An arc PQ of a circle subtends a right angle | vedclass

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Question

(A) Let the radius of the circle be $r$. Then $|\vec{u}| = |\vec{v}| = |\vec{OQ}| = r$.Since $\angle POQ = 90^{\circ}$ and $R$ is the midpoint of arc

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$x^2-x-2=0$

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(A) Let the radius of the circle be $r$. Then $|\vec{u}| = |\vec{v}| = |\vec{OQ}| = r$.Since $\angle POQ = 90^{\circ}$ and $R$ is the midpoint of arc $PQ$,$\angle POR = \angle ROQ = 45^{\circ}$.Given $\vec{OQ} = \alpha \vec{u} + \beta \vec{v}$.Taking dot product with $\vec{u}$:$\vec{u} \cdot \vec{OQ} = \alpha |\vec{u}|^2 + \beta (\vec{u} \cdot \vec{v})$Since $\angle POQ = 90^{\circ}$,$\vec{u} \cdot \vec{OQ} = 0$. Also $\vec{u} \cdot \vec{v} = r^2 \cos 45^{\circ} = \frac{r^2}{\sqrt{2}}$.$0 = \alpha r^2 + \beta \frac{r^2}{\sqrt{2}} \implies \alpha = -\frac{\beta}{\sqrt{2}} \implies \alpha^2 = \frac{\beta^2}{2}$.Now,$|\vec{OQ}|^2 = r^2 = |\alpha \vec{u} + \beta \vec{v}|^2 = \alpha^2 r^2 + \beta^2 r^2 + 2\alpha \beta (\vec{u} \cdot \vec{v})$.$1 = \alpha^2 + \beta^2 + 2\alpha \beta \frac{1}{\sqrt{2}} = \alpha^2 + \beta^2 + \sqrt{2} \alpha \beta$.Substituting $\alpha = -\frac{\beta}{\sqrt{2}}$:$1 = \frac{\beta^2}{2} + \beta^2 + \sqrt{2} (-\frac{\beta}{\sqrt{2}}) \beta = \frac{3\beta^2}{2} - \beta^2 = \frac{\beta^2}{2} \implies \beta^2 = 2$.Then $\alpha^2 = \frac{2}{2} = 1$,so $\alpha = -1$ (since $\alpha = -\frac{\beta}{\sqrt{2}}$ and $\beta = \sqrt{2}$).The roots are $\alpha = -1$ and $\beta^2 = 2$.The quadratic equation is $(x - (-1))(x - 2) = (x+1)(x-2) = x^2 - x - 2 = 0$.

An arc $PQ$ of a circle subtends a right angle at its centre $O$. The midpoint of the arc $PQ$ is $R$. If $\vec{OP}=\vec{u}$,$\vec{OR}=\vec{v}$ and $\vec{OQ}=\alpha \vec{u}+\beta \vec{v}$,then $\alpha, \beta^2$ are the roots of the equation

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