Let the arc AC of a circle subtend a right angle at the centre O. If the point B on the arc AC, divides the arc AC such that $\frac{\text{length of arc AB}}{\text{length of arc BC}} = \frac{1}{5}$ and $\vec{OC} = \alpha \vec{OA} + \beta \vec{OB}$, then $\alpha + \sqrt{2}(\sqrt{3}-1)\beta$ is equal to
- (1) $2-\sqrt{3}$
- (2) $2\sqrt{3}$
- (3) $5\sqrt{3}$
- (4) $2+\sqrt{3}$
Solution:
Let $|\vec{OA}| = |\vec{OB}| = |\vec{OC}| = R$.
Angle $\angle AOC = 90^\circ$. Given arc ratio $1:5 \implies \angle AOB = 15^\circ$ and $\angle BOC = 75^\circ$.
Given equation: $\vec{c} = \alpha \vec{a} + \beta \vec{b}$.
Taking dot product with $\vec{a}$:
$$\vec{a}\cdot\vec{c} = \alpha |\vec{a}|^2 + \beta \vec{a}\cdot\vec{b}$$
$$0 = \alpha R^2 + \beta R^2 \cos 15^\circ \implies \alpha = -\beta \cos 15^\circ$$
Taking dot product with $\vec{b}$:
$$\vec{b}\cdot\vec{c} = \alpha \vec{a}\cdot\vec{b} + \beta |\vec{b}|^2$$
$$R^2 \cos 75^\circ = \alpha R^2 \cos 15^\circ + \beta R^2$$
Substitute $\alpha = -\beta \cos 15^\circ$:
$$\sin 15^\circ = (-\beta \cos 15^\circ)\cos 15^\circ + \beta$$
$$\sin 15^\circ = \beta (1 – \cos^2 15^\circ) = \beta \sin^2 15^\circ$$
$$\therefore \beta = \frac{1}{\sin 15^\circ} = \frac{2\sqrt{2}}{\sqrt{3}-1}$$
Then, $\alpha = -\frac{\cos 15^\circ}{\sin 15^\circ} = -\cot 15^\circ = -(2+\sqrt{3})$.
Required Value:
$$\alpha + \sqrt{2}(\sqrt{3}-1)\beta = -(2+\sqrt{3}) + \sqrt{2}(\sqrt{3}-1) \cdot \frac{2\sqrt{2}}{\sqrt{3}-1}$$
$$= -2-\sqrt{3} + 4 = 2-\sqrt{3}$$
Ans.(1)