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. The mid point of the arc PQ is R. If $$\overrightarrow {OP}  = \overrightarrow u ,\overrightarrow {OR}  = \overrightarrow v $$

Accepted Answer

An arc $PQ$ of a circle subtends a right angle at its centre $O$. The mid-point of an $arcPQ$ is $R$. So, $PR=RQ$

Also given that, $\overset{	o \leftarrow }{OP}=\overset{	o }{u},\overset{	o \leftarrow }{OR}=\overset{	o }{v}$, and $\overset{	o \leftarrow }{OQ}=\alpha \overset{	o }{u}+\beta \overset{	o }{v}$

Let $\overset{	o \leftarrow }{OP}=\overset{	o }{u}$

$\begin{gathered} \overset{	o \leftarrow }{OQ}=\overset{	o }{q}=\alpha \overset{	o }{u}+\beta \overset{	o }{v} \ \overset{	o \leftarrow }{OR}=\overset{	o }{v} \end{gathered}$

Clearly, $\overset{	o }{q}=\hat{j}$

$\overset{	o }{u}=\hat{i},\overset{	o }{v}=\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

Now, given, $\overset{	o }{q}=\alpha \overset{	o }{u}+\beta \overset{	o }{v}$

$\begin{gathered} \Rightarrow \hat{j}=\alpha (\hat{i})+\beta (\frac{\hat{i}+\hat{j}}{\sqrt{2}}) \ \ \Rightarrow \hat{j}=i(\alpha +\frac{\beta }{\sqrt{2}})+\frac{\beta }{\sqrt{2}}\hat{j} \end{gathered}$

On comparing both sides, we get

$\begin{gathered} \alpha +\frac{\beta }{\sqrt{2}}=0\frac{\beta }{\sqrt{2}}=1 \ \Rightarrow \alpha =-\frac{\beta }{\sqrt{2}}\, 	herefore \beta =\sqrt{2} \ \Rightarrow \alpha =-1,{\beta }^{2}=2 \end{gathered}$

Now, $\alpha $ and ${\beta }^{2}$ are roots of the quadratic equation is ${x}^{2}-x-2=0$

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