An arc P Q of a circle subtends a right angle at its centre O . The mid point of the arc P Q is R . If O P…

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 $\overrightarrow{OP}=\overrightarrow{u}, \overrightarrow{OR}=\overrightarrow{v}$ and $\overrightarrow{OQ}=\alpha \overrightarrow{u}+\beta \overrightarrow{v}$, 

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

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

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

Now $\left|\overrightarrow{OP }\right|=\left|\overrightarrow{OQ }\right|=\text{radius}$, so $R$ will lie on angle bisector of $\overrightarrow{OQ}$ and $\overrightarrow{OP}$

Now using the perpendicular condition we get,

$\overrightarrow{OQ}·\overrightarrow{OP}=0$

$\Rightarrow \overrightarrow{OQ}·\overrightarrow{OP}=\alpha {\left|\overrightarrow{u}\right|}^2+\beta ·\left(\overrightarrow{v}·\overrightarrow{u}\right)=0$

$\Rightarrow \alpha {\left|\overrightarrow{u}\right|}^2+\beta ·\left(\left|\overrightarrow{v}\right|\left|\overrightarrow{u}\right|\cos {45}^{°}\right)=0$

$\Rightarrow \alpha +\beta ·\cos {45}^{°}=0 \left\{\text{as} \left|\overrightarrow{u}\right|=\left|\overrightarrow{v}\right|=\text{radius}\right\}$

$\Rightarrow \alpha =\frac{-\beta }{\sqrt{2}}$

Now solving, $\overrightarrow{OQ}·\overrightarrow{OR}=\left|\overrightarrow{OQ}\right|\left|\overrightarrow{OR}\right|\cos {45}^{°}=\frac{1}{\sqrt{2}}{r}^2$

$\Rightarrow \left(\alpha \overrightarrow{v}+\beta \overrightarrow{v}\right)·\left(\overrightarrow{v}\right)=\frac{r^2}{\sqrt{2}}$

$\Rightarrow \alpha ·\frac{r^2}{\sqrt{2}}+\beta ·r^2=\frac{r^2}{\sqrt{2}}$

$\Rightarrow \frac{\alpha }{\sqrt{2}}+\beta =\frac{1}{\sqrt{2}}$

$\Rightarrow \beta =\sqrt{2}$ $\left\{\text{as} \alpha =\frac{-\beta }{\sqrt{2}}\right\}$

So, $\alpha =-1$

Now finding the equation with roots $\alpha =-1, {\beta }^2=2$ we get, $x^2-x-2=0$