Question
Using vectors, prove that the parallelogram on the same base and between the same parallels are equal in area.
✓
Answer
Let ABCD and ABFE are parallelograms on the same base AB and between the same parallel line AB and DF.
Let $\overrightarrow{\text{AB}}=\vec{\text{a}}$ and $\overrightarrow{\text{AD}}=\vec{\text{b}}$
$\therefore$ Area of parallelogram $\overrightarrow{\text{ABCD}}=\vec{\text{a}}\times\vec{\text{b}}$
Now, area of parallelogram of $\overrightarrow{\text{ABFE}}=\overrightarrow{\text{AB}}\times\overrightarrow{\text{AE}}$
$=\overrightarrow{\text{AB}}\times(\overrightarrow{\text{AD}}+\overrightarrow{\text{DE}})$
$=\overrightarrow{\text{AB}}\times(\vec{\text{b}}+\text{k}\vec{\text{a}})$ $[\text{let}\overrightarrow{\text{ DE}}=\text{k}\vec{\text{a}},\text{where k is a scalar}]$
$=\vec{\text{a}}\times(\vec{\text{b}}+\text{k}\vec{\text{a}})$
$=(\vec{\text{a}}\times\vec{\text{b}})+(\vec{\text{a}}\times\text{k}\vec{\text{a}})$
$=(\vec{\text{a}}\times\vec{\text{b}})+\text{k}(\vec{\text{a}}\times\vec{\text{a}})$
$=(\vec{\text{a}}\times\vec{\text{b}})\ [\because\vec{\text{a}}\times\vec{\text{a}}=0]$
Area of parallelogram ABCD.
Hence proved.
Let $\overrightarrow{\text{AB}}=\vec{\text{a}}$ and $\overrightarrow{\text{AD}}=\vec{\text{b}}$
$\therefore$ Area of parallelogram $\overrightarrow{\text{ABCD}}=\vec{\text{a}}\times\vec{\text{b}}$
Now, area of parallelogram of $\overrightarrow{\text{ABFE}}=\overrightarrow{\text{AB}}\times\overrightarrow{\text{AE}}$
$=\overrightarrow{\text{AB}}\times(\overrightarrow{\text{AD}}+\overrightarrow{\text{DE}})$
$=\overrightarrow{\text{AB}}\times(\vec{\text{b}}+\text{k}\vec{\text{a}})$ $[\text{let}\overrightarrow{\text{ DE}}=\text{k}\vec{\text{a}},\text{where k is a scalar}]$
$=\vec{\text{a}}\times(\vec{\text{b}}+\text{k}\vec{\text{a}})$
$=(\vec{\text{a}}\times\vec{\text{b}})+(\vec{\text{a}}\times\text{k}\vec{\text{a}})$
$=(\vec{\text{a}}\times\vec{\text{b}})+\text{k}(\vec{\text{a}}\times\vec{\text{a}})$
$=(\vec{\text{a}}\times\vec{\text{b}})\ [\because\vec{\text{a}}\times\vec{\text{a}}=0]$
Area of parallelogram ABCD.
Hence proved.
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