Given a rectangle ABCD . Express the side vectors in terms…

Given a rectangle $ABCD$. Express the side vectors in terms of the diagonal vectors.

Explanation of Method

Expressing side vectors of a rectangle through its diagonal vectors.

Step 1: Setup

Denote the diagonal vectors by

\[ \vec{p} = \overrightarrow{AC}, \qquad \vec{q} = \overrightarrow{BD}. \]

By the triangle rule, \[\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}, \qquad \overrightarrow{BD} = \overrightarrow{BA} + \overrightarrow{AD},\] so \[\vec{p} = \overrightarrow{AB} + \overrightarrow{BC}, \qquad \vec{q} = -\overrightarrow{AB} + \overrightarrow{AD}.\]

In a rectangle opposite sides are equal and parallel, hence $\overrightarrow{BC} = \overrightarrow{AD}$, and both equations become

\[ \vec{p} = \overrightarrow{AB} + \overrightarrow{AD}, \qquad \vec{q} = -\overrightarrow{AB} + \overrightarrow{AD}. \]

$A$ $B$ $C$ $D$ $\vec p$ $\vec q$

$\vec p=\overrightarrow{AC}$
$\vec q=\overrightarrow{BD}$

Rectangle $ABCD$ with diagonal vectors $\vec p$ and $\vec q$.

Step 2: Solve for the side vectors

Adding the two equations: $\vec{p} + \vec{q} = 2\,\overrightarrow{AD}$.

\[\overrightarrow{AD} = \frac{1}{2}(\vec{p} + \vec{q})\]

Subtracting the second equation from the first: $\vec{p} - \vec{q} = 2\,\overrightarrow{AB}$.

\[\overrightarrow{AB} = \frac{1}{2}(\vec{p} - \vec{q})\]

Step 3: Remaining sides

Along the traversal $A \to B \to C \to D$, opposite sides are equal in length and point in opposite directions: $\overrightarrow{BC} = \overrightarrow{AD} = \frac{1}{2}(\vec{p} + \vec{q})$.

\[\overrightarrow{CD} = -\overrightarrow{AB} = \frac{1}{2}(\vec{q} - \vec{p})\]

Result

All four side vectors expressed through $\vec{p} = \overrightarrow{AC}$ and $\vec{q} = \overrightarrow{BD}$:

Side vector	Expression
$\overrightarrow{AB}$	$\dfrac{1}{2}(\vec{p} - \vec{q})$
$\overrightarrow{BC}$	$\dfrac{1}{2}(\vec{p} + \vec{q})$
$\overrightarrow{CD}$	$\dfrac{1}{2}(\vec{q} - \vec{p})$
$\overrightarrow{DA}$	$-\dfrac{1}{2}(\vec{p} + \vec{q})$

Remark

Check. The four side vectors must add up to the zero vector, since the polygon is closed:

\[ \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA} = \tfrac12\big[(\vec p-\vec q)+(\vec p+\vec q)+(\vec q-\vec p)-(\vec p+\vec q)\big]=\vec 0. \]