Question

Let √3i + j,i + √3j and βi + (1-β)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 3√2 then the sum of all possible values of b is​

Answer 1

Let √3i + j,i + √3j and βi + (1-β)j respectively be the position vectors of the points A, B and C with respect to the origin O.

The distance of C from the bisector of the acute angle between OA and OB is 3√2, then the sum of all possible values of b is​ given by,

From given, we have,

$y - 1=\dfrac{1}{\sqrt 3}(x-\sqrt 3)$

$\sqrt 3 y - \sqrt 3 = x -\sqrt 3$

$x- \sqrt 3 y =0$

$y - \sqrt 3 = \sqrt 3 (x-1)$

$y -\sqrt 3 = \sqrt 3 (x-1)$

$\sqrt 3 x - y = 0$

$\dfrac{x-\sqrt 3 y}{2} = \pm \dfrac{\sqrt 3x-y}{2}$

$x-\sqrt 3 x = \sqrt 3 y -y$

$x(1-\sqrt 3) = - y(1-\sqrt 3)$

$x + y = 0$

$x - y = 0$

$\dfrac{|\beta - (1-\beta)|}{\sqrt 2} = \dfrac{3}{\sqrt 2}$

$|2\beta - 1|}= 3$

$\beta = -1,2$

-1 + 2 = 1

The sum of all possible values of b is​ 1.

Answer 2

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