Let a⃗ and b⃗ be the vectors of the same magnitude such that
(|a⃗ + b⃗| + |a⃗ - b⃗|)/(|a⃗ + b⃗| - |a⃗ - b⃗|) = √2 + 1. Then (|a⃗ + b⃗|²)/|a⃗|² is :

Solution

Maths sir · 5 steps

Componendo-dividendo gives the diagonal ratio, then the parallelogram law links them to |a⃗|.

1. Name the lengths. Let p = |a⃗+b⃗| and q = |a⃗-b⃗|. The given condition is (p+q)/(p-q) = √2+1.
2. Componendo-dividendo. Applying it gives p/q = ((√2+1)+1)/((√2+1)-1) = (√2+2)/√2 = 1+√2. Squaring, p² = (1+√2)² q² = (3+2√2) q².
3. Parallelogram law. Expand both squares with the dot product. The 2 a⃗·b⃗ terms cancel when you add them, and |b⃗| = |a⃗|, which gives the relation below.

   p² + q² = 2|a⃗|² + 2|b⃗|² = 4|a⃗|²

4. Eliminate q. Here q² = p²/(3+2√2) = (3-2√2) p². So p²(1 + 3 - 2√2) = 4|a⃗|², which gives p²/|a⃗|² = 4/(4-2√2) = 2/(2-√2).
5. So. Rationalising, 2(2+√2)/(4-2) = 2+√2. The answer is (a).

The idea to keep. Whenever you see (x+y)/(x-y), reach for componendo-dividendo to get x/y. And |a⃗+b⃗|² + |a⃗-b⃗|² = 2|a⃗|² + 2|b⃗|² links the diagonals to the sides.

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