Let a⃗ ,b⃗ ,c⃗ three vectors mutually perpendicular to each other and have same magnitude. If a vector r⃗ satisfies.
a⃗ × { (r⃗ - b⃗ ) × a⃗ } + b⃗ × { (r⃗ - c⃗ ) × b⃗ } + c⃗ × { (r⃗ - a⃗ ) × c⃗ } = 0⃗ , then r⃗ is equal to :

Solution

Maths sir · 4 steps

Expand each triple product, then use the fact that three perpendicular vectors rebuild r⃗ for you.

1. Expand one term. For any u⃗, a⃗ × (u⃗ × a⃗) = |a⃗|² u⃗ - (a⃗ · u⃗) a⃗. Let the common magnitude be k. Since a⃗ ⊥ b⃗, (r⃗ - b⃗)· a⃗ = r⃗ · a⃗, and the same happens in the other two terms.

   3k² r⃗ - k²(a⃗ + b⃗ + c⃗) - [(r⃗·a⃗)a⃗ + (r⃗·b⃗)b⃗ + (r⃗·c⃗)c⃗] = 0⃗

2. Rebuild r. Because a⃗, b⃗, c⃗ are mutually perpendicular with magnitude k, any vector splits along them: r⃗ = 1/k²[(r⃗·a⃗)a⃗ + (r⃗·b⃗)b⃗ + (r⃗·c⃗)c⃗]. So the bracket equals k² r⃗.
3. Solve. The equation becomes 3k²r⃗ - k²r⃗ = k²(a⃗ + b⃗ + c⃗), that is, 2r⃗ = a⃗ + b⃗ + c⃗.
4. So. r⃗ = 1/2(a⃗ + b⃗ + c⃗), which is option (c).

The idea to keep. With three mutually perpendicular vectors of equal length k, Σ (r⃗·a⃗)a⃗ = k²r⃗. Spot this and every triple-product sum collapses.

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