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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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Answer» 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)` |
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