## University Calculus: Early Transcendentals (3rd Edition)

$\theta =\dfrac{\pi}{2}$ or $90 ^{\circ}$; $\overrightarrow{CA} \perp\overrightarrow{DB}$
We have two diagonals, let us say $\overrightarrow{CA}$ and $\overrightarrow{DB}$, of a rhombus ABCD . The formula to calculate the angle between two diagonals $\overrightarrow{CA}$ and $\overrightarrow{DB}$ of a rhombus is: $\theta = \cos ^{-1} (\dfrac{\overrightarrow{CA} \cdot \overrightarrow{DB}}{|\overrightarrow{CA}||\overrightarrow{DB}|})=\cos ^{-1} (\dfrac{(\overrightarrow{DA})^2-(\overrightarrow{AB})^2}{ |\overrightarrow{CA}||\overrightarrow{DB}|})$ As ABCD is a rhombus, thus $|\overrightarrow{DA}|=|\overrightarrow{AB}|$ or, $\theta=\cos ^{-1} (0)$ or, $\theta =\dfrac{\pi}{2}$ or $90 ^{\circ}$ Hence, $\overrightarrow{CA} \perp\overrightarrow{DB}$