Let A, B, C be three points in x y -plane, whose position vector are given by 3 i + j , i + 3 j and a i +(1-a) j respectively with respect to the origin O . If the distance of the point C from the line bisecting the

Correct answer: C. 1 Equation of line in the internal bisector of O A and O B is ( 3 +1) i +( 3 +1) j line will be y=x x-y=0 D=| a-(1-a) a^2+(1-a)^2 |= 9 2 aligned & (2 a-1)^2=(81)/(2)(a^2+(1-a)^2) & 2(4 a^2-4 a+1)=81 a^2+81 a^2-162 a-81 & 162 a^2-162 a+81-8 a^2+8 a-2=0 & 154 a^2-154 a+79=0 & Sum of values =-((-154))/(154)=1 aligned

JEE Main 202528 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2025 · 28 January · Evening Shift · Mathematics · Vector Algebra

Let A, B, C be three points in x y -plane, whose position vector...

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}

and

\mathrm{a} \hat{i}+(1-\mathrm{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{\mathrm{OA}}

and

\overrightarrow{\mathrm{OB}}

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

, then the sum of all the possible values of

a

is :

Options

A $2$
B $9 / 2$
C $1$
D $0$

Correct answer

C. $1$

Step-by-step solution

Equation of line in the internal bisector of

O \dot{A}

and O B is

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

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}}

\begin{aligned} & (2 a-1)^2=\frac{81}{2}\left(a^2+(1-a)^2\right) \\ & \Rightarrow 2\left(4 a^2-4 a+1\right)=81 a^2+81 a^2-162 a-81 \\ & \Rightarrow 162 a^2-162 a+81-8 a^2+8 a-2=0 \\ & \Rightarrow 154 a^2-154 a+79=0 \\ & \text{ Sum\text{ of } values }=-\frac{(-154)}{154}=1\end{aligned}

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