Let a⃗, b⃗, c⃗ be three non-coplanar vectors such that a⃗ × b⃗ = 4c⃗, b⃗ × c⃗ = 9a⃗ and c⃗ × a⃗ = αb⃗, α > 0. If | a⃗ | + | b⃗ | + | c⃗ | = 36, then α is equal to __________.

Solution

Maths sir · 4 steps

Each cross product is perpendicular to both factors, so all three vectors are mutually perpendicular.

1. Mutually perpendicular. a⃗ × b⃗ = 4c⃗ means c⃗ is perpendicular to both a⃗ and b⃗. Likewise b⃗ × c⃗ = 9a⃗ makes a⃗ perpendicular to b⃗. So every angle between the vectors is 90° and |x⃗ × y⃗| = |x⃗||y⃗|.
2. Magnitude equations. Taking magnitudes gives |a⃗||b⃗| = 4|c⃗|, |b⃗||c⃗| = 9|a⃗| and |c⃗||a⃗| = α|b⃗|. Multiply them in pairs and the unwanted lengths cancel.

   |b⃗|² = 36, |a⃗|² = 4α, |c⃗|² = 9α

3. Use the sum. So |b⃗| = 6, |a⃗| = 2√α and |c⃗| = 3√α. Then 2√α + 6 + 3√α = 36 gives 5√α = 30, so √α = 6.
4. So. α = 36, with |a⃗| = 12, |b⃗| = 6, |c⃗| = 18. Check: 18 × 12 = 36 × 6 = 216. The answer is 36.

The common slip: 6. Stopping at √α = 6 gives 6. The question asks for α itself, so square it.

The idea to keep. When a⃗ × b⃗ is a multiple of c⃗ and so on cyclically, the vectors are mutually perpendicular. Multiply the magnitude equations in pairs to isolate each length.

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