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)        \[\frac{A{{B}^{2}}+A{{C}^{2}}}{(A{{B}^{2}}){{(AC)}^{2}}}\]

Correct Answer: C

Solution :

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)]