The vectors A→1 and A→2 each of magnitude A are inclined to each other such that their resultant is equal to 3A. Then the resultant of A→1 and −A→2 is.

Let θ be the angle between, A1 and A2.Then, A12+A22+2A1A2cos⁡θ=R2or A2+A2+2AAcosθ=3A2 or cos⁡θ=12=cos⁡60∘ or θ=60∘The angle between A1 and -A2 is 180∘−60∘=120∘therefore resultant of A1 and -A2 is ‘R’A12+A22+2A1A2cos⁡180∘−60∘]1/2=A2+A2+2AAcos⁡120∘1/2=A