Let a and b be two unit vectors such that |(a + b) + 2(a × b)| = 2. If θ ∈ (0, π) is the angle between a and b, then among the statements :
(S1) : 2|a × b| = |a - b|
(S2) : The projection of a on (a + b) is 1/2

Solution

Maths sir · 3 steps

Square the given modulus: the cross product is perpendicular to both vectors, so mixed terms drop out.

1. Square the modulus. The cross product a × b is perpendicular to both a and b, so its dot product with a + b is zero. Squaring the given modulus therefore leaves only two terms: |a + b|² = 2 + 2 cos θ and |2(a × b)|² = 4 sin² θ.

   2 + 2 cos θ + 4 sin² θ = 4 ⇒ 2 cos² θ - cos θ - 1 = 0

2. Find the angle. This factorises as (2 cos θ + 1)(cos θ - 1) = 0. Since θ ∈ (0, π), cos θ = 1 is not allowed. So cos θ = -1/2 and θ = 2π/3.

   Statement	Left side	Right side	Verdict
   (S1)	2 sin 2π/3 = √3	|a - b| = √(2 - 2 cos θ) = √3	true
   (S2)	(a · (a + b))/(|a + b|) = (1 + cos θ)/√(2 + 2 cos θ)	(1/2)/1 = 1/2	true

3. So. With θ = 2π/3, both (S1) and (S2) hold, so the answer is (c).

The idea to keep. When a modulus mixes a ± b with a × b, square it. The mixed terms vanish, leaving an equation in cos θ alone.

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