Question

JEE Mains Maths question is attached

Answer 1

Hey! met see ur full answer is in attachment

Answer 2

it has given that a/cosθ = b/cos(θ + 2π/3) = c/cos(θ + 4π/3) , here θ = 2π/9 and a² + b² + c² = 1

we have to find the angle between vectors a i + b j + c k and b i + c j + a k

solution : angle between two vectors A and B is given by, θ = cos¯¹ (A.B)/|A||B|

so angle between a i + b j + c k and b i + c j + a k , θ = cos¯¹ [(a i + b j + c k)(b i + c j + a k)]/√(a² + b² + c²)√(b² + c² + a²)

as a² + b² + c³ = 1

√(a² + b² + c²) = 1 = √(b² + c² + a²)

= cos¯¹ (ab + bc + ca)/1

= cos¯¹(ab + bc + ca) ........(1)

now a/cosθ = b/cos(θ + 2π/3) = c/cos(θ + 4π/3)

a + b + c = [cosθ + cos(θ + 2π/3) + cos(θ + 4π/3)]

= [{cosθ + cos(θ + 4π/3)} + cos(θ + 2π/3)]

= [2cos(θ + 2π/3) cos(2π/3) + cos(θ + 2π/3)]

= [2 cos(θ + 2π/3) × -1/2 + cos(θ + 2π/3)]

= 0

so, a + b + c = 0

squaring both sides we get,

a² + b² + c² + 2(ab + bc + ca) = 0

⇒1 + 2(ab + bc + ca) = 0

⇒ab + bc + ca = -1/2

now from equation (1) we get, cos¯¹ (-1/2) = 2π/3

Therefore the angle between vectors is 2π/3.