Key Concepts and Formulas

• Position Vectors and Coordinates: A position vector $\vec{r} = x\widehat{i} + y\widehat{j}$ corresponds to a point $(x, y)$ in the Cartesian plane.
• Dot Product and Angle Between Vectors: For vectors $\vec{u}$ and $\vec{v}$, $\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta$, where $\theta$ is the angle between them.
• Unit Vector: A unit vector in the direction of $\vec{u}$ is $\widehat{u} = \frac{\vec{u}}{|\vec{u}|}$.
• Angle Bisector of Two Vectors: The direction vector of the angle bisector of two vectors $\vec{u}$ and $\vec{v}$ is given by $\widehat{u} + \widehat{v}$ for the acute angle bisector and $\widehat{u} - \widehat{v}$ (or $\widehat{v} - \widehat{u}$) for the obtuse angle bisector.
• Equation of a Line Through Origin: A line passing through the origin with direction vector $d_x \widehat{i} + d_y \widehat{j}$ has the Cartesian equation $d_y x - d_x y = 0$.
• Distance from a Point to a Line: The distance $D$ from a point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is $D = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

Step-by-Step Solution

Step 1: Define the given position vectors and calculate their magnitudes. We are given the position vectors of points A, B, and C with respect to the origin O. $\vec{a} = \vec{OA} = \sqrt{3} \widehat{i} + \widehat{j}$ $\vec{b} = \vec{OB} = \widehat{i} + \sqrt{3} \widehat{j}$ $\vec{c} = \vec{OC} = \beta \widehat{i} + (1 - \beta) \widehat{j}$

We need to calculate the magnitudes of $\vec{a}$ and $\vec{b}$ to find their unit vectors. $|\vec{a}| = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2$. $|\vec{b}| = \sqrt{(1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$.

Step 2: Determine the angle between OA and OB and identify the acute angle bisector. To find the angle between $\vec{OA}$ and $\vec{OB}$, we use the dot product. $\vec{a} \cdot \vec{b} = (\sqrt{3})(1) + (1)(\sqrt{3}) = \sqrt{3} + \sqrt{3} = 2\sqrt{3}$. The cosine of the angle $\theta$ between $\vec{OA}$ and $\vec{OB}$ is: $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{2\sqrt{3}}{(2)(2)} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$. Since $\cos \theta = \frac{\sqrt{3}}{2}$, the angle $\theta = 30^\circ$. This is an acute angle. For the acute angle bisector, the direction vector is the sum of the unit vectors along $\vec{OA}$ and $\vec{OB}$.

Step 3: Find the unit vectors and the direction vector of the acute angle bisector. The unit vector along $\vec{OA}$ is: $\widehat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{\sqrt{3} \widehat{i} + \widehat{j}}{2}$. The unit vector along $\vec{OB}$ is: $\widehat{b} = \frac{\vec{b}}{|\vec{b}|} = \frac{\widehat{i} + \sqrt{3} \widehat{j}}{2}$. The direction vector of the acute angle bisector is: $\vec{d} = \widehat{a} + \widehat{b} = \frac{\sqrt{3} \widehat{i} + \widehat{j}}{2} + \frac{\widehat{i} + \sqrt{3} \widehat{j}}{2} = \frac{(\sqrt{3}+1)\widehat{i} + (1+\sqrt{3})\widehat{j}}{2}$. We can simplify this direction vector by factoring out $\frac{\sqrt{3}+1}{2}$, giving a simpler direction vector $\widehat{i} + \widehat{j}$.

Step 4: Write the Cartesian equation of the acute angle bisector. The acute angle bisector passes through the origin $O(0,0)$ and has a direction vector $\widehat{i} + \widehat{j}$. If $(x, y)$ is a point on the line, then the vector $x\widehat{i} + y\widehat{j}$ is parallel to $\widehat{i} + \widehat{j}$. This implies that the slope of the line is $\frac{y}{x} = \frac{1}{1}$, so $y = x$. The Cartesian equation of the bisector line is $x - y = 0$.

Step 5: Determine the coordinates of point C and apply the distance formula. The position vector of point C is $\vec{c} = \beta \widehat{i} + (1 - \beta) \widehat{j}$. So, the coordinates of point C are $(x_C, y_C) = (\beta, 1 - \beta)$. The equation of the bisector line is $x - y = 0$. Here, $A=1$, $B=-1$, and $C=0$. The distance $D$ from point C to the bisector is given by the formula: $D = \frac{|Ax_C + By_C + C|}{\sqrt{A^2 + B^2}} = \frac{|(1)(\beta) + (-1)(1 - \beta) + 0|}{\sqrt{(1)^2 + (-1)^2}}$. $D = \frac{|\beta - 1 + \beta|}{\sqrt{1 + 1}} = \frac{|2\beta - 1|}{\sqrt{2}}$.

Step 6: Set the distance equal to the given value and solve for $\beta$. We are given that the distance of C from the bisector is $\frac{3}{\sqrt{2}}$. Therefore, we have the equation: $\frac{|2\beta - 1|}{\sqrt{2}} = \frac{3}{\sqrt{2}}$. Multiplying both sides by $\sqrt{2}$, we get: $|2\beta - 1| = 3$.

This absolute value equation leads to two possible cases:

Case 1: $2\beta - 1 = 3$ $2\beta = 3 + 1$ $2\beta = 4$ $\beta = 2$.

Case 2: $2\beta - 1 = -3$ $2\beta = -3 + 1$ $2\beta = -2$ $\beta = -1$.

The possible values of $\beta$ are $2$ and $-1$.

Step 7: Calculate the sum of all possible values of $\beta$. The sum of the possible values of $\beta$ is $2 + (-1) = 1$.

Common Mistakes & Tips

• Acute vs. Obtuse Angle Bisector: Carefully determine if the angle between the two vectors is acute or obtuse. The problem specifies the acute angle bisector, which is crucial. If the angle were obtuse, the direction vector would be $\widehat{a} - \widehat{b}$ or $\widehat{b} - \widehat{a}$.
• Equation of the Line: Ensure the equation of the line is in the form $Ax + By + C = 0$ for the distance formula. A line through the origin with direction $d_x \widehat{i} + d_y \widehat{j}$ has equation $d_y x - d_x y = 0$.
• Absolute Value Equation: Remember that $|x| = k$ implies $x = k$ or $x = -k$. Both possibilities must be considered when solving for $\beta$.

Summary

The problem required finding the equation of the acute angle bisector of the vectors $\vec{OA}$ and $\vec{OB}$. This involved calculating unit vectors, summing them to get the direction of the bisector, and then converting this direction into a Cartesian line equation. Subsequently, the distance from point C to this line was calculated using the standard formula and set equal to the given distance. Solving the resulting absolute value equation yielded two possible values for $\beta$, and their sum was computed.

The final answer is $\boxed{1}$.