What is the angle between two vector forces of equal magnitude such that the result is one third as much as either of the original forces?

Hi Praveen,

Let the force vectors be A and B.
Then 
|A| =|B| ….......................(1)   {Given}

And  also,

|A +B| = |A|
Squaring both sides,
|A|^2 + |B|^2 + 2.A.B = |A|^2
|B|^2 + 2A.B = 0..........................(2)

Now Angle between the tqo vectors is:
cos@ = A.B/|A|.|B| = A.B/|B|^2 (Given = |A| =|B|).
 From eqn. (2)

A.B/|B|^2 = -1/2 = cos@
@ = 120 degree.