• question_answer ABC is a triangle, right singled at A. The resultant of the forces acting along $\overrightarrow{AB},\,\overrightarrow{AC}$ with magnitudes$\frac{1}{AB}$and$\frac{1}{AC}$respectively is the force along $\overrightarrow{AD},$ where D is the foot of the perpendicular from A to BC. The magnitude of the resultant is     AIEEE  Solved  Paper-2006 A) $\frac{(AB)(AC)}{AB+AC}$        B)        $\frac{1}{AB}+\frac{1}{AC}$     C)        $\frac{1}{AD}$                                D)        $\frac{A{{B}^{2}}+A{{C}^{2}}}{(A{{B}^{2}}){{(AC)}^{2}}}$

Let $|BC|=l$ In $\Delta ABC,$ $l=\sqrt{A{{B}^{2}}+A{{C}^{2}}}$ and   $\tan \theta =\frac{AB}{AC}$ $\Rightarrow$ $\sin \theta =\frac{AB}{l}$and $\cos \theta =\frac{AC}{l}$ $\therefore$Resultant vector $=\frac{1}{AB}\hat{i}+\frac{1}{AC}\hat{j}=\left( \frac{1}{l\sin \theta }\hat{i}+\frac{1}{l\cos \theta }\hat{j} \right)$ $=k\,AD$ Now,$AD=AC\text{ }\sin \theta =l\,\cos \theta \sin \theta$ $=\frac{AB.AC}{l}$                                            ...(i) Magnitude of resultant vector $=\sqrt{\frac{1}{{{t}^{2}}}\left( \frac{1}{{{\sin }^{2}}\theta }+\frac{1}{{{\cos }^{2}}\theta } \right)}$ $=\frac{l}{(AB)(AC)}=\frac{1}{AD}$                       [from Eq.(i)]