Let , and respectively be the position vectors of the points , and with respect to the origin . If the distance of from the bisector of the acute angle between and is , then the sum of all possible values of is
- A.1
- B.3
- C.4
- D.2
Show answer and solution
Correct option: A
Solution
The unit vectors along OA and OB: OA has magnitude 2, so unit vector = (√3/2)î + (1/2)ĵ. OB has magnitude 2, so unit vector = (1/2)î + (√3/2)ĵ.
The bisector of the acute angle between OA and OB is along the sum of unit vectors: (√3/2 + 1/2)î + (1/2 + √3/2)ĵ, which simplifies to direction î + ĵ. So the bisector line is y = x.
Distance of C = (β, 1−β) from the line x − y = 0 is |β − (1−β)|/√2 = |2β − 1|/√2 = 3/√2.
So |2β − 1| = 3, giving 2β − 1 = 3 → β = 2, or 2β − 1 = −3 → β = −1.
Sum of all possible values = 2 + (−1) = 1.
The bisector of the acute angle between OA and OB is along the sum of unit vectors: (√3/2 + 1/2)î + (1/2 + √3/2)ĵ, which simplifies to direction î + ĵ. So the bisector line is y = x.
Distance of C = (β, 1−β) from the line x − y = 0 is |β − (1−β)|/√2 = |2β − 1|/√2 = 3/√2.
So |2β − 1| = 3, giving 2β − 1 = 3 → β = 2, or 2β − 1 = −3 → β = −1.
Sum of all possible values = 2 + (−1) = 1.
[JEE Main 2019 · 11 Jan · Shift II]