begin{matrix} -4\\-3\\2\end{matrix}\right)\]

The magnitudes are

\[\left| \vec{A} \right| = \sqrt{a_x^2 +a_y^2 +a_z^2 } = \sqrt{2^2 + (-3)^2 + 1^2 } = \sqrt{14}\]

\[\left| \vec{B} \right| = \sqrt{b_x^2 +b_y^2 +b_z^2 } = \sqrt{(-4)^2 + (-3)^2 + 2^2 } = \sqrt{29}\]

The dot product is

\[\vec{A} \cdot \vec{B} = a_x b_x + a_y b_y + a_z b_z = 2 (-4) + (-3) (-3) +2 = 3\]

The angle between the two vectors is calculated as follows:

\[\vec{A} \cdot \vec{B} = \left| \vec{A} \right|\left| \vec{B} \right| \cos \varphi\]

\[\Rightarrow \cos \varphi = \frac{ \vec{A} \cdot \vec{B}} { \left| \vec{A} \right|\left| \vec{B} \right|}\]

\[\Rightarrow \varphi = \arccos \frac{ \vec{A} \cdot \vec{B}} { \left| \vec{A} \right|\left| \vec{B} \right|} = \arccos \frac{3}{\sqrt{14} \sqrt{29}} \approx 1,42 \approx 81,44^\circ\]