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Step 1: Interpret the given vectors and the angle bisector
• The position vector of A is given by $ \sqrt{3}\,\hat{i} + \hat{j} $.
• The position vector of B is given by $ \hat{i} + \sqrt{3}\,\hat{j} $.
• The position vector of C is $ \beta \,\hat{i} + (1-\beta)\,\hat{j} $.
• We are told that the distance of C from the bisector of the acute angle between OA and OB is $ \frac{3}{\sqrt{2}} $.
Step 2: Find the equation of the bisector of the acute angle
The vectors of OA and OB suggest that the lines OA and OB meet at the origin, with directions A($\sqrt{3}$, 1) and B(1, $\sqrt{3}$). The bisector of the acute angle between them in the $xy$-plane can be found as the line passing through the origin with equal angles to vectors A and B.
The acute angle bisector in this case simplifies to the line:
$$
x - y = 0 \quad \text{or} \quad x = y.
$$
Step 3: Express the coordinates of point C
Point C, with position vector $ \beta \,\hat{i} + (1-\beta)\,\hat{j} $, has coordinates $(\beta,\,1-\beta)$.
Step 4: Formula for the distance from a point to a line
The distance $d$ of a point $(x_1, y_1)$ from a line $ax + by + c = 0$ is given by:
$$
d = \frac{\lvert a x_1 + b y_1 + c \rvert}{\sqrt{a^2 + b^2}}.
$$
Here, the line is $x - y = 0$, which can be written as $1 \cdot x + (-1) \cdot y + 0 = 0$.
Thus, for the line $x - y = 0$, $a = 1, b = -1, c = 0$.
Step 5: Apply the distance formula
Substitute $x_1 = \beta$ and $y_1 = 1 - \beta$ into the distance formula:
$$
\frac{\big\lvert \beta - (1-\beta)\big\rvert}{\sqrt{1^2 + (-1)^2}} \;=\; \frac{3}{\sqrt{2}}.
$$
So,
$$
\frac{\big\lvert \beta - 1 + \beta \big\rvert}{\sqrt{2}}
= \frac{3}{\sqrt{2}}
\quad\Longrightarrow\quad
\frac{\big\lvert 2\beta - 1 \big\rvert}{\sqrt{2}}
= \frac{3}{\sqrt{2}}.
$$
Cancelling the common factor $1/\sqrt{2}$, we get:
$$
\lvert 2\beta - 1 \rvert = 3.
$$
Step 6: Solve for β
The equation $|2\beta - 1| = 3$ gives two possible solutions:
$$
2\beta - 1 = 3
\quad\text{or}\quad
2\beta - 1 = -3.
$$
Solving these separately:
If $2\beta - 1 = 3$, then $2\beta = 4 \implies \beta = 2.$
If $2\beta - 1 = -3$, then $2\beta = -2 \implies \beta = -1.$
Step 7: Find the sum of all possible values of β
The possible values of β are $2$ and $-1$. Their sum is:
$$
2 + (-1) = 1.
$$
Final Answer
The sum of all possible values of $ \beta $ is 1.
