Consider three vectors a⃗, b⃗, c⃗. Let |a⃗|=2,|b⃗|=3 and a⃗=b⃗ × c⃗. If α ∈[0, π/3] is the angle between the vectors b⃗ and c⃗, then the minimum value of 27|c⃗-a⃗|² is equal to:

Solution

Maths sir · 4 steps

Since a⃗ = b⃗ × c⃗, a⃗ is perpendicular to c⃗, so the cross term vanishes.

1. Cross term vanishes. A cross product is perpendicular to both of its factors, so a⃗ ⊥ c⃗ and a⃗ · c⃗ = 0. Expanding gives |c⃗-a⃗|² = |c⃗|² + |a⃗|² - 2 a⃗·c⃗ = |c⃗|² + 4.
2. Link the magnitudes. Take magnitudes in a⃗ = b⃗ × c⃗. This gives |a⃗| = |b⃗||c⃗|sin α, that is 2 = 3|c⃗|sin α.

   |c⃗| = 2/(3 sin α)

3. Minimise. |c⃗| is smallest when sin α is largest. On [0, π/3] that happens at α = π/3, where sin α = √3/2. Then |c⃗| = 4/3√3 and |c⃗|² = 16/27.
4. So. The minimum is 27(16/27 + 4) = 16 + 108 = 124, which is option (a).

The idea to keep. Whenever one vector is a cross product involving another, use the perpendicularity first: the dot product is zero, and the magnitude equation |b⃗||c⃗|sin α does the rest.

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