[ Vectors help solve tasks] [ Scalar product $]

Given a rectangular parallelepiped $A B C D A 1 B 1 C 1 D 1$, where $A B=2, A D=4, B B 1=12$. Points $M$ and $K$ are located on the edges $C C 1$ and $A D$ respectively, such that $C M: M C 1=1: 2, A K=K D$. Find the angle between the lines АМ and $K B 1$.

## Answer

 $\arccos \frac{11}{3 \sqrt{38}}$.Given a tetrahedron $A B C D$, where $A B=B D=3, A C=C D=5, A D=B C=4$. Find $A M$, where $M$ is the point of intersection of the medians of the face $B C D$.

#
Since $M$ is the point of intersection of the medians of triangle $BCD$,

![](https://cdn.mathpix.com/cropped/2024_05_06_cdc67dfeb844af03d94fg-12.jpg?height=117&width=514&top_left_y=410&top_left_x=772)

therefore,

$$
\begin{gathered}
\overrightarrow{AM}^2 = \frac{1}{9}(\overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD})^2 = \\
= \frac{1}{9}(AB^2 + AC^2 + AD^2 + 2 \overrightarrow{AB} \cdot \overrightarrow{AC} + 2 \overrightarrow{AB} \cdot \overrightarrow{AD} + 2 \overrightarrow{AC} \cdot \overrightarrow{AD})
\end{gathered}
$$

From the equality $\overrightarrow{BD} = \overrightarrow{AD} - \overrightarrow{AB}$, it follows that

$$
BD^2 = AD^2 - 2 \overrightarrow{AD} \cdot \overrightarrow{AB} + AB^2
$$

from which we find that

$$
2 \overrightarrow{AD} \cdot \overrightarrow{AB} = AD^2 + AB^2 - BD^2 = 16 + 9 - 9 = 16
$$

Similarly,

$$
\begin{aligned}
& \overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB}, BC^2 = AC^2 - 2 \cdot \overrightarrow{AC} \cdot \overrightarrow{AB} + AB^2 \\
& \overrightarrow{DC} = \overrightarrow{AC} - \overrightarrow{AD}, AC^2 = DC^2 - 2 \overrightarrow{AC} \cdot \overrightarrow{AD} + AD^2
\end{aligned}
$$

from which

$$
\begin{aligned}
& 2 \overrightarrow{AC} \cdot \overrightarrow{AB} = AC^2 + AB^2 - BC^2 = 25 + 9 - 16 = 18 \\
& 2 \overrightarrow{AC} \cdot \overrightarrow{AD} = AC^2 + AD^2 - DC^2 = 25 + 16 - 25 = 16
\end{aligned}
$$

Therefore,

$$
\begin{aligned}
\overrightarrow{AM}^2 = & \frac{1}{9}(AB^2 + AC^2 + AD^2 + 2 \overrightarrow{AB} \cdot \overrightarrow{AC} + 2 \overrightarrow{AB} \cdot \overrightarrow{AD} + 2 \overrightarrow{AC} \cdot \overrightarrow{AD}) = \\
& = \frac{1}{9}(9 + 16 + 25 + 16 + 18 + 16) = \frac{1}{9} \cdot 100 \Rightarrow AM = \frac{10}{3}
\end{aligned}
$$

## Answer

$\frac{10}{3}$.