Answer:
A+B| = |A| = |B|
On the square of all… |A+B|^2 = |A|^2 = |B|^2
|A|^2 + |B|^2 + 2A.B = |A|^2 = |B|^2
We can also write this to….
|A|^2 + |B|^2 + 2|A||B| cosθ = |A|^2 = |B|^2 .. [where θ is angle b/w Vec(A) & Vec(B)]
=====2|A|^2 + 2|A|^2 cosθ = |A|^2 ….(because |A| = |B|)
From this… cosθ = -1/2
So, …. θ = 2π/3
ITO NA PO YUNF ANSWER
