Given a square ABCD . Let A_1, B_1, C_1, D_1 be the…

Given a square $ABCD$. Let $A_1, B_1, C_1, D_1$ be the midpoints of the sides of the square. Express in terms of vectors $\overrightarrow{AB}$ and $\overrightarrow{AD}$ the vectors $\overrightarrow{DA},\ \overrightarrow{BC},\ \overrightarrow{AC},\ \overrightarrow{BD},\ \overrightarrow{A_1B_1},\ \overrightarrow{B_1C_1},\ \overrightarrow{C_1D_1},\ \overrightarrow{A_1C}$. Calculate their magnitudes if $AB = 2$.

Step 1: Setup

Denote $\vec{a} = \overrightarrow{AB}$ and $\vec{b} = \overrightarrow{AD}$. Since $ABCD$ is a square, $|\vec{a}| = |\vec{b}| = AB = 2$ and $\vec{a} \perp \vec{b}$. The vertices are positioned so that

\[ B = A + \vec{a}, \qquad D = A + \vec{b}, \qquad C = A + \vec{a} + \vec{b}. \]

The midpoints of the sides are

\[ A_1 = A + \tfrac{1}{2}\vec{a}\ (\text{on } AB), \qquad B_1 = A + \vec{a} + \tfrac{1}{2}\vec{b}\ (\text{on } BC), \]

\[ C_1 = A + \tfrac{1}{2}\vec{a} + \vec{b}\ (\text{on } CD), \qquad D_1 = A + \tfrac{1}{2}\vec{b}\ (\text{on } DA). \]

Step 2: Express each vector

$\overrightarrow{DA}$

\[ \overrightarrow{DA} = A - D = A - (A + \vec{b}) = -\vec{b} \]

\[\overrightarrow{DA} = -\vec{b}, \qquad |\overrightarrow{DA}| = |\vec{b}| = 2\]

$\overrightarrow{BC}$

\[ \overrightarrow{BC} = C - B = (A + \vec{a} + \vec{b}) - (A + \vec{a}) = \vec{b} \]

\[\overrightarrow{BC} = \vec{b}, \qquad |\overrightarrow{BC}| = 2\]

$\overrightarrow{AC}$

\[ \overrightarrow{AC} = C - A = \vec{a} + \vec{b} \]

Because $\vec a\perp\vec b$, the Pythagorean theorem gives the length of the diagonal: \[\overrightarrow{AC} = \vec{a} + \vec{b}, \qquad |\overrightarrow{AC}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2} = \sqrt{4+4} = 2\sqrt{2}\]

$\overrightarrow{BD}$

\[ \overrightarrow{BD} = D - B = (A + \vec{b}) - (A + \vec{a}) = -\vec{a} + \vec{b} \]

\[\overrightarrow{BD} = -\vec{a} + \vec{b}, \qquad |\overrightarrow{BD}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2} = 2\sqrt{2}\]

$\overrightarrow{A_1B_1}$

\[ \overrightarrow{A_1B_1} = B_1 - A_1 = \left(A + \vec{a} + \tfrac{1}{2}\vec{b}\right) - \left(A + \tfrac{1}{2}\vec{a}\right) = \tfrac{1}{2}\vec{a} + \tfrac{1}{2}\vec{b} \]

\[\overrightarrow{A_1B_1} = \tfrac{1}{2}\vec{a} + \tfrac{1}{2}\vec{b}, \qquad |\overrightarrow{A_1B_1}| = \tfrac{1}{2}\sqrt{4+4} = \sqrt{2}\]

$\overrightarrow{B_1C_1}$

\[ \overrightarrow{B_1C_1} = C_1 - B_1 = \left(A + \tfrac{1}{2}\vec{a} + \vec{b}\right) - \left(A + \vec{a} + \tfrac{1}{2}\vec{b}\right) = -\tfrac{1}{2}\vec{a} + \tfrac{1}{2}\vec{b} \]

\[\overrightarrow{B_1C_1} = -\tfrac{1}{2}\vec{a} + \tfrac{1}{2}\vec{b}, \qquad |\overrightarrow{B_1C_1}| = \tfrac{1}{2}\sqrt{4+4} = \sqrt{2}\]

$\overrightarrow{C_1D_1}$

\[ \overrightarrow{C_1D_1} = D_1 - C_1 = \left(A + \tfrac{1}{2}\vec{b}\right) - \left(A + \tfrac{1}{2}\vec{a} + \vec{b}\right) = -\tfrac{1}{2}\vec{a} - \tfrac{1}{2}\vec{b} \]

\[\overrightarrow{C_1D_1} = -\tfrac{1}{2}\vec{a} - \tfrac{1}{2}\vec{b}, \qquad |\overrightarrow{C_1D_1}| = \sqrt{2}\]

$\overrightarrow{A_1C}$

\[ \overrightarrow{A_1C} = C - A_1 = \left(A + \vec{a} + \vec{b}\right) - \left(A + \tfrac{1}{2}\vec{a}\right) = \tfrac{1}{2}\vec{a} + \vec{b} \]

The components $\tfrac12\vec a$ (length $1$) and $\vec b$ (length $2$) are perpendicular, so

\[\overrightarrow{A_1C} = \tfrac{1}{2}\vec{a} + \vec{b}, \qquad |\overrightarrow{A_1C}| = \sqrt{\left(\tfrac{1}{2}\cdot 2\right)^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}\]