Let a⃗=2 î+ĵ+k̂, and b⃗ and c⃗ be two nonzero vectors such that |a⃗+b⃗+c⃗|=|a⃗+b⃗-c⃗| and b⃗ · c⃗=0. Consider the following two statements:
(A) |a⃗+λ c⃗| ≥|a⃗| for all λ ∈ ℝ.
(B) a⃗ and c⃗ are always parallel.
Then,

Solution

Maths sir · 4 steps

Square both sides of the modulus condition and it tells you that a⃗ is perpendicular to c⃗.

1. Square the condition. Let u⃗ = a⃗+b⃗. Then |u⃗+c⃗|² - |u⃗-c⃗|² = 4 u⃗·c⃗. The two moduli are equal, so this difference is 0 and u⃗·c⃗ = 0.

   (a⃗+b⃗)·c⃗ = 0 ⇒ a⃗·c⃗ + b⃗·c⃗ = 0 ⇒ a⃗·c⃗ = 0

2. Check statement A. Since a⃗·c⃗ = 0, we get |a⃗+λc⃗|² = |a⃗|² + 2λ a⃗·c⃗ + λ²|c⃗|² = |a⃗|² + λ²|c⃗|². The extra term λ²|c⃗|² is never negative, so |a⃗+λc⃗| ≥ |a⃗| for every real λ. A is true.
3. Check statement B. Both a⃗ and c⃗ are nonzero, and their dot product is 0, so they are perpendicular. Two nonzero perpendicular vectors can never be parallel. B is false.
4. So. Only statement (A) is correct, which is option (c).

The idea to keep. If |p⃗+q⃗| = |p⃗-q⃗|, then p⃗·q⃗ = 0. Equal diagonals of a parallelogram mean the sides are perpendicular.

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