A cube with edge length 2 cm has vertices ABCD and…

A cube with edge length $2$ cm has vertices $ABCD$ and $A_1B_1C_1D_1$. Let $M$ and $N$ be the centers of faces $A_1B_1C_1D_1$ and $BCC_1B_1$ respectively. Express vectors $\overrightarrow{AM}$, $\overrightarrow{AN}$, $\overrightarrow{MN}$ and $\overrightarrow{CA_1}$ in terms of unit vectors $\vec{i}$, $\vec{j}$, $\vec{k}$ of directions $\overrightarrow{AB}$, $\overrightarrow{AD}$ and $\overrightarrow{AA_1}$. Show that $\overrightarrow{MN}$ is parallel to $\overrightarrow{A_1B}$.

Step 1: Setup and coordinate system

Let the edge be $a = 2$ cm and define the unit vectors

\[ \vec{i} = \frac{\overrightarrow{AB}}{|\overrightarrow{AB}|}, \qquad \vec{j} = \frac{\overrightarrow{AD}}{|\overrightarrow{AD}|}, \qquad \vec{k} = \frac{\overrightarrow{AA_1}}{|\overrightarrow{AA_1}|}, \] so $|\vec{i}| = |\vec{j}| = |\vec{k}| = 1$, mutually perpendicular, and

\[ \overrightarrow{AB} = 2\vec{i}, \qquad \overrightarrow{AD} = 2\vec{j}, \qquad \overrightarrow{AA_1} = 2\vec{k}. \]

Coordinates of all vertices

Placing $A$ at the origin: \[A = (0,0,0), \quad B = (2,0,0), \quad C = (2,2,0), \quad D = (0,2,0),\]

\[ A_1 = (0,0,2), \quad B_1 = (2,0,2), \quad C_1 = (2,2,2), \quad D_1 = (0,2,2). \]

Centers of faces

The center of a square face is the average of its four vertices: \[M = \frac{A_1+B_1+C_1+D_1}{4} = (1,1,2), \qquad N = \frac{B+C+C_1+B_1}{4} = (2,1,1).\]

Step 2: Express each vector

$\overrightarrow{AM}$

\[ \overrightarrow{AM} = M - A = (1,1,2) \]

\[\overrightarrow{AM} = \vec{i} + \vec{j} + 2\vec{k}, \qquad |\overrightarrow{AM}| = \sqrt{1+1+4} = \sqrt{6}\ \text{cm}\]

$\overrightarrow{AN}$

\[ \overrightarrow{AN} = N - A = (2,1,1) \]

\[\overrightarrow{AN} = 2\vec{i} + \vec{j} + \vec{k}, \qquad |\overrightarrow{AN}| = \sqrt{4+1+1} = \sqrt{6}\ \text{cm}\]

$\overrightarrow{MN}$

\[ \overrightarrow{MN} = N - M = (2,1,1) - (1,1,2) = (1,0,-1) \]

\[\overrightarrow{MN} = \vec{i} - \vec{k}, \qquad |\overrightarrow{MN}| = \sqrt{2}\ \text{cm}\]

$\overrightarrow{CA_1}$

\[ \overrightarrow{CA_1} = A_1 - C = (0,0,2) - (2,2,0) = (-2,-2,2) \]

\[\overrightarrow{CA_1} = -2\vec{i} - 2\vec{j} + 2\vec{k}, \qquad |\overrightarrow{CA_1}| = \sqrt{12} = 2\sqrt{3}\ \text{cm}\]

Step 3: Proof that $\overrightarrow{MN} \parallel \overrightarrow{A_1B}$

\[ \overrightarrow{A_1B} = B - A_1 = (2,0,0) - (0,0,2) = (2,0,-2) = 2\vec{i} - 2\vec{k}. \]

Comparing with $\overrightarrow{MN} = \vec{i} - \vec{k}$: $\overrightarrow{A_1B} = 2(\vec{i} - \vec{k}) = 2\,\overrightarrow{MN}$.

$\overrightarrow{A_1B}$ is a positive scalar multiple of $\overrightarrow{MN}$, therefore

\[\overrightarrow{MN} \parallel \overrightarrow{A_1B}, \qquad |\overrightarrow{A_1B}| = 2|\overrightarrow{MN}| = 2\sqrt2\ \text{cm}\]

Geometrically: $M$ is the midpoint of the top-face diagonal $A_1C_1$ and $N$ is the midpoint of the side-face diagonal $BC_1$. So $MN$ is the midline of triangle $A_1BC_1$, parallel to $A_1B$ and half its length.