Let a⃗, b⃗, c⃗ be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is 14 and (a⃗ × b⃗) ·(b⃗ × c⃗)+(b⃗ × c⃗) ·(c⃗ × a⃗)+(c⃗ × a⃗) ·(a⃗ × b⃗)=168, then |a⃗|+|b⃗|+|c⃗| is equal to :

Solution

Maths sir · 4 steps

Equal angles in a plane means 120° each, then Lagrange's identity turns each dot product into magnitudes.

1. Find the angle. Three concurrent vectors in one plane with equal angles between every pair must split the full turn equally. So each angle is 120°, and a⃗·b⃗ = -1/2|a⃗||b⃗|, and the same for the other pairs.

   (a⃗×b⃗)·(b⃗×c⃗) = (a⃗·b⃗)(b⃗·c⃗) - (a⃗·c⃗)|b⃗|²

2. One term. Write a, b, c for the magnitudes. The first product is ab/2·bc/2 = ab²c/4, and the second is -ac/2· b². Their difference is ab²c/4 + ab²c/2 = 3/4ab²c.
3. Add all three. By symmetry the other two terms are 3/4abc² and 3/4a²bc. The sum is 3/4abc (a+b+c) = 168.
4. So. With abc = 14, we get 3/4× 14×(a+b+c) = 168, so a+b+c = (168× 4)/42 = 16, option (c).

The idea to keep. A dot product of two cross products always opens up with Lagrange's identity, (a⃗×b⃗)·(c⃗×d⃗) = (a⃗·c⃗)(b⃗·d⃗) - (a⃗·d⃗)(b⃗·c⃗).

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