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

Step 2: Solve for \( a \)

Equating both sides yields: \[ \left| a(1-a) \right| = 9 \] Removing the absolute value, consider both possibilities: \[ a(1-a) = 9 \quad \text{or} \quad a(1-a) = -9 \] Solving the first equation: \[ a^2 - a - 9 = 0 \] Applying the quadratic formula: \[ a = \frac{1 \pm \sqrt{(1)^2 + 4 \times 9}}{2} = \frac{1 \pm \sqrt{37}}{2} \] Solving the second equation: \[ a^2 - a + 9 = 0 \] Applying the quadratic formula: \[ a = \frac{1 \pm \sqrt{(1)^2 - 4 \times 9}}{2} = \frac{1 \pm \sqrt{-35}}{2} \] The solutions involving \( \sqrt{-35} \) are imaginary. Therefore, only the real roots are considered: \[ a = 5 \quad \text{or} \quad a = -4 \]

Step 3: Calculate the Sum of Values

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

Final Answer: 1