Let A, B, C be three points in the xy-plane,w | vedclass

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(A) The position vectors are $\vec{OA} = \sqrt{3}\hat{i} + \hat{j}$ and $\vec{OB} = \hat{i} + \sqrt{3}\hat{j}$.Since $|\vec{OA}| = \sqrt{(\sqrt{3})^2

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(A) The position vectors are $\vec{OA} = \sqrt{3}\hat{i} + \hat{j}$ and $\vec{OB} = \hat{i} + \sqrt{3}\hat{j}$.Since $|\vec{OA}| = \sqrt{(\sqrt{3})^2 + 1^2} = 2$ and $|\vec{OB}| = \sqrt{1^2 + (\sqrt{3})^2} = 2$,the angle bisector of $\angle AOB$ is the line passing through the origin with the direction vector $\vec{OA} + \vec{OB} = (\sqrt{3} + 1)\hat{i} + (1 + \sqrt{3})\hat{j}$.This simplifies to the line $y = x$ or $x - y = 0$.The distance of point $C(a, 1 - a)$ from the line $x - y = 0$ is given by $d = \frac{|a - (1 - a)|}{\sqrt{1^2 + (-1)^2}} = \frac{|2a - 1|}{\sqrt{2}}$.Given $d = \frac{9}{\sqrt{2}}$,we have $\frac{|2a - 1|}{\sqrt{2}} = \frac{9}{\sqrt{2}}$,which implies $|2a - 1| = 9$.Thus,$2a - 1 = 9 \Rightarrow 2a = 10 \Rightarrow a = 5$,or $2a - 1 = -9 \Rightarrow 2a = -8 \Rightarrow a = -4$.The sum of all possible values of $a$ is $5 + (-4) = 1$.

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