JEE Main · 2006 · 30 Apr 2006 · Offline · Vector Algebra

Place the right angle at $A$ so that the two forces, of magnitudes $\frac{1}{AB}$ and $\frac{1}{AC}$, act along the mutually perpendicular directions $\overrightarrow{AB}$ and $\overrightarrow{AC}$. Being perpendicular, their resultant has magnitude $\sqrt{{\left(\frac{1}{AB}\right)}^2+{\left(\frac{1}{AC}\right)}^2}=\frac{\sqrt{A{B}^2+A{C}^2}}{AB\cdot AC}$. Now $\sqrt{A{B}^2+A{C}^2}=BC$ by Pythagoras, and comparing the two expressions for the area of the triangle, $\frac{1}{2}AB\cdot AC=\frac{1}{2}BC\cdot AD$, gives $AB\cdot AC=BC\cdot AD$. Substituting, the magnitude becomes $\frac{BC}{AB\cdot AC}=\frac{BC}{BC\cdot AD}=\frac{1}{AD}$, which is exactly the size of the resultant directed along $\overrightarrow{AD}$.