Question

If $ABC$ is a traingle , right angled at $A$ the resultant forces acting along $\overrightarrow{AB}$ and $\overrightarrow{AC}$ with magnitude $\frac{1}{AB}$ and $\frac{1}{AC}$ respectively is the force along $AD$.where $D$ is the foot of the perpendicular from $A$ onto $BC$ the square of the magnitude of the resultant is

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

Answer 1

The correct option is C $\frac{A{B}^2+A{C}^2}{{\left(AB\right)}^2{\left(AC\right)}^2}$

Let $\hat{i}$ and $\hat{j}$ be unit vectors along $AB$ and $AC$.
$\overrightarrow{AD}=\frac{\hat{i}}{AB}+\frac{\hat{j}}{AC}$
$\overrightarrow{AD}=AD\cos \theta \hat{i}+AD\sin \theta \hat{j}$
$\therefore AD\cos \theta =\frac{1}{AB},AD\sin \theta =\frac{1}{AC}$
squaring and adding
${AD}^2{\cos }^2\theta +{AD}^2{\sin }^2\theta =\frac{1}{{AB}^2}+\frac{1}{{AC}^2}$
$\Rightarrow {AD}^2({\cos }^2\theta +{\sin }^2\theta )=\frac{1}{{AB}^2}+\frac{1}{{AC}^2}$
$\Rightarrow {AD}^2=\frac{1}{{AB}^2}+\frac{1}{{AC}^2}$
$\Rightarrow {AD}^2=\frac{A{B}^2+A{C}^2}{{\left(AB\right)}^2{\left(AC\right)}^2}$