JEE Main 2023 — Mathematics Vectors & 3D Geometry

Given,

An arc $PQ$PQ of a circle subtends a right angle at its centre $O$O. The mid point of the arc $PQ$PQ is $R$R,

And $\vec{OP}=\vec{u},\vec{OR}=\vec{v}$OP=u,OR=v and $\vec{OQ}=\alpha \vec{u}+\beta \vec{v}$OQ​=αu+βv,

So, plotting the diagram of the given value we get,

Now given, $\vec{OQ}=\alpha \vec{u}+\beta \vec{v}$OQ​=αu+βv

Where, $\vec{OR}=\vec{v}$OR=v and $\vec{OP}=\vec{u}$OP=u

Now $|\vec{OP}|=|\vec{OQ}|=\text{radius}$∣OP∣=∣OQ​∣=radius, so $R$R will lie on angle bisector of $\vec{OQ}$OQ​ and $\vec{OP}$OP

Now using the perpendicular condition we get,

$$\vec{OQ}\cdot \vec{OP}=0$$OQ​⋅OP=0

$$\Rightarrow \vec{OQ}\cdot \vec{OP}=\alpha {|\vec{u}|}^{2}+\beta \cdot (\vec{v}\cdot \vec{u})=0$$⇒OQ​⋅OP=α∣u∣2+β⋅(v⋅u)=0

$$\Rightarrow \alpha {|\vec{u}|}^{2}+\beta \cdot (|\vec{v}||\vec{u}|\mathrm{cos}{45}^{^{\circ}})=0$$⇒α∣u∣2+β⋅(∣v∣∣u∣cos45∘)=0

$$\Rightarrow \alpha +\beta \cdot \mathrm{cos}{45}^{^{\circ}}=0{\text{as}|\vec{u}|=|\vec{v}|=\text{radius}}$$⇒α+β⋅cos45∘=0as∣u∣=∣v∣=radius

$$\Rightarrow \alpha =\frac{-\beta }{\sqrt{2}}$$⇒α=2​−β​

Now solving, $\vec{OQ}\cdot \vec{OR}=|\vec{OQ}||\vec{OR}|\mathrm{cos}{45}^{^{\circ}}=\frac{1}{\sqrt{2}}{r}^{2}$OQ​⋅OR=∣OQ​∣∣OR∣cos45∘=2​1​r2

$$\Rightarrow (\alpha \vec{v}+\beta \vec{v})\cdot (\vec{v})=\frac{{r}^{2}}{\sqrt{2}}$$⇒(αv+βv)⋅(v)=2​r2​

$$\Rightarrow \alpha \cdot \frac{{r}^{2}}{\sqrt{2}}+\beta \cdot {r}^{2}=\frac{{r}^{2}}{\sqrt{2}}$$⇒α⋅2​r2​+β⋅r2=2​r2​

$$\Rightarrow \frac{\alpha }{\sqrt{2}}+\beta =\frac{1}{\sqrt{2}}$$⇒2​α​+β=2​1​

$\Rightarrow \beta =\sqrt{2}$⇒β=2​ ${\text{as}\alpha =\frac{-\beta }{\sqrt{2}}}$asα=2​−β​

So, $\alpha =-1$α=−1

Now finding the equation with roots $\alpha =-1,{\beta }^{2}=2$α=−1,β2=2 we get, ${x}^{2}-x-2=0$x2−x−2=0