If the system of equations
x+(√2 sin α) y+(√2 cos α) z=0 ; x+(cos α) y+(sin α) z=0 ; x+(sin α) y-(cos α) z=0
has a non-trivial solution, then α ∈(0, π/2) is equal to :

Solution

Maths sir · 5 steps

A non-trivial solution means the coefficient determinant is zero, so expand it and solve the trig equation.

1. Set determinant zero. All three equations are homogeneous, so a non-trivial solution exists only when the coefficient determinant is zero. Write s = sin α and c = cos α to keep it short.

   |1, √2s, √2c; 1, c, s; 1, s, -c| = -(c²+s²) + √2s(c+s) + √2c(s-c) = 0

2. Tidy it up. Since c² + s² = 1, this becomes -1 + √2(2sc - (c² - s²)) = 0. That is sin 2α - cos 2α = 1/√2.
3. One sine term. Use sin 2α - cos 2α = √2 sin(2α - π/4). This gives sin(2α - π/4) = 1/2.
4. Use the interval. For α ∈ (0, π/2), the angle 2α - π/4 lies in (-π/4, 3π/4). The only angle there with sine 1/2 is π/6, because 5π/6 is too large.
5. So. 2α = π/6 + π/4 = 5π/12, so α = 5π/24, which is option (a).

The idea to keep. Homogeneous system with a non-trivial solution means determinant equals zero. Then combine a sin θ + b cos θ into one sine and let the given interval pick the root.

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