For any vector a⃗=a₁ î+a₂ ĵ+a₃ k̂, with 10|aᵢ|<1, i=1,2,3, consider the following statements :
(A): max {|a₁|,|a₂|,|a₃|} ≤|a⃗|
(B) : |a⃗| ≤ 3 max {|a₁|,|a₂|,|a₃|}

Solution

Maths sir · 4 steps

Call the biggest component M, then square |a⃗| and bound it from both sides.

1. Name the largest. Let M = max{|a₁|, |a₂|, |a₃|}. Then |a⃗|² = a₁² + a₂² + a₃², and each aᵢ² satisfies 0 ≤ aᵢ² ≤ M².
2. Lower bound gives (A). One of the three squares equals M², and the other two are never negative. So |a⃗|² ≥ M², which gives M ≤ |a⃗|. (A) is true.
3. Upper bound gives (B). Each of the three squares is at most M², so |a⃗|² ≤ 3M². Since √3 < 3 and M ≥ 0, this gives the chain below. (B) is true.

   M ≤ |a⃗| ≤ √3 M ≤ 3M

4. So. Both statements hold for every vector. The condition 10|aᵢ| < 1 only keeps the components small and plays no part in either inequality. The answer is (d).

The idea to keep. To compare a length with its components, square it. Every component is at least 0 and at most the largest one, so max|aᵢ| ≤ |a⃗| ≤ √n max|aᵢ| in n dimensions.

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