Question

Let A, B, C be three points in the xy-plane, whose position vectors are given by \( \sqrt{3} \hat{i} + \hat{j} \), \( \hat{i} + \sqrt{3} \hat{j} \), and \( a\hat{i} + (1-a) \hat{j }\) respectively with respect to the origin \( O \). If the distance of the point C from the line bisecting the angle between the vectors \( \overrightarrow{OA} \) and \( \overrightarrow{OB} \) is \( \frac{9}{\sqrt{2}} \), then the sum of all possible values of \( a \) is:

• A. 1
• B. \( \frac{9}{2} \)
• C. 0
• D. 2

Hint:

The equation of an angle bisector can help in finding distances between points and lines.

Answer 1

The Correct Option is A

Step 1: Identify the Given Information

The equation of the angle bisector is: \[ x - y = 0 \] From the given condition: \[ \left| \frac{a(1-a)}{\sqrt{2}} \right| = \frac{9}{\sqrt{2}} \]

Step 2: Solve for \( a \)

Equating both sides: \[ \left| a(1-a) \right| = 9 \] Removing the absolute value and setting both possible conditions: \[ a(1-a) = 9 \quad \text{or} \quad a(1-a) = -9 \] Solving the first equation: \[ a^2 - a - 9 = 0 \] Using the quadratic formula: \[ a = \frac{1 \pm \sqrt{(1)^2 + 4 \times 9}}{2} = \frac{1 \pm \sqrt{37}}{2} \] Solving the second equation: \[ a^2 - a + 9 = 0 \] Using the quadratic formula: \[ a = \frac{1 \pm \sqrt{(1)^2 - 4 \times 9}}{2} = \frac{1 \pm \sqrt{-35}}{2} \] Since roots with \( \sqrt{-35} \) are imaginary, only the real roots remain: \[ a = 5 \quad \text{or} \quad a = -4 \]

Step 3: Calculate the Sum of Values

The sum of the valid \( a \)-values is: \[ 5 + (-4) = 1 \]

Final Answer: 1

Answer 2

Step 1: Write down the given vectors.
\[ \overrightarrow{OA} = \sqrt{3}\hat{i} + \hat{j}, \quad \overrightarrow{OB} = \hat{i} + \sqrt{3}\hat{j}, \quad \overrightarrow{OC} = a\hat{i} + (1 - a)\hat{j} \] We are told that the distance of point \( C \) from the line bisecting the angle between \( \overrightarrow{OA} \) and \( \overrightarrow{OB} \) is \( \frac{9}{\sqrt{2}} \).

Step 2: Find the equation of the angle bisector.
Let the direction ratios of \( \overrightarrow{OA} \) be proportional to \( (\sqrt{3}, 1) \) and for \( \overrightarrow{OB} \), they are proportional to \( (1, \sqrt{3}) \).
The line bisecting the angle between these two lines will make equal angles with both directions. Hence, its slope \( m \) satisfies: \[ \tan \theta = \frac{m_1 + m_2}{1 - m_1 m_2} \] where \( m_1 = \frac{1}{\sqrt{3}} \) and \( m_2 = \sqrt{3} \).
\[ m = \frac{\tan^{-1}(m_1) + \tan^{-1}(m_2)}{2} \Rightarrow \text{but we can find it by symmetry: since OA and OB are symmetric about the line } y = x. \] Thus, the bisector is \( y = x \).

Step 3: Distance of C from the bisector.
The distance of a point \( (x_1, y_1) \) from the line \( y = x \) is given by: \[ \frac{|x_1 - y_1|}{\sqrt{2}} \] For point \( C(a, 1 - a) \): \[ \text{Distance} = \frac{|a - (1 - a)|}{\sqrt{2}} = \frac{|2a - 1|}{\sqrt{2}} \] Given that this distance = \( \frac{9}{\sqrt{2}} \): \[ \frac{|2a - 1|}{\sqrt{2}} = \frac{9}{\sqrt{2}} \] \[ |2a - 1| = 9 \] \[ 2a - 1 = 9 \text{ or } 2a - 1 = -9 \] \[ a = 5 \text{ or } a = -4 \]

Step 4: Sum of all possible values of \( a \).
\[ a_1 + a_2 = 5 + (-4) = 1 \]

Final Answer:
\[ \boxed{1} \]