Between the following two statements:
Statement I : Let a⃗=î+2 ĵ-3 k̂ and b⃗=2 î+ĵ-k̂. Then the vector r⃗ satisfying a⃗ × r⃗=a⃗ × b⃗ and a⃗ · r⃗=0 is of magnitude √10.
Statement II : In a triangle A B C, cos 2 A+cos 2 B+cos 2 C ≥-3/2.

Solution

Maths sir · 5 steps

Equal cross products force r⃗=b⃗+λa⃗, and the dot condition then fixes λ.

1. Shape of r. From a⃗×r⃗=a⃗×b⃗ we get a⃗×(r⃗-b⃗)=0⃗. So r⃗-b⃗ is parallel to a⃗, which means r⃗=b⃗+λa⃗ for some scalar λ.
2. Find lambda. Put this into a⃗·r⃗=0 to get a⃗·b⃗+λ|a⃗|²=0. Here a⃗·b⃗=2+2+3=7 and |a⃗|²=1+4+9=14, so λ=-1/2.
3. Magnitude of r. r⃗=(2î+ĵ-k̂)-1/2(î+2ĵ-3k̂)=3/2î+1/2k̂. Then |r⃗|=√(9/4+1/4)=√10/2, not √10. Statement I is incorrect.
4. Statement II. Let O be the circumcentre, with OA=OB=OC=R. The angle between OB⃗ and OC⃗ has cosine cos 2A, and similarly for the other pairs. Expanding a square that can never be negative gives the inequality.

   0 ≤ |OA⃗+OB⃗+OC⃗|² = 3R² + 2R²(cos 2A+cos 2B+cos 2C)

5. So. Dividing by 2R² gives cos 2A+cos 2B+cos 2C ≥ -3/2, so Statement II is correct, with equality for an equilateral triangle. Statement I is incorrect and Statement II is correct, which is option (d).

The idea to keep. If a⃗×r⃗=a⃗×b⃗, then r⃗=b⃗+λa⃗. Use the second condition to fix λ. For sums of cos 2A, square the sum of the circumradius vectors.

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