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If $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of corners $A, B, C$ of a parallelogram $\mathrm{ABCD}$, then what is the position vector of the corner $\mathrm{D}$ ?
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Verified Answer
The correct answer is:
$\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}$
Let O be the origin and $\mathrm{ABCD}$ be the parallelogram. $\operatorname{In} \Delta \mathrm{ODC}$,
$\overrightarrow{\mathrm{OD}}=\overrightarrow{\mathrm{OC}}+\overrightarrow{\mathrm{CD}}$
$\overrightarrow{\mathrm{CD}}=-\overrightarrow{\mathrm{AB}}$
and $\operatorname{In} \Delta \mathrm{AOB}, \quad \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$
Thus, $\overrightarrow{\mathrm{CD}}=-\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$
So, $\overrightarrow{\mathrm{OD}}=\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}[$ since, $\overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{C}}$ and $\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}]$
$\overrightarrow{\mathrm{OD}}=\overrightarrow{\mathrm{OC}}+\overrightarrow{\mathrm{CD}}$
$\overrightarrow{\mathrm{CD}}=-\overrightarrow{\mathrm{AB}}$
and $\operatorname{In} \Delta \mathrm{AOB}, \quad \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$
Thus, $\overrightarrow{\mathrm{CD}}=-\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$
So, $\overrightarrow{\mathrm{OD}}=\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}[$ since, $\overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{C}}$ and $\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}]$
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