Question

Diagonals of rhombus are at right angles. Prove by vector method.

Answer 1

Given $\left|\overrightarrow{AB}\right|=\left|\overrightarrow{BC}\right|=\left|\overrightarrow{CD}\right|=\left|\overrightarrow{DA}\right|$

Diagonals are $\overrightarrow{AC}and\overrightarrow{DB}$

We know diagonals bisect each other,

From triangular law of addition:

$\Rightarrow \overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}(\because diagonalsbisect)$

and $\overrightarrow{DB}=\overrightarrow{AB}-\overrightarrow{AD}$

$\overrightarrow{AC}.\overrightarrow{DB}=(\overrightarrow{AB}-\overrightarrow{AD}).(\overrightarrow{AB}-\overrightarrow{AD})$

$\Rightarrow \overrightarrow{AC}.\overrightarrow{DB}={\left|\overrightarrow{AB}\right|}^2-{\left|\overrightarrow{AD}\right|}^2=0$

$\therefore \overrightarrow{AC}and\overrightarrow{DB}$ are perpendicular $(\because dotproduct=0)$

$\therefore$ Diagonals are at right angle

II method:

$\overrightarrow{AB}=\overrightarrow{DC}$

and $\overrightarrow{AD}=\overrightarrow{BC}$ ($\because$ all sides of rhombus are equal and opposite sides are parallel)

Diagonal $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}$

Diagonal $\overrightarrow{BD}=\overrightarrow{BA}+\overrightarrow{AD}=\overrightarrow{BA}+\overrightarrow{BC}(\because \overrightarrow{AD}=\overrightarrow{BC})$

$\therefore \overrightarrow{AC}=\overrightarrow{BC}+\overrightarrow{AB}$

and $\overrightarrow{BD}=\overrightarrow{BC}-\overrightarrow{AB}$

$\overrightarrow{AC}.\overrightarrow{BD}=(\overrightarrow{BC}+\overrightarrow{AB}).(\overrightarrow{BC}-\overrightarrow{AB})={\left|BC\right|}^2-{\left|AB\right|}^2=0$

$\therefore AC\perp BD$

Hence, diagonals intersect at ${90}^0$