Question:

Let \(\sqrt{3}\,\hat{i} + \hat{j}\), \(\hat{i} + \sqrt{3}\,\hat{j}\) and \(\beta\hat{i} + (1-\beta)\hat{j}\) respectively be the position vectors of the points A, B and C with respect to the origin O. If the distance of C from the bisector of the acute angle between OA and OB is \(\frac{3}{\sqrt{2}}\), then the sum of all possible values of \(\beta\) is :

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OA and OB make 30 and 60 degrees with the x-axis, so the acute-angle bisector is the line y = x. Use the point-to-line distance for C.
Updated On: Nov 5, 2026
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The Correct Option is B

Solution and Explanation

A = \((\sqrt3, 1)\) makes \(30^\circ\) with the x-axis and B = \((1, \sqrt3)\) makes \(60^\circ\). The bisector of the acute angle AOB is at \(45^\circ\), the line \(y = x\), i.e. \(x - y = 0\).
C = \((\beta, 1-\beta)\). Distance from C to the line: \(\frac{|\beta - (1-\beta)|}{\sqrt2} = \frac{|2\beta - 1|}{\sqrt2}\).
Setting equal to \(\frac{3}{\sqrt2}\): \(|2\beta - 1| = 3\), so \(2\beta - 1 = 3\) or \(-3\), giving \(\beta = 2\) or \(\beta = -1\).
Sum of all possible values = \(2 + (-1) = 1\).
Answer: (2) 1
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