In the cube ABCDA_1B_1C_1D_1 , M is the centre of the face…

In the cube $ABCDA_1B_1C_1D_1$, $M$ is the centre of the face $A_1B_1C_1D_1$ and $N$ is the centre of the face $BCC_1B_1$. Find the angle between the vectors $\overrightarrow{AM}$ and $\overrightarrow{AN}$.

Step 1: Coordinates

Let the edge be $a$, put $A$ at the origin and take $\vec i,\vec j,\vec k$ along $AB$, $AD$, $AA_1$. Then

\[ M=\left(\tfrac a2,\tfrac a2,a\right),\qquad N=\left(a,\tfrac a2,\tfrac a2\right), \]

\[ \overrightarrow{AM}=\tfrac a2(1,1,2),\qquad \overrightarrow{AN}=\tfrac a2(2,1,1). \]

Step 2: The angle

The factor $\tfrac a2$ cancels, so we may use $(1,1,2)$ and $(2,1,1)$: $\cos\varphi=\frac{2+1+2}{\sqrt6\,\sqrt6}=\frac56$.

\[\varphi=\arccos\frac56\approx33.6^\circ\]

The figure uses $a=4$, so $\overrightarrow{AM}=2\vec i+2\vec j+4\vec k$ and $\overrightarrow{AN}=4\vec i+2\vec j+2\vec k$.