Question

Two vectors a and b have equal magnitudes. If magnitude of a+b is equal to n times the magnitude of a-b, then the angle between a and b is

Answer 1

Answer:

Explanation:

=> Here, Two vectors a and b have equal magnitudes. So,

 

=>  |A+B|² = (A+B)•(A+B)

= A•A + B•B + 2A•B  

= |A|² + |B|² + 2|A||B|cosθ  

= 2|A|²(1+cosθ)  

=> Similarly,  

|A−B|² =2|A|²(1−cosθ)  

=> But, magnitude of A+B is equal to n times magnitude of A-B. Thus,  

2|A|²(1+cosθ) = n² * 2|A|²(1−cosθ)  

1+cosθ = n²(1−cosθ)

1 + cosθ =n² - n² cosθ  

n² cosθ + cosθ = n² - 1

cosθ (n² + 1) = n² - 1

cosθ = n² - 1 / n² + 1

Thus,  the angle cosθ between a and b is n² - 1 / n² + 1.