JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Vector Algebra

Let

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

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

and

β \hat{i} + (1 - β) \hat{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

\frac{3}{2}

, then the sum of all possible values of

β

Answer: (D)

Solution

Since

∣ O A ∣ = ∣ O B ∣ = 2

, the bisector of the acute angle between

O A

and

O B

has direction

\hat{O A} + \hat{O B} = (\frac{3 + 1}{2}) (\hat{i} + \hat{j})

, i.e. the line

y = x

x - y = 0

. The distance of

C (β, 1 - β)

from this line is

\frac{∣ β - ( 1 - β ) ∣}{2} = \frac{∣2 β - 1∣}{2}

. Setting this equal to

\frac{3}{2}

gives

∣2 β - 1∣ = 3

, so

β = 2

β = - 1

; the sum of all possible values of

β

1

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option