Question

∣ A ∣=∣ B ∣=∣ C ∣. Further, \vec{\textbf{A}} + \vec{\textbf{B}} + \vec{\textbf{C}} = 0 A + B + C =0. What is the angle between \vec{\textbf{A}} A and \vec{\textbf{B}} B ?

Answer 1

Given: ∣ A ∣=∣ B ∣=∣ C ∣ and A + B + C = 0.

To find:  The angle between vector A and B?

Solution:

• Now we have given A + B + C = 0

             A + B = -C

• Squaring on both sides, we get:

             (A + B)^2 = (-C)^2

             |A|^2 + |B|^2 + 2 A.B = |C|^2

• Now we have given ∣ A ∣=∣ B ∣=∣ C ∣, so replacing it, we get:

             |A|^2 + |A|^2 + 2 A.A = |A|^2

• Let the angle be theta, then:

             |A|^2 + |A|^2 + 2 |A| |A| cos theta = |A|^2

             |A|^2 + |A|^2 + 2 |A|^2 cos theta = |A|^2

             2 |A|^2 cos theta =  - |A|^2

             cos theta = -1/2

             theta = 2π / 3

Answer:

          So the angle between vector A and B is 2π / 3.