An arc PQ of a circle subtends a right angle at its centre O. The mid point of the arc PQ is R. If O P = u , O R = v and O Q= u+ v , then 2 are the roots of the equation [2023]

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 $\vec{O P} = \vec{u}$ , $\vec{O R} = \vec{v}$ and $\vec{O Q} = α \vec{u} + β \vec{v}$ , then $α, β^{2}$ are the roots of the equation [2023]

Smart Achievers · Accepted Answer

(3)

$We have, | \vec{u} | = | \vec{v} | = | α \vec{u} + β \vec{v} | = r$

$Also, (\vec{u}) \cdot (α \vec{u} + β \vec{v}) = 0$

$Now, \vec{u} \cdot \vec{v} = | \vec{u} | | \vec{v} | \cos 45 ° = \frac{r^{2}}{\sqrt{2}}$

$∴$ $α | \vec{u} |^{2} + β \vec{u} \cdot \vec{v} = 0$

$\Rightarrow r^{2} α + β \frac{r^{2}}{\sqrt{2}} = 0$

$\Rightarrow α + β \times \frac{1}{\sqrt{2}} = 0 \Rightarrow α = - \frac{β}{\sqrt{2}} [∵ r \neq 0]$

$Also, | α \vec{u} + β \vec{v} |^{2} = r^{2}$ $\Rightarrow α^{2} | \vec{u} |^{2} + β^{2} | \vec{v} |^{2} + 2 α β \vec{u} \cdot \vec{v} = r^{2}$

$\Rightarrow α^{2} + β^{2} + \sqrt{2} α β = 1$ $\Rightarrow β^{2} = 2 and α = - 1$

$∴$ $α, β^{2} are roots of equation x^{2} - x - 2 = 0$

Solution

Q.
An arc PQ of a circle subtends a right angle at its centre O. The mid point of the arc PQ is R. If $\vec{O P} = \vec{u}$ , $\vec{O R} = \vec{v}$ and $\vec{O Q} = α \vec{u} + β \vec{v}$ , then $α, β^{2}$ are the roots of the equation [2023]

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

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

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

4 $3 x^{2} - 2 x - 1 = 0$

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Solution

Q. An arc PQ of a circle subtends a right angle at its centre O. The mid point of the arc PQ is R. If OP→=u→, OR→=v→ and OQ→=αu→+βv→, then α,β2 are the roots of the equation [2023]

1 x2+x-2=0

2 3x2+2x-1=0

3 x2-x-2=0