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  • An arc PQ of a circle subtends a right angle at its centre O. The mid point of the arc PQ is R. If <img alt="\overrightarrow{\mathrm{OP}}=\overrightarrow{\mathrm{u}}, \overrightarrow{\mathrm{OR}}=\

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{\mathrm{OP}}=\overrightarrow{\mathrm{u}}, \overrightarrow{\mathrm{OR}}=\overrightarrow{\mathrm{v}} and  \overrightarrow{\mathrm{OQ}}=\alpha \overrightarrow{\mathrm{u}}+\beta \overrightarrow{\mathrm{v}},, then \alpha ,\beta ^{2} are the roots of the equation :

Option: 1

3x^{2}-2x-1=0


Option: 2

3x^{2} +2x-1=0


Option: 3

x^{2}-x-2=0


Option: 4

x^{2}+x-2=0


Answers (1)

best_answer

Let \begin{aligned} & \text { Let } \overrightarrow{\mathrm{OP}}=\overrightarrow{\mathrm{u}}=\hat{\mathrm{i}} \\ & \overrightarrow{\mathrm{OQ}}=\overrightarrow{\mathrm{q}}=\hat{\mathrm{j}} \end{aligned}

\because R is the mid point of  \overrightarrow{\mathrm{PQ}}

Then \overrightarrow{O R}=\vec{v}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}

Now

\begin{aligned} & \overrightarrow{\mathrm{OQ}}=\alpha \vec{u}+\beta \overrightarrow{\mathrm{v}} \\ & \hat{\mathrm{j}}=\alpha \hat{\mathrm{i}}+\beta\left(\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{j}}\right) \\ & \beta=\sqrt{2}, \alpha+\frac{\beta}{\sqrt{2}}=0 \Rightarrow \alpha=-1 \end{aligned}

Now equation

\begin{aligned} & x^2-\left(\alpha+\beta^2\right) x+\alpha \beta^2=0 \\ & x^2-(-1+2) x+(-1)(2)=0 \\ & x^2-x-2=0 \end{aligned}

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