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If G is the Intersection of Diagonals of a Parallelogram Abcd and O is Any Point, Then O → a + O → B + O → C + O → D = - Mathematics

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प्रश्न

If G is the intersection of diagonals of a parallelogram ABCD and O is any point, then \[O \vec{A} + O \vec{B} + O \vec{C} + O \vec{D} =\] 

विकल्प

  • \[2 \overrightarrow{OG}\]

     

  • \[4 \overrightarrow{OG}\]

     

  • \[5 \overrightarrow{OG}\]

     

  • \[3 \overrightarrow{OG}\]
MCQ

उत्तर

\[4 \overrightarrow{OG}\]
Let us consider the point O as origin.
G is the mid point of AC.

\[\therefore \overrightarrow{OG} = \frac{\overrightarrow{OA} + \overrightarrow{OC}}{2}\]
\[2 \overrightarrow{OG} = \overrightarrow{OA} + \overrightarrow{OC} . . . . . \left( 1 \right)\]
Also, G is the mid point BD
\[\therefore \overrightarrow{OG} = \frac{\overrightarrow{OB} + \overrightarrow{OD}}{2}\]
\[2 \overrightarrow{OG} = \overrightarrow{OB} + \overrightarrow{OD} . . . . . \left( 2 \right)\]
On adding (1) and (2) we get,
\[2 \overrightarrow{OG} + 2 \overrightarrow{OG} = \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} \]
\[4 \overrightarrow{OG} = \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} \]
\[ \therefore \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} = 4 \overrightarrow{OG}\]
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अध्याय 23: Algebra of Vectors - MCQ [पृष्ठ ७८]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 23 Algebra of Vectors
MCQ | Q 6 | पृष्ठ ७८

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