Question

In a triangle OAB, $\angle$AOB = 90º. If P and Q are points of trisection of AB, prove that ${\mathrm{OP}}^2+{\mathrm{OQ}}^2=\frac{5}{9}{\mathrm{AB}}^2$.

Answer 1

In triangle OAB, $\angle$AOB = 90º. P and Q are points of trisection of AB.

Taking O as the origin, let the position vectors of A and B be $\overrightarrow{a}$ and $\overrightarrow{b}$, respectively.

Since P and Q are the points of trisection of AB, so AP : PB = 1 : 2 and AQ : QB = 2 : 1.

Position vector of P, $\overrightarrow{\mathrm{OP}}=\frac{2\overrightarrow{a}+\overrightarrow{b}}{3}$ (Using section formula)
Position vector of Q, $\overrightarrow{\mathrm{OQ}}=\frac{\overrightarrow{a}+2\overrightarrow{b}}{3}$

Also, $\overrightarrow{\mathrm{OA}}\perp \overrightarrow{\mathrm{OB}}$.

$\therefore \overrightarrow{a}.\overrightarrow{b}=0$ .....(1)

Now,

$$\begin{gathered} {\mathrm{OP}}^2+{\mathrm{OQ}}^2 \\ ={\left|\overrightarrow{\mathrm{OP}}\right|}^2+{\left|\overrightarrow{\mathrm{OQ}}\right|}^2 \\ =\left(\frac{2\overrightarrow{a}+\overrightarrow{b}}{3}\right).\left(\frac{2\overrightarrow{a}+\overrightarrow{b}}{3}\right)+\left(\frac{\overrightarrow{a}+2\overrightarrow{b}}{3}\right).\left(\frac{\overrightarrow{a}+2\overrightarrow{b}}{3}\right) \\ =\frac{4{\left|\overrightarrow{a}\right|}^2+4\overrightarrow{a}.\overrightarrow{b}+{\left|\overrightarrow{b}\right|}^2+{\left|\overrightarrow{a}\right|}^2+4\overrightarrow{a}.\overrightarrow{b}+4{\left|\overrightarrow{b}\right|}^2}{9} \end{gathered}$$
$$\begin{gathered} =\frac{5{\left|\overrightarrow{a}\right|}^2+5{\left|\overrightarrow{b}\right|}^2}{9}\left[\mathrm{Using}\left(1\right)\right] \\ =\frac{5}{9}\left({\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2\right) \\ =\frac{5}{9}{\left|\overrightarrow{\mathrm{AB}}\right|}^2\left[\mathrm{Using}\mathrm{Pythagoras}\mathrm{Theorem}\right] \\ =\frac{5}{9}{\mathrm{AB}}^2 \end{gathered}$$