JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Vector Algebra

Let

A, B, C

be three points in xy -plane, whose position vector are given by

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

OA

and

OB

\frac{9}{2}

, then the sum of all the possible values of

a

is :

Answer: (A)

Solution

The vectors

O A = 3 \hat{i} + \hat{j}

and

O B = \hat{i} + 3 \hat{j}

both have magnitude 2, so the bisector of the angle between them is along the sum of their unit vectors, which is proportional to

(1, 1)

; the bisector is the line

x - y = 0

.

For

C = (a, 1 - a)

, the distance from this line is

d = \frac{∣ a - ( 1 - a ) ∣}{2} = \frac{∣2 a - 1∣}{2}

Setting

d = \frac{9}{2}

gives

∣2 a - 1∣ = 9

, so

2 a - 1 = 9

2 a - 1 = - 9

, i.e.

a = 5

a = - 4

.

Sum of all possible values of

a

5 + (- 4) = 1

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Single correct option