JEE Main & Advanced Physics Vectors Triangle Law of Vector Addition of Two Vectors

If two non zero vectors are represented by the two sides of a triangle taken in same order then the resultant is given by the closing side of triangle in opposite order. i.e.\[\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}\]
Q \[\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}\]    

(1) Magnitude of resultant vector

In \[\Delta \,ABN,\]\[\cos \theta =\frac{AN}{B}\]\[\therefore \]\[AN=B\cos \theta \]           

\[\sin \theta =\frac{BN}{B}\]  \[\therefore \] \[BN=B\sin \theta \]

In \[\Delta OBN,\] we have \[O{{B}^{2}}=O{{N}^{2}}+B{{N}^{2}}\]

\[\Rightarrow \]\[{{R}^{2}}={{(A+B\cos \theta )}^{2}}+{{(B\sin \theta )}^{2}}\] 

\[\Rightarrow \]\[{{R}^{2}}={{A}^{2}}+{{B}^{2}}{{\cos }^{2}}\theta +2AB\cos \theta +{{B}^{2}}{{\sin }^{2}}\theta \]

\[\Rightarrow \]\[{{R}^{2}}={{A}^{2}}+{{B}^{2}}({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )+2AB\cos \theta \]

\[\Rightarrow \]\[{{R}^{2}}={{A}^{2}}+{{B}^{2}}+2AB\cos \theta \] \[\Rightarrow \]

\[R=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }\]

(2) Direction of resultant vectors : If \[\theta \] is angle between \[\overrightarrow{A}\] and \[\overrightarrow{B,}\] then

\[\,|\overrightarrow{A}+\overrightarrow{B}|\,=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }\]

If \[\overrightarrow{R}\]makes an angle a with \[\overrightarrow{A},\] then in \[\Delta OBN,\]

\[\tan \alpha =\frac{BN}{ON}=\frac{BN}{OA+AN}\]

\[\tan \alpha =\frac{B\sin \theta }{A+B\cos \theta }\]