Question

$A B C$ is a triangle, right angled at $A .$ The resultant of the forces acting along $\overline{A B}, \overline{B C}$ with magnitudes $\frac{1}{A B}$ and $\frac{1}{A C}$ respectively is the force along $\overline{A D},$ where $D$ is the foot of the perpendicular from $A$ onto $B C$. The magnitude of the resultant is

• A. $\frac{\left(AB\right)\left(AC\right)}{AB+AC}$
• B. $\frac{1}{AB}+\frac{1}{AC}$
• C. $\frac{1}{AD}$
• D. $\frac{AB^{2}+AC^{2}}{\left(AB\right)^{2}\left(AC\right)^{2}}$

Answer 1

Answer: (C)

$\frac{1}{AD}$

Answer 2

Let $\left|\overrightarrow{BC}\right|=l$ $\therefore In\,\Delta\Lambda ABC,$ $l=\sqrt{AB^{2}+AC^{2}}$ and $tan \,\theta=\frac{AB}{AC}$ $\Rightarrow sin\,\theta=\frac{AB}{l}$ and $cos\,\theta=\frac{AC}{l}$ $\therefore$ Resultant vector $\frac{1}{AB} \hat{i}+\frac{1}{AC} \hat{j}=\left(\frac{1}{l\,sin\,\theta} \hat{i}+\frac{1}{l\,cos\,\theta} \hat{j}\right)$ $=k\,\overrightarrow{AD}$ Now, $AD=AC\,sin\,\theta=l\,sin\,\theta\,cos\,\theta$ $=\frac{AB\cdot AC}{l}$ Magnitude of resultant vector $=\sqrt{\frac{1}{l^{2}}\left(\frac{1}{sin^{2}\,\theta}+\frac{1}{cos^{2}\,\theta}\right)}$ $=\frac{l}{\left(AB\right)\left(AC\right)}=\frac{1}{AD}$