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Vector Algebra Part 3

Section: Mathematics

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Question : 33 of 72

Marks: +1, -0

Let

A, B, C

be three points in

x y

-plane, whose position vector are given by

\sqrt{3} \hat{i} + \hat{j}, \hat{i} + \sqrt{3} \hat{j}

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

\overrightarrow{O A}

\overrightarrow{O B}

\;\frac{9}{\sqrt{2}}

, then the sum of all the possible values of

a

[28 Jan 2025 Shift 2]

2
0
1
$\;\frac{9}{2}$

Solution: 👈: Video Solution

Equation of line in the internal bisector of

\overrightarrow{O A}

and

\overrightarrow{O B}

is

(\sqrt{3}+1) \hat{i} + (\sqrt{3}+1) \hat{j}

⇒ line will be

y=x \Rightarrow x-y=0

\;D = \left| \;\frac{a-(1-a)}{\sqrt{a^2+(1-a)^2}} \right| = \;\frac{9}{\sqrt{2}}

\;(2a-1)^2 = \;\frac{81}{2} (a^2+(1-a)^2)

\; \Rightarrow 2(4a^2-4a+1) = 81a^2+81a^2-162a-81

\; \Rightarrow 162a^2-162a+81-8a^2+8a-2=0

\; \Rightarrow 154a^2-154a+79=0

\;\; \text{Sum of values}\; = -\;\frac{-154}{154} = 1

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