Let $A$, $B$, $C$ be three points in $xy$ -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 $\vec{OA}$ and $\vec{OB}$ is $\frac{\theta}{\sqrt{2}}$, then the sum of all the possible values of $a$ is:

Step 1: Find the unit vectors and angle bisector

First, we find the unit vectors along $\vec{OA}$ and $\vec{OB}$. The magnitude of $\vec{OA}$ is $\sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2$, and similarly for $\vec{OB}$. The vector along the angle bisector of $\vec{OA}$ and $\vec{OB}$ is given by the sum of their unit vectors. We denote this bisector vector as $\vec{d}$.

Step 2: Determine the equation of the angle bisector line

From the previous step, we see that the angle bisector vector $\vec{d}$ is proportional to $\hat{i} + \hat{j}$. This means the line passes through the origin and has a slope of 1. Therefore, the equation of the line bisecting the angle between $\vec{OA}$ and $\vec{OB}$ is $x - y = 0$.

Step 3: Calculate the distance from point C to the line

The position vector of point $C$ is $a\hat{i} + (1-a)\hat{j}$, so its coordinates are $(a, 1-a)$. We use the formula for the distance from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$, which is $\frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$. Substituting the coordinates of $C$ and the line equation $x - y = 0$, we get the distance as $\frac{|a - (1-a)|}{\sqrt{1^2 + (-1)^2}}$.

Step 4: Solve for 'a' using the given distance

The problem states that the distance of point $C$ from the line is $\frac{3}{\sqrt{2}}$. We equate our calculated distance to this given value. This leads to an absolute value equation, $|2a - 1| = 3$, which can be split into two linear equations: $2a - 1 = 3$ and $2a - 1 = -3$.

Step 5: Find the possible values of 'a' and their sum

Solving the first equation, $2a - 1 = 3$, we get $2a = 4$, so $a = 2$. Solving the second equation, $2a - 1 = -3$, we get $2a = -2$, so $a = -1$. The sum of all possible values of $a$ is $2 + (-1) = 1$.

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