91Ó°ÊÓ

For two vectors $\stackrel{\mathbf{→}}{\mathbf{x}}$and$\mathbf{}\stackrel{\mathbf{→}}{\mathbf{y}}$ ,

role="math" localid="1658991640641" $$\begin{gathered} {\mathbf{|}\stackrel{\mathbf{→}}{\mathbf{x}}\mathbf{-}\mathbf{}\stackrel{\mathbf{→}}{\mathbf{y}}\mathbf{|}}^{\mathbf{2}}\mathbf{=}\mathbf{(}\stackrel{\mathbf{→}}{\mathbf{x}}\mathbf{×}\stackrel{\mathbf{→}}{\mathbf{x}}\mathbf{)}\mathbf{-}\mathbf{\left(}\mathbf{2}\stackrel{\mathbf{→}}{\mathbf{x}}\stackrel{\mathbf{→}}{\mathbf{y}}\mathbf{\right)}\mathbf{+}\mathbf{(}\stackrel{\mathbf{→}}{\mathbf{y}}\mathbf{×}\stackrel{\mathbf{→}}{\mathbf{y}}\mathbf{)} \\ \mathbf{=}{\mathbf{\left|}\stackrel{\mathbf{→}}{\mathbf{x}}\mathbf{\right|}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\stackrel{\mathbf{→}}{\mathbf{x}}\stackrel{\mathbf{→}}{\mathbf{y}}\mathbf{+}{\mathbf{\left|}\stackrel{\mathbf{→}}{\mathbf{y}}\mathbf{\right|}}^{\mathbf{2}} \end{gathered}$$

Let Q A B C be a kite in which adjacent sides O A and O C are equal. Other two adjacent sides C B and A B also are equal.

Let $\stackrel{→}{a},\stackrel{→}{b}$and $\stackrel{→}{c}$are the position vectors of points A, B, and C respectively as shown in figure here.

Therefore,

$$\begin{gathered} \mathrm{OA}=\mathrm{OC} \\ ⇒\stackrel{→}{\mathrm{OA}}=\stackrel{→}{\mathrm{OC}} \\ ⇒\left|\stackrel{→}{\mathrm{a}}\right|=\left|\stackrel{→}{\mathrm{c}}\right|…..\left(1\right) \\ \mathrm{AB}=\mathrm{CB} \\ ⇒\stackrel{→}{\mathrm{AB}}=\stackrel{→}{\mathrm{CB}} \\ ⇒|\stackrel{→}{\mathrm{b}}-\stackrel{→}{\mathrm{a}}|=|\stackrel{→}{\mathrm{b}}-\stackrel{→}{\mathrm{c}}|…..\left(2\right) \end{gathered}$$

The vector of diagonal OB is $\stackrel{→}{\mathrm{OB}}=\stackrel{→}{\mathrm{b}},$and vector of diagonal CA is $\stackrel{→}{\mathrm{CA}}=(\stackrel{→}{\mathrm{a}}-\stackrel{→}{\mathrm{c}})$.

The scalar product of vectors $\stackrel{→}{OB}$ and $\stackrel{→}{CA}$

$$\begin{gathered} \stackrel{→}{OB}×\stackrel{→}{CA}=\stackrel{→}{b}×(\stackrel{→}{a}-\stackrel{→}{c}) \\ \stackrel{→}{OB}×\stackrel{→}{CA}=\stackrel{→}{b}×\stackrel{→}{a}-\stackrel{→}{b}×\stackrel{→}{c}…..\left(3\right) \end{gathered}$$

Squaring both the sides in equation (2)

$$\begin{gathered} |\stackrel{→}{\mathrm{b}}-\stackrel{→}{\mathrm{a}}{|}^2=|\stackrel{→}{\mathrm{b}}-\mathrm{c}⃗{|}^2 \\ ⇒|\stackrel{→}{\mathrm{b}}{|}^2-2\mathrm{b}⃗×\mathrm{a}⃗+|\mathrm{a}⃗{|}^2=|\mathrm{b}⃗{|}^2-2\stackrel{→}{\mathrm{b}}×\mathrm{c}⃗+|\mathrm{c}⃗{|}^2 \end{gathered}$$

But as known that,

$\left|\stackrel{→}{c}\right|=\left|\stackrel{→}{a}\right|$

So,

role="math" localid="1658993663930" $\stackrel{→}{b}×\stackrel{→}{a}=\stackrel{→}{b}×\stackrel{→}{c}$ ….. (4)

Substituting the above equation in result (3), and you have

$$\begin{gathered} \stackrel{→}{OB}×\stackrel{→}{CA}=\stackrel{→}{b}×\stackrel{→}{a}×\stackrel{→}{b}×\stackrel{→}{c} \\ =0 \end{gathered}$$

Since, the scalar product of two vectors $\stackrel{→}{OB}$ and $\stackrel{→}{CA}$ is zero, hence, they are perpendicular to each other.

Thus, the diagonals are perpendicular in the kite.