There are two force vectors,one of 5, N and t | vedclass

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(A) The resultant $R$ of two vectors $A$ and $B$ is given by $R = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$.$1$. For $R = 17\, N$: Since $17 = 5 + 12$,the v

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$0^o, 180^o$ and $90^o$

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(A) The resultant $R$ of two vectors $A$ and $B$ is given by $R = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$.$1$. For $R = 17\, N$: Since $17 = 5 + 12$,the vectors must be in the same direction. Thus,$	heta = 0^o$.$2$. For $R = 7\, N$: Since $7 = 12 - 5$,the vectors must be in opposite directions. Thus,$	heta = 180^o$.$3$. For $R = 13\, N$: Using the Pythagorean theorem,$\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\, N$. This occurs when the vectors are perpendicular,so $	heta = 90^o$.

There are two force vectors,one of $5\, N$ and the other of $12\, N$. At what angle should the two vectors be added to get a resultant vector of $17\, N$,$7\, N$,and $13\, N$ respectively?

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