Question

Let the arc AC of a circle subtend a right angle at the center 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 \[ \overrightarrow{OC} = \alpha \overrightarrow{OA} + \beta \overrightarrow{OB}, \] then \( \alpha = \sqrt{2} (\sqrt{3}-1) \beta \) is equal to:

• A. \( 2 - \sqrt{3} \)
• B. \( 2 \sqrt{3} \)
• C. \( 5 \sqrt{3} \)
• D. \( 2 + \sqrt{3} \)

Hint:

Vector projections and trigonometric identities are essential tools in solving geometric vector problems.

Answer 1

The Correct Option is A

 

Step 1: Expressing the relation. \[ \overrightarrow{c} = \alpha \overrightarrow{a} + \beta \overrightarrow{b} \] 

Step 2: Dot product condition. \[ \overrightarrow{a} \cdot \overrightarrow{c} = \alpha (\overrightarrow{a} \cdot \overrightarrow{a}) + \beta (\overrightarrow{b} \cdot \overrightarrow{a}) \] \[ 0 = \alpha + \beta \cos 15^\circ \] \[ \Rightarrow \alpha = -\beta \cos 15^\circ. \] 

Step 3: Solving for \( \alpha \) and \( \beta \). \[ \cos 75^\circ = \alpha \cos 15^\circ + \beta \] \[ \beta = \frac{\cos 75^\circ}{\sin 15^\circ} = \frac{1}{\sqrt{3}-1} \frac{2\sqrt{2}}{\sqrt{3}-1} \] 

Step 4: Computing \( \alpha + \sqrt{2} (\sqrt{3}-1) \beta \). \[ \alpha + \sqrt{2} (\sqrt{3}-1) \beta = (-\frac{\sqrt{3}+1}{\sqrt{3}-1}) + \frac{\sqrt{2} (\sqrt{3}-1) 2\sqrt{2}}{\sqrt{3}-1} \] \[ = -\frac{\sqrt{3}+1}{2} + 4 \] \[ = 2 - \sqrt{3}. \]