JEE Main 2025 Jan 23 Shift 1, Mathematics Q7: Scalar (or Dot) Product of Two Vectors

The total arc subtends $90^{\circ }$ at $O$. The ratio $1:5$ splits this into $\angle AOB=15^{\circ }$ and $\angle BOC=75^{\circ }$. Let $\vec{a}=\overrightarrow{OA}$, $\vec{b}=\overrightarrow{OB}$, $\vec{c}=\overrightarrow{OC}$, with $|\vec{a}|=|\vec{b}|=|\vec{c}|=r$.

Take dot products of $\vec{c}=\alpha \vec{a}+\beta \vec{b}$ with $\vec{a}$ and $\vec{b}$, using $\vec{a}\cdot \vec{c}=r^2\cos 90^{\circ }=0$ and $\vec{a}\cdot \vec{b}=r^2\cos 15^{\circ }$, $\vec{b}\cdot \vec{c}=r^2\cos 75^{\circ }$:

$0=\alpha +\beta \cos 15^{\circ }$ ... (i)

$\cos 75^{\circ }=\alpha \cos 15^{\circ }+\beta$ ... (ii)

From (i), $\alpha =-\beta \cos 15^{\circ }$. Substituting into (ii) and using $\cos 75^{\circ }=\sin 15^{\circ }$:

$\sin 15^{\circ }=-\beta {\cos }^{2}15^{\circ }+\beta =\beta {\sin }^{2}15^{\circ }\Rightarrow \beta =\frac{1}{\sin 15^{\circ }}$.

With $\sin 15^{\circ }=\frac{\sqrt{3}-1}{2\sqrt{2}}$: $\beta =\frac{2\sqrt{2}}{\sqrt{3}-1}$, and $\alpha =-\beta \cos 15^{\circ }=-\frac{\sqrt{3}+1}{\sqrt{3}-1}$.

Now compute $\alpha +\sqrt{2}(\sqrt{3}-1)\beta =\frac{-(\sqrt{3}+1)}{\sqrt{3}-1}+\frac{\sqrt{2}(\sqrt{3}-1)\cdot 2\sqrt{2}}{\sqrt{3}-1}=\frac{-(\sqrt{3}+1{)}^2+8}{2}=\frac{-(4+2\sqrt{3})+8}{2}=2-\sqrt{3}$.