JEE Main 2025 — 23 January, Morning Shift — Question 7

$\vec{c}=\alpha \vec{a}+\beta \vec{b}$

$\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=\alpha \overrightarrow{\mathrm{a}} \cdot \vec{a}+\beta \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}$

$0=\alpha+\beta \cos 15^{\circ}$ (1) $\Rightarrow \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\alpha \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\beta \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{b}}$

$\Rightarrow \cos 75^{\circ}=\alpha \cos 15^{\circ}+\beta$

(2) & (3) $\Rightarrow \cos 75^{\circ}=-\beta \cos ^{2} 15^{\circ}+\beta$

$\beta=\frac{\cos 75^{\circ}}{\sin ^{2} 15^{\circ}}=\frac{1}{\sin 15^{\circ}}=\frac{2 \sqrt{2}}{\sqrt{3}-1}$

(2) $\Rightarrow \alpha=\frac{-\cos 15^{\circ}}{\sin 15^{\circ}}=\frac{-(\sqrt{3}+1)}{(\sqrt{3}-1)}$

$\therefore \overrightarrow{\mathrm{c}}=\frac{-(\sqrt{3}+1)}{(\sqrt{3}-1)} \overrightarrow{\mathrm{a}}+\left(\frac{2 \sqrt{2}}{\sqrt{3}-1}\right) \overrightarrow{\mathrm{b}}$

Now $\alpha+\sqrt{2}(\sqrt{3}-1) \beta=\frac{-(\sqrt{3}+1)}{(\sqrt{3}-1)}+\frac{\sqrt{2}(\sqrt{3}-1) \cdot 2 \sqrt{2}}{\sqrt{3}-1}$

$=\frac{-(\sqrt{3}+1)^{2}}{2}+4$

$=\frac{-3-1-2 \sqrt{3}+8}{2}$

$=2-\sqrt{3}$