Given that A + B + C = 0. Out of three vectors two are equal in magnitude and the magnitude of the third vector is 2 times that of either of the two having equal magnitude. Then the angles between vectors are given by

Let |A|=|B|=athen |C|=2aGiven that A+B+C = 0or A+B = -CTaking self Product, we have(A+B)⋅(A+B)=(−C)⋅(−C)A2+B2+2A⋅B=C2a2+a2+2a2cos⁡θ=2a2cos⁡θ=0 or θ=90∘So, angle between A and B - 90"Again B+C = -Aor (B+C)⋅(B+C)=(−A)⋅(−A)or B2+C2+2B⋅C=A2a2+2a2+22a2cos⁡ϕ=a2cos⁡ϕ=−1/2 or ϕ=135∘So, angle between B and C is 135°'similarly, angle between c and A is 135°.