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It is known that A + R = B and R perpendicular A . The angle between A and B is

It is known that A + R = B and R perpendicular
A . The angle between A and B is

Grade:12th pass

2 Answers

Arun
25750 Points
4 years ago
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°
 
 
Vikas TU
14149 Points
4 years ago
Dear student
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°

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