Question

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

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 \[\vec{a}\] and \[\vec{b}\] 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, \[\vec{OP} = \frac{2 \vec{a} + \vec{b}}{3}\]  ..................(Using section formula)
Position vector of Q,  

\[\vec{OQ} = \frac{\vec{a} + 2 \vec{b}}{3}\]  

\[\therefore \vec{a} . \vec{b} = 0\] ................(1) 

Now, 

\[{OP}^2 + {OQ}^2 \]

\[ = \left| \vec{OP} \right|^2 + \left| \vec{OQ} \right|^2 \]

\[ = \left( \frac{2 \vec{a} + \vec{b}}{3} \right) . \left( \frac{2 \vec{a} + \vec{b}}{3} \right) + \left( \frac{\vec{a} + 2 \vec{b}}{3} \right) . \left( \frac{\vec{a} + 2 \vec{b}}{3} \right)\]

\[ = \frac{4 \left| \vec{a} \right|^2 + 4 \vec{a} . \vec{b} + \left| \vec{b} \right|^2 + \left| \vec{a} \right|^2 + 4 \vec{a} . \vec{b} + 4 \left| \vec{b} \right|^2}{9}\] 

\[= \frac{5 \left| \vec{a} \right|^2 + 5 \left| \vec{b} \right|^2}{9}.............. \left[ \text{ Using } \left( 1 \right) \right]\]

\[ = \frac{5}{9}\left( \left| \vec{a} \right|^2 + \left| \vec{b} \right|^2 \right)\]

\[ = \frac{5}{9} \left| \vec{AB} \right|^2 ........................\left[ \text{ Using Pythagoras Theorem } \right]\]

\[ = \frac{5}{9} {AB}^2\]