Let sqrt{3} hat{i} + hat{j},hat{i} + sqrt{3} | vedclass

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(D) The position vectors are $\vec{OA} = \sqrt{3} \hat{i} + \hat{j}$ and $\vec{OB} = \hat{i} + \sqrt{3} \hat{j}$.The unit vectors along $OA$ and $OB$

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$1$

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(D) The position vectors are $\vec{OA} = \sqrt{3} \hat{i} + \hat{j}$ and $\vec{OB} = \hat{i} + \sqrt{3} \hat{j}$.The unit vectors along $OA$ and $OB$ are $\hat{u}_A = \frac{\sqrt{3} \hat{i} + \hat{j}}{2}$ and $\hat{u}_B = \frac{\hat{i} + \sqrt{3} \hat{j}}{2}$.The angle bisector of $\angle AOB$ is along the vector $\vec{u}_A + \vec{u}_B = \frac{(\sqrt{3}+1) \hat{i} + (1+\sqrt{3}) \hat{j}}{2}$,which simplifies to the line $y = x$.The point $C$ has coordinates $(\beta, 1 + \beta)$.The distance of point $C(x_0, y_0)$ from the line $x - y = 0$ is given by $d = \frac{|x_0 - y_0|}{\sqrt{1^2 + (-1)^2}}$.Substituting the coordinates of $C$,we get $d = \frac{|\beta - (1 + \beta)|}{\sqrt{2}} = \frac{|-1|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.Wait,re-evaluating the problem statement: the distance is $\frac{3}{\sqrt{2}}$.Let's re-check the bisector: The vectors are $\vec{OA} = (\sqrt{3}, 1)$ and $\vec{OB} = (1, \sqrt{3})$. The angle bisector is indeed $y = x$.Distance of $C(\beta, 1+\beta)$ from $x-y=0$ is $\frac{|\beta - (1+\beta)|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.If the distance is $\frac{3}{\sqrt{2}}$,there might be a typo in the question's provided solution or coordinates. Assuming the distance is $\frac{|\beta - (1+\beta)|}{\sqrt{2}} = \frac{3}{\sqrt{2}}$ is impossible as it results in $1=3$. Re-reading: Perhaps the vector is $\beta \hat{i} + (1-\beta) \hat{j}$? If $C = (\beta, 1-\beta)$,then distance is $\frac{|\beta - (1-\beta)|}{\sqrt{2}} = \frac{|2\beta - 1|}{\sqrt{2}} = \frac{3}{\sqrt{2}}$.Then $|2\beta - 1| = 3$,so $2\beta - 1 = 3$ or $2\beta - 1 = -3$.$\beta = 2$ or $\beta = -1$.Sum of values $= 2 + (-1) = 1$.

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