Let a⃗, b⃗ and c⃗ be three vectors such that a⃗ = b⃗ × (b⃗ × c⃗). If magnitudes of the vectors a⃗, b⃗ and c⃗ are √2, 1 and 2 respectively and the angle between b⃗ and c⃗ is θ ( 0 < θ < π/2 ), then the value of 1 + tanθ is equal to :

Solution

Maths sir · 4 steps

b⃗ is perpendicular to b⃗ × c⃗, so |a⃗| is a plain product of lengths.

1. Spot the right angle. The vector b⃗ × c⃗ is perpendicular to b⃗. So in b⃗ × (b⃗ × c⃗) the angle between the two factors is 90°, and sin 90° = 1.

   |a⃗| = |b⃗| |b⃗ × c⃗| = |b⃗| (|b⃗| |c⃗|sin θ)

2. Put in numbers. √2 = 1 × (1 × 2 sin θ) = 2 sin θ, so sin θ = 1/√2. With 0 < θ < π/2, this gives θ = π/4.
3. Check by expansion. a⃗ = (b⃗ · c⃗)b⃗ - |b⃗|² c⃗ gives |a⃗|² = 4 - (b⃗ · c⃗)² = 4 - 4 cos² θ. Setting this to 2 gives cos² θ = 1/2, the same angle.
4. So. tan θ = tan π/4 = 1, so 1 + tan θ = 2, which is option (b).

The idea to keep. In b⃗ × (b⃗ × c⃗), the two factors are always at right angles. So the magnitude is a plain product of lengths, and you do not need to expand it.

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