Question

Five forces $\overrightarrow{AB,}\overrightarrow{AC,}\overrightarrow{AD,}\overrightarrow{AE}$ and $\overrightarrow{AF}$ act at the vertex of a regular hexagon ABCDEF. Prove that the resultant is
6 $\overrightarrow{AO,}$ where O is the centre of hexagon.

Answer 1

$\overrightarrow{\mathit{A}\mathit{B}}\mathit{+}\overrightarrow{\mathit{A}\mathit{C}}\mathit{+}\mathit{}\overrightarrow{\mathit{A}\mathit{D}}\mathit{+}\overrightarrow{\mathit{A}\mathit{E}}\mathit{+}\overrightarrow{\mathit{A}\mathit{F}}$

Consider ∆ADE,

$\begin{array}{l}\overrightarrow{AD}\mathit{+}\overrightarrow{DE}\mathit{+}\overrightarrow{EA}\mathit{=}\mathit{0}\\ \overrightarrow{AD}\mathit{+}\overrightarrow{DE}\mathit{=}\overrightarrow{AE}\\ \mathit{2}\overrightarrow{AO}\mathit{+}\left(\mathit{-}\overrightarrow{AB}\right)\mathit{=}\overrightarrow{AE}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\overrightarrow{AD}\mathit{=}\mathit{2}\overrightarrow{AO}\mathit{\hspace{0.17em}}\mathit{\hspace{0.17em}}\mathrm{and}\mathit{}ED\mathit{}\mathit{\left|}\mathit{\right|}\hspace{0.17em}AB\mathit{\hspace{0.17em}}\mathit{\hspace{0.17em}}\overrightarrow{DE}\mathit{=}\mathit{-}\overrightarrow{AB}\right]\\ \mathit{\therefore }\mathit{\hspace{0.17em}}\mathit{\hspace{0.17em}}\overrightarrow{AE}\mathit{+}\overrightarrow{AB}\mathit{=}\mathit{2}\overrightarrow{AO}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\left(\mathrm{1}\right)\end{array}$

Now, consider ∆ADC

$$\begin{gathered} \overrightarrow{AC}+\overrightarrow{CD}+\overrightarrow{DA}=0 \\ \overrightarrow{AC}+\overrightarrow{CD}=\overrightarrow{AD}\left[\because \overrightarrow{CD}=\overrightarrow{AF}\right] \\ \overrightarrow{AC}+\overrightarrow{AF}=2\overrightarrow{AO}.....\left(2\right) \end{gathered}$$

Using (1) and (2),

$$\begin{gathered} \overrightarrow{AB}+\overrightarrow{AE}+\overrightarrow{AC}+\overrightarrow{AF}+\overrightarrow{AD} \\ 2\overrightarrow{AO}+2\overrightarrow{AO}+2\overrightarrow{AO} \\ =6\overrightarrow{AO} \end{gathered}$$