Question

the resultant of two vectors A and B is perpendicular to A and it's magnitude is equal to half of vector B .Find out the angle between them​

Answer 1

Hola mate :p

Magnitude of resultant of A and B is given by,

R = √(A² + B² + 2ABcosФ)

According to question ,

Magnitude of resultant = half of Magnitude of B

e.g., √(A² + B² + 2A.BcosФ) = B/2

Taking square both sides,

A² + B² + 2A.BcosФ = B²/4

A² + 2ABcosФ + 3B²/4 = 0 --------(1)

Also, A and R is perpendicular upon each other ,

e.g., A.(A + B) = 0

A.A + A.B = 0

A² + A.BcosФ = 0

cosФ = - A/B , put it in equation (1)

A² + 2A.B(-A/B) + 3B²/4 = 0

A² - 2A² + 3B²/4 = 0

A = √3B/2

Now, cosФ = -A/B = -√3B/2B = -√3/2

cosФ = cos150° ⇒Ф = 150°

Hence angle between A and B = 150°

Answer 2

Answer:

the best answer is given