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

Question

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

Accepted Answer

Take $\vec u=\hat i$, $\vec{OQ}=\hat j$ (right angle at $O$); $R$ (midpoint of arc) gives $\vec v = \dfrac1{\sqrt2}\hat i+\dfrac1{\sqrt2}\hat j$.

$$\hat j = \alpha\hat i + \beta\left(\frac1{\sqrt2}\hat i+\frac1{\sqrt2}\hat j\right) \implies \frac{\beta}{\sqrt2}=1 \Rightarrow \beta=\sqrt2,\qquad \alpha+\frac{\beta}{\sqrt2}=0 \Rightarrow \alpha=-1$$

Roots $\alpha=-1$, $\beta^2=2$: equation $x^2-(\alpha+\beta^2)x+\alpha\beta^2=0 \Rightarrow x^2-(1)x+(-2)=0 \Rightarrow x^2-x-2=0$.

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