Let A, B, C be three points in xy-plane, whose position vector are given by √3 î+ĵ, î+√3 ĵ and a î+(1-a) ĵ respectively with respect to the origin O . If the distance of the point C from the line bisecting the angle between the vectors OA⃗ and OB⃗ is 9/√2, then the sum of all the possible values of a is :

Solution

Maths sir · 5 steps

OA and OB have the same length, so the bisector is just the line y = x. Then use the distance formula.

1. Equal lengths. |OA⃗| = √(3 + 1) = 2 and |OB⃗| = √(1 + 3) = 2. When the two vectors are equally long, their sum points along the angle bisector.
2. Find the bisector. OA⃗ + OB⃗ = (√3+1)î + (√3+1)ĵ, which points along î + ĵ. So the bisector through O is the line x - y = 0.
3. Distance of C. C is (a, 1-a). Its perpendicular distance from x - y = 0 must equal 9/√2.

   (|a - (1-a)|)/√2 = (|2a - 1|)/√2 = 9/√2

4. Solve. |2a - 1| = 9, so 2a - 1 = 9 or 2a - 1 = -9. That gives a = 5 or a = -4.
5. So. The sum of all possible values is 5 + (-4) = 1, option (d).

The idea to keep. For two vectors of equal length, their sum points along the angle bisector. If the lengths differ, add the unit vectors instead. A modulus equation usually gives two values, so keep both.

Answer from the NTA final answer key. Solution by PrepWiser.