Let a, b c ∈ R be such that a² + b² + c² = 1. If
a cos θ = b cos ( θ + 2π/3 ) = c cos ( θ + 4π/3 ),
where θ = π/9, then the angle between the vectors aî + bĵ + ck̂ and bî + cĵ + ak̂ is :

Solution

Maths sir · 4 steps

Call the common value k, write each of a, b, c as k over a cosine, and the dot product collapses to zero.

1. Dot product needed. Both vectors have length √(a²+b²+c²) = 1. So the cosine of the angle between them is just their dot product, ab + bc + ca.
2. Name the common value. Let a cos θ = b cos(θ+2π/3) = c cos(θ+4π/3) = k. At θ = π/9 none of the three cosines is zero. So k ≠ 0, because k = 0 would force a = b = c = 0, which breaks a²+b²+c² = 1. Hence 1/a = (cos θ)/k, and similarly for b and c.

   ab+bc+ca = abc(1/a+1/b+1/c) = abc/k[cos θ + cos(θ+2π/3) + cos(θ+4π/3)]

3. Bracket is zero. Pair the last two terms with sum-to-product: cos(θ+2π/3) + cos(θ+4π/3) = 2 cos(θ+π)cos π/3 = -cos θ. Adding cos θ gives 0.
4. So. ab+bc+ca = 0, so the cosine of the angle is 0 and the angle is π/2, option (d).

The idea to keep. Cosines of three angles spaced 2π/3 apart always add to zero. When equal products hide behind such cosines, rewrite ab+bc+ca as abc(1/a+1/b+1/c).

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