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Algebra of Vectors — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Vectors3 Marks
Question
ABCDE is a pentagon, prove that,

$\overrightarrow{\text{AB}}+\overrightarrow{\text{AE}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{ED}}+\overrightarrow{\text{AC}}=3\ \overrightarrow{\text{AC}}$

✓

Answer

It is given that ABCDE is a pentagon, So
$\overrightarrow{\text{AB}}+\overrightarrow{\text{AE}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{ED}}+\overrightarrow{\text{AC}}$
$=\Big(\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}\Big)+\overrightarrow{\text{AE}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{ED}}+\overrightarrow{\text{AC}}$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{DC}}+\Big(\overrightarrow{\text{AE}}+\overrightarrow{\text{ED}}\Big)+\overrightarrow{\text{AC}}$ $\Big[$Using triangle law in $\triangle\text{ABC},\ \overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}\Big]$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{DC}}+\Big(\overrightarrow{\text{AD}}\Big)+\overrightarrow{\text{AC}}$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{DC}}-\overrightarrow{\text{DA}}+\overrightarrow{\text{AC}}$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{AD}}+\overrightarrow{\text{AC}}$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{AD}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{AC}}$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{AC}}+\overrightarrow{\text{AC}}$
$=3\ \overrightarrow{\text{AC}}$
So,
$\overrightarrow{\text{AB}}+\overrightarrow{\text{AE}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{ED}}+\overrightarrow{\text{AC}}=3\ \overrightarrow{\text{AC}}$

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