Let A, B, C be three points in xy-plane, whose position vector are given by and respectively with respect to the origin O. If the distance of the point C from the line bisecting the angle between the vectors and is , then the sum of all the possible v

Question

Let A, B, C be three points in xy-plane, whose position vector are given by $\sqrt{3} \hat{i} + \hat{j}, \hat{i} + \sqrt{3} \hat{j}$ and $a \hat{i} + (1 - a) \hat{j}$ respectively with respect to the origin O. If the distance of the point C from the line bisecting the angle between the vectors $\vec{O A}$ and $\vec{O B}$ is $\frac{9}{\sqrt{2}}$ , then the sum of all the possible values of a is : [2025]

Smart Achievers · Accepted Answer

(1)

Equation of line OA is $x - \sqrt{3} y = 0$

Equation of line OB is $\sqrt{3} x - y = 0$

Equation of angle bisector is $\frac{x - \sqrt{3} y}{\sqrt{1 + 3}} + \frac{\sqrt{3} x - y}{\sqrt{3 + 1}} = 0$

$\Rightarrow x - y = 0$

$| \frac{a - (1 - a)}{\sqrt{2}} |$ $= \frac{9}{\sqrt{2}} \Rightarrow a = 5 or - 4$

Required sum = 5 + (–4) = 1.

Solution

1 1

2 0

3 9/2

4 2

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Solution

1 1

2 0

3 9/2

4 2

Ans. (1) Equation of line OA is x–3y=0 Equation of line OB is 3x–y=0 Equation of angle bisector is x–3y1+3+3x–y3+1=0 ⇒ x–y=0 |a–(1–a)2|=92 ⇒ a=5 or –4 Required sum = 5 + (–4) = 1.

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