Let A, B, C be three points in xy-plane, whose position vector are given by sqrt3hati+hatj$\sqrt{3}\hat{i}+\hat{j}$, hati+sqrt3hatj$\hat{i}+\sqrt{3}\hat{j}$ and ahati+(1-a)hatj$a\hat{i}+(1-a)\hat{j}$ respectively with respect to the origin O. If the distance of the point C from the line bisecting the angle between the vectors overlineOA$\overline{OA}$ and overlineOB$\overline{OB}$ is frac9sqrt2$\frac{9}{\sqrt{2}}$ then the sum of all the possible values of a is:

### Related Formula The line bisecting the angle between two symmetric vectors passing through the origin in the first quadrant is given by y = x$y = x$ or x - y = 0$x - y = 0$. Distance from point (x_1, y_1)$(x_1, y_1)$ to line Ax + By + C = 0$Ax + By + C = 0$ is: d = frac|Ax_1 + By_1 + C|sqrtA^2 + B^2$$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$ ### Core Logic Vectors overlineOA = sqrt3hati+hatj$\overline{OA} = \sqrt{3}\hat{i}+\hat{j}$ and overlineOB = hati+sqrt3hatj$\overline{OB} = \hat{i}+\sqrt{3}\hat{j}$ are symmetric about the line y = x$y = x$. Therefore, the angle bisector of overlineOA$\overline{OA}$ and overlineOB$\overline{OB}$ is the line x - y = 0$x - y = 0$. Point C has coordinates (a, 1 - a)$(a, 1 - a)$. ### Step 1: Calculate Distance and Solve for a The perpendicular distance from C(a, 1 - a)$C(a, 1 - a)$ to x - y = 0$x - y = 0$ is: d = frac|a - (1 - a)|sqrt1^2 + (-1)^2 = frac|2a - 1|sqrt2$$d = \frac{|a - (1 - a)|}{\sqrt{1^2 + (-1)^2}} = \frac{|2a - 1|}{\sqrt{2}}$$ Given that this distance is frac9sqrt2$\frac{9}{\sqrt{2}}$: frac|2a - 1|sqrt2 = frac9sqrt2 implies |2a - 1| = 9$$\frac{|2a - 1|}{\sqrt{2}} = \frac{9}{\sqrt{2}} \implies |2a - 1| = 9$$ This gives two solutions: 1) 2a - 1 = 9 implies 2a = 10 implies a = 5$2a - 1 = 9 \implies 2a = 10 \implies a = 5$ 2) 2a - 1 = -9 implies 2a = -8 implies a = -4$2a - 1 = -9 \implies 2a = -8 \implies a = -4$ ### Step 2: Find the Sum of Values Sum of all possible values of a$a$: textSum = 5 + (-4) = 1$$\text{Sum} = 5 + (-4) = 1$$ ### Pattern Recognition Notice that overlineOA$\overline{OA}$ and overlineOB$\overline{OB}$ have swapped coordinates, meaning they are symmetric with respect to y=x$y=x$. Thus, the angle bisector equation is immediate (x-y=0$x-y=0$), simplifying the problem to a standard point-to-line distance calculation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines Class 12 Mathematics: Vector Algebra