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Algebra of Vectors — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Vectors5 Marks
Question
Five forces $\overrightarrow{\text{AB}},\ \overrightarrow{\text{AC}},\ \overrightarrow{\text{AD}},\ \overrightarrow{\text{AE}}\text{ and }\overrightarrow{\text{AF}}$ act at the vertex of a regular hexagon ABCDEF. Prove that the resultant is:$6\ \overrightarrow{\text{AO}}$, where o is the center of hexagon.
✓

Answer


$\overrightarrow{\text{AB}}+\overrightarrow{\text{AC}}+\overrightarrow{\text{AD}}+\overrightarrow{\text{AE}}+\overrightarrow{\text{AF}}$
Consider $\triangle\text{ADE}$,
$\overrightarrow{\text{AD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}=0$
$\overrightarrow{\text{AD}}+\overrightarrow{\text{DE}}=\overrightarrow{\text{AE}}$
$2\ \overrightarrow{\text{AO}}+\Big(-\overrightarrow{\text{AB}}\Big)=\overrightarrow{\text{AE}}$ $\Big[\overrightarrow{\text{AD}}=2\ \overrightarrow{\text{AO}}\text{ and }\overrightarrow{\text{ED}}\ \Big|\Big|\ \overrightarrow{\text{AB}}+\overrightarrow{\text{DE}}=-\overrightarrow{\text{AB}}\Big]$
$\therefore\ \overrightarrow{\text{AE}}+\overrightarrow{\text{AB}}=2\ \overrightarrow{\text{AO}}\ \dots(1)$
Now, consider $\triangle\text{ADC}$,
$\overrightarrow{\text{AC}}+\overrightarrow{\text{CD}}+\overrightarrow{\text{DA}}=0$
$\overrightarrow{\text{AC}}+\overrightarrow{\text{CD}}=\overrightarrow{\text{AD}}$ $\Big[\because\ \overrightarrow{\text{CD}}=\overrightarrow{\text{AF}}\Big]$
$\overrightarrow{\text{AC}}+\overrightarrow{\text{AF}}=2\ \overrightarrow{\text{AO}}\ \dots(2)$
Using (1) and (2),
$\overrightarrow{\text{AB}}+\overrightarrow{\text{AE}}+\overrightarrow{\text{AC}}+\overrightarrow{\text{AF}}+\overrightarrow{\text{AD}}$
$2\ \overrightarrow{\text{AO}}+2\ \overrightarrow{\text{AO}}+2\ \overrightarrow{\text{AO}}$
$=6\ \overrightarrow{\text{AO}}$

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