Question

If D is the mid-point of side BC of $∆ABC\mathrm{and}\overrightarrow{AB}+\overrightarrow{AC}=k\overrightarrow{AD}$ , then k = ______________.

Answer 1

Given:
D is the mid-point of side BC
$\overrightarrow{AB}+\overrightarrow{AC}=k\overrightarrow{AD}$


$$\begin{gathered} \mathrm{in}∆ABC, \\ \mathrm{using}\mathrm{triangle}\mathrm{law}\mathrm{of}\mathrm{vector}\mathrm{addition}, \\ \overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}...\left(1\right) \\ \Rightarrow \overrightarrow{BD}=\frac{1}{2}\overrightarrow{BC}\left(\because D\mathrm{is}\mathrm{the}\mathrm{mid}-\mathrm{point}\mathrm{of}BC\right) \\ \Rightarrow \overrightarrow{BD}=\frac{1}{2}\overrightarrow{AC}-\frac{1}{2}\overrightarrow{AB}...\left(2\right) \\ \mathrm{Now}, \\ \mathrm{in}∆ABD, \\ \mathrm{using}\mathrm{triangle}\mathrm{law}\mathrm{of}\mathrm{vector}\mathrm{addition}, \\ \overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD} \\ \Rightarrow \overrightarrow{AD}=\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}-\frac{1}{2}\overrightarrow{AB}\left(\mathrm{From}\left(2\right)\right) \\ \Rightarrow \overrightarrow{AD}=\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC} \\ \Rightarrow 2\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{AC} \end{gathered}$$


Hence, k = $\underline{2}.$