Let √3 hati + hatj , 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 (3/√2) , then the sum of all possible values of β is :

Question

Let √3 hati + hatj , 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 (3/√2) , then the sum of all possible values of β is : (A) 2 (B) 1 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

Angle bisector is x - y = 0 ⇒ (|β -(1-β)|/√2) = (3/√2) ⇒ |2β -1|=3 ⇒ β = 2 or - 1

Q. Let 3​i^+j^​,i^+3j^​ and βi^+(1−β)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 2​3​ , then the sum of all possible values of β is :

2

1

3

4

Angle bisector is x−y=0 ⇒2​∣β−(1−β)∣​=2​3​ ⇒∣2β−1∣=3 ⇒β=2 or −1

Angle bisector is $x - y = 0$
$\Rightarrow \frac{∣ β - ( 1 - β ) ∣}{2} = \frac{3}{2}$
$\Rightarrow ∣ 2 β - 1 ∣ = 3$
$\Rightarrow β = 2$ or $- 1$