Key Concepts and Formulas

• Unit Vector: A unit vector in the direction of a vector $\vec{v}$ is $\hat{v} = \frac{\vec{v}}{|\vec{v}|}$.
• Direction of Angle Bisector: The direction vector of the internal angle bisector of two vectors $\vec{u}$ and $\vec{v}$ originating from the same point is proportional to $\frac{\vec{u}}{|\vec{u}|} + \frac{\vec{v}}{|\vec{v}|}$.
• Equation of a Line Through Origin: A line passing through the origin $(0,0)$ with a direction vector $d_x \hat{i} + d_y \hat{j}$ has the Cartesian equation $d_y x - d_x y = 0$.
• Distance of a Point from a Line: The distance $D$ of a point $(x_0, y_0)$ from a line $Ax+By+C=0$ is $D = \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$.

Step-by-Step Solution

Step 1: Calculate the magnitudes of $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OB}}$. This is necessary to find the unit vectors along these directions. $|\overrightarrow{\mathrm{OA}}| = |\sqrt{3}\hat{i} + \hat{j}| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = \sqrt{4} = 2$ $|\overrightarrow{\mathrm{OB}}| = |\hat{i} + \sqrt{3}\hat{j}| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1+3} = \sqrt{4} = 2$

Step 2: Find the unit vectors along $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OB}}$. The unit vectors are found by dividing each vector by its magnitude. $\hat{u}_{OA} = \frac{\overrightarrow{\mathrm{OA}}}{|\overrightarrow{\mathrm{OA}}|} = \frac{\sqrt{3}\hat{i} + \hat{j}}{2} = \frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}$ $\hat{u}_{OB} = \frac{\overrightarrow{\mathrm{OB}}}{|\overrightarrow{\mathrm{OB}}|} = \frac{\hat{i} + \sqrt{3}\hat{j}}{2} = \frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j}$

Step 3: Determine the direction vector of the angle bisector. The direction vector of the angle bisector is the sum of the unit vectors. $\vec{d}_{bisector} = \hat{u}_{OA} + \hat{u}_{OB} = \left(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\right) + \left(\frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j}\right)$ $\vec{d}_{bisector} = \left(\frac{\sqrt{3}+1}{2}\right)\hat{i} + \left(\frac{1+\sqrt{3}}{2}\right)\hat{j}$ We can simplify this direction vector by taking out a common factor of $\frac{\sqrt{3}+1}{2}$. The simplified direction vector is $\hat{i} + \hat{j}$.

Step 4: Find the equation of the angle bisector line. Since the original vectors are position vectors originating from O, the angle bisector line passes through the origin $(0,0)$. With a direction vector $\hat{i} + \hat{j}$ (i.e., $d_x=1, d_y=1$), the equation of the line is $1 \cdot y - 1 \cdot x = 0$, which simplifies to $y-x=0$, or $x-y=0$.

Step 5: Identify the coordinates of point C. The position vector of C is $\overrightarrow{\mathrm{OC}} = a \hat{i} + (1-a)\hat{j}$. Thus, the coordinates of C are $(x_C, y_C) = (a, 1-a)$.

Step 6: Calculate the distance of point C from the angle bisector line. The line is $x-y=0$, so $A=1$, $B=-1$, $C=0$. The point is $(a, 1-a)$. The distance $D$ is given by the formula: $D = \frac{|A x_C + B y_C + C|}{\sqrt{A^2+B^2}} = \frac{|1 \cdot a + (-1) \cdot (1-a) + 0|}{\sqrt{1^2 + (-1)^2}}$ $D = \frac{|a - (1-a)|}{\sqrt{1+1}} = \frac{|a - 1 + a|}{\sqrt{2}} = \frac{|2a-1|}{\sqrt{2}}$

Step 7: Set up and solve the equation for 'a'. We are given that the distance is $\frac{9}{\sqrt{2}}$. $\frac{|2a-1|}{\sqrt{2}} = \frac{9}{\sqrt{2}}$ Multiplying both sides by $\sqrt{2}$, we get: $|2a-1| = 9$ This absolute value equation yields two possibilities: Case 1: $2a-1 = 9$ $2a = 10$ $a = 5$

Case 2: $2a-1 = -9$ $2a = -8$ $a = -4$

Step 8: Calculate the sum of all possible values of 'a'. The possible values of $a$ are $5$ and $-4$. Sum of possible values of $a = 5 + (-4) = 1$.

Common Mistakes & Tips

• Internal vs. External Bisector: The problem typically implies the internal angle bisector. If the external bisector were intended, the direction vector would be $\frac{\vec{u}}{|\vec{u}|} - \frac{\vec{v}}{|\vec{v}|}$.
• Absolute Value: Always remember to solve absolute value equations by considering both positive and negative cases.
• Line Equation: Ensure the line equation is in the standard form $Ax+By+C=0$ before applying the distance formula.

Summary

The problem requires finding the equation of the line that bisects the angle between two given vectors originating from the origin. This involves calculating unit vectors, summing them to get the direction of the bisector, and then forming the Cartesian equation of the line. Subsequently, the distance of point C from this line is used to form an absolute value equation in terms of 'a'. Solving this equation yields the possible values of 'a', and their sum is then computed. The derived possible values of $a$ are $5$ and $-4$, leading to a sum of $1$.

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