An arc PQ of a circle subtends a right angle at its centre O . The...

Given,

An arc $PQ$ of a circle subtends a right angle at its centre $O$. The mid point of the arc $PQ$ is $R$,

And $\overset{\to}{OP} = \overset{\to}{u}, \overset{\to}{OR} = \overset{\to}{v}$ and $\overset{\to}{OQ} = \alpha\overset{\to}{u} + \beta\overset{\to}{v}$,

So, plotting the diagram of the given value we get,

Now given, $\overset{\to}{OQ} = \alpha\overset{\to}{u} + \beta\overset{\to}{v}$

Where, $\overset{\to}{OR} = \overset{\to}{v}$ and $\overset{\to}{OP} = \overset{\to}{u}$

Now \left|\overset{\to}{OP}\right| = \left|\overset{\to}{OQ}\right| = \text{radius}$, so$R$will lie on angle bisector of$\overset{\to}{OQ} and \overset{\to}{OP}$</p><p>Now using the perpendicular condition we get,</p><p>$\overset{\to}{OQ} \cdot \overset{\to}{OP} = 0$</p><p>$⇒ \overset{\to}{OQ} \cdot \overset{\to}{OP} = \alpha{\left|\overset{\to}{u}\right|}^{2} + \beta \cdot \left(\overset{\to}{v} \cdot \overset{\to}{u}\right) = 0$</p><p>$⇒ \alpha{\left|\overset{\to}{u}\right|}^{2} + \beta \cdot \left(\left|\overset{\to}{v}\right|\left|\overset{\to}{u}\right|cos{45}^{°}\right) = 0$</p><p>$⇒ \alpha + \beta \cdot cos{45}^{°} = 0\left\{\text{as}\left|\overset{\to}{u}\right| = \left|\overset{\to}{v}\right| = \text{radius}\right\}$</p><p>$⇒ \alpha = \dfrac{- \beta}{\sqrt{2}}$</p><p>Now solving,$\overset{\to}{OQ} \cdot \overset{\to}{OR} = \left|\overset{\to}{OQ}\right|\left|\overset{\to}{OR}\right|cos{45}^{°} = \dfrac{1}{\sqrt{2}}{r}^{2}$</p><p>$⇒ \left(\alpha\overset{\to}{v} + \beta\overset{\to}{v}\right) \cdot \left(\overset{\to}{v}\right) = \dfrac{{r}^{2}}{\sqrt{2}}$</p><p>$⇒ \alpha \cdot \dfrac{{r}^{2}}{\sqrt{2}} + \beta \cdot {r}^{2} = \dfrac{{r}^{2}}{\sqrt{2}}$</p><p>$⇒ \dfrac{\alpha}{\sqrt{2}} + \beta = \dfrac{1}{\sqrt{2}}$</p><p>$⇒ \beta = \sqrt{2} \left\{\text{as}\alpha = \dfrac{- \beta}{\sqrt{2}}\right\}$</p><p>So,$\alpha = - 1$</p><p>Now finding the equation with roots$\alpha = - 1, {\beta}^{2} = 2$we get,${x}^{2} - x - 2 = 0