If a⃗ and b⃗ are unit vectors, then the greatest value of
√3 | a⃗ + b⃗ | + | a⃗ - b⃗ | is_____.

Solution

Maths sir · 4 steps

Write both lengths with the half-angle, and the whole thing becomes a cos t + b sin t.

1. Use the angle. Let θ be the angle between the unit vectors. Then |a⃗+b⃗|² = 2 + 2 cos θ = 4 cos² θ/2 and |a⃗-b⃗|² = 2 - 2 cos θ = 4 sin² θ/2.
2. Take square roots. Since 0 ≤ θ ≤ π, the half-angle lies in [0, π/2], so both sine and cosine are non-negative. This gives |a⃗+b⃗| = 2 cos θ/2 and |a⃗-b⃗| = 2 sin θ/2.

   2√3 cos θ/2 + 2 sin θ/2 = 4 cos(θ/2 - π/6)

3. Find the maximum. The cosine reaches 1 when θ/2 = π/6, that is θ = π/3. This angle is allowed, so the maximum is reached.
4. So. The greatest value of the expression is 4.

The common slip: 5.46. Maximising each term on its own gives 2√3 + 2 ≈ 5.46. That cannot happen, because |a⃗+b⃗| = 2 forces a⃗ = b⃗, which makes |a⃗-b⃗| = 0.

The idea to keep. For unit vectors, |a⃗+b⃗| = 2 cos θ/2 and |a⃗-b⃗| = 2 sin θ/2. The maximum of p cos t + q sin t is √(p²+q²).

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