An arc PQ of a circle subtends a right angle at its centre O. The mid point of the arc PQ i R. If OP→=u→,OR¯=v→ and OQ¯=αu→+βv→, then α,β2 are the roots of the equation

OQ→=αu→+βv→OR=v→OP→=u→|u→|=|v→|=|αu→+βv→|(u→)⋅(αu→+βv→)=0u→.v→=|u||v|cos⁡45∘α=−β2=|αu→+βv→|=rα2+β2+2αβ=1α=−1,β2=2∴ α,β2 are the roots of the equation x2−x−2=0

An arc PQ of a circle subtends a right angle at its centre O. The mid point of the arc PQ i R. If OP→=u→,OR¯=v→ and OQ¯=αu→+βv→, then α,β2 are the roots of the equation

OQ→=αu→+βv→OR=v→OP→=u→|u→|=|v→|=|αu→+βv→|(u→)⋅(αu→+βv→)=0u→.v→=|u||v|cos⁡45∘α=−β2=|αu→+βv→|=rα2+β2+2αβ=1α=−1,β2=2∴ α,β2 are the roots of the equation x2−x−2=0

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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 i $R. If \vec{OP} = \vec{u}, \bar{OR} = \vec{v}$
$and \bar{OQ} = α \vec{u} + β \vec{v}, then α, β^{2} are the roots of the equation$

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

$x^{2} + x - 2 = 0$

$x^{2} - x - 2 = 0$

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

answer is B.

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Detailed Solution

$\begin{aligned} \vec{O Q} = α \vec{u} + β \vec{v} \\ O R = \vec{v} \\ \vec{O P} = \vec{u} \\ \begin{aligned} | \vec{u} | = | \vec{v} | = | α \vec{u} + β \vec{v} | \\ (\vec{u}) \cdot (α \vec{u} + β \vec{v}) = 0 \\ \vec{u} . \vec{v} = | u | | v | \cos 45^{\circ} \\ α = - \frac{β}{\sqrt{2}} \\ = | α \vec{u} + β \vec{v} | = r \\ α^{2} + β^{2} + \sqrt{2} α β = 1 \\ α = - 1, β^{2} = 2 \end{aligned} \end{aligned}$