Question

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 \(\frac{\text { length of } \operatorname{arc} \mathrm{AB}}{\text { length of } \operatorname{arc} \mathrm{BC}}=\frac{1}{5},\) and \(\overrightarrow{\mathrm{OC}}=\alpha \overrightarrow{\mathrm{OA}}+\beta \overrightarrow{\mathrm{OB}},\) then \(\alpha=\sqrt{2}(\sqrt{3}-1) \beta\) is equal to

(1) \(2-\sqrt{3}\)

(2) \(2 \sqrt{3}\)

(3) \(5 \sqrt{3}\)

(4) \(2+\sqrt{3}\)

Answer 1

Correct option is (1) \(2-\sqrt{3}\)  

\(\vec{c}=\alpha \vec{a}+\beta \vec{b}\)    ...(1)

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

\(0=\alpha+\beta \cos 15^{\circ}\)     ....(2)

\((1) \Rightarrow \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\alpha \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\beta \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{b}}\)      

\(\Rightarrow \cos 75^{\circ}=\alpha \cos 15^{\circ}+\beta\)    ...(3)

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

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

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

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

Now  

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

\(=\frac{-(\sqrt{3}+1)^{2}}{2}+4\)   

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

\(=2-\sqrt{3}\)