The magnitude of vectors `vec(OA), vec(OB) and vec (OC)` in figure are equal. Find the direction of `vec(OA)+vec(OB)-vec(OC)`.
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Explanation

Let `OA=OB=OC=F`.
x - component of `vec(OA)=F cos 30^0= F sqrt(3)/2`
x-component of `vec(OB)=F cos60^0 = F/2`
`x-component of `vec(OC)=F cos 135^0= - F/sqrt2`
`x-component of `vec(OA)+vec(OB)-vec(OC)`
`-((Fsqrt(3))/2)+(F/2)+(-F/sqrt(2))`
`= F/2 (sqrt3+1+sqrt2)`.
y-component of `vec(OA)=F cos 60^0=F/2`
y- component of `vec(OB)=F cos 150^0 = (Fsqrt(3)/2`
y- component of `vec(OC)=F cos 45^0=F/sqrt2`.
y-component of `vec(OA)+vec(OB)+vec(OC)`
`=(F/2)(sqrt3+1+sqrt2)`
y-component of `vec(OA)=F cos 60^0= F/2`
y-component of `vec(OC)=F cos 45^0=F/sqrt2`
y-component of `vec(OA)+vec(OB)+vec(OC)`
`=(F/2)+(-(Fsqrt3)/2-(F/sqrt2))`
=`F/2 (1-sqrt3-sqrt2)`
Angle of `vec(OA)+vec(OB)-vec(OC)` with the X - axis
`=tan^-1 (F/2 (1-sqrt3-sqrt2))/(F/2 (1+sqrt3+sqrt2))= tan^-1 (1-sqrt3-sqrt2)/(1+sqrt3+sqrt2)`