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If Ad is the Median of ∆Abc, Using Vectors, Prove that a B 2 + a C 2 = 2 ( a D 2 + C D 2 ) - Mathematics

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प्रश्न

If AD is the median of ∆ABC, using vectors, prove that \[{AB}^2 + {AC}^2 = 2\left( {AD}^2 + {CD}^2 \right)\] 

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उत्तर

 

Taking A as the origin, let the position vectors of B and C be \[\vec{b}\] and \[\vec{c}\] respectively.  

It is given that AD is the median of ∆ABC. 
∴ Position vector of mid-point of BC = \[\vec{AD} = \frac{\vec{b} + \vec{c}}{2}\]................(Mid-point formula) 

Now, 

\[{AB}^2 + {AC}^2 = \left| \vec{AB} \right|^2 + \left| \vec{AC} \right|^2 = \left| \vec{b} \right|^2 + \left| \vec{c} \right|^2\] 

Also, 

\[2\left( {AD}^2 + {CD}^2 \right)\]
\[ = 2\left( \left| \vec{AD} \right|^2 + \left| \vec{CD} \right|^2 \right)\]
\[ = 2\left[ \left( \frac{\vec{b} + \vec{c}}{2} \right) . \left( \frac{\vec{b} + \vec{c}}{2} \right) + \left( \frac{\vec{b} + \vec{c}}{2} - \vec{c} \right) . \left( \frac{\vec{b} + \vec{c}}{2} - \vec{c} \right) \right]\]
\[ = 2\left[ \left( \frac{\vec{b} + \vec{c}}{2} \right) . \left( \frac{\vec{b} + \vec{c}}{2} \right) + \left( \frac{\vec{b} - \vec{c}}{2} \right) . \left( \frac{\vec{b} - \vec{c}}{2} \right) \right]\]
\[ = \frac{\left| \vec{b} \right|^2 + 2 \vec{b} . \vec{c} + \left| \vec{c} \right|^2}{2} + \frac{\left| \vec{b} \right|^2 - 2 \vec{b} . \vec{c} + \left| \vec{c} \right|^2}{2}\]
\[ = \frac{2 \left| \vec{b} \right|^2 + 2 \left| \vec{c} \right|^2}{2}\]
\[ = \left| \vec{b} \right|^2 + \left| \vec{c} \right|^2 . . . . . \left( 2 \right)\] 

From (1) and (2), we have 

\[{AB}^2 + {AC}^2 = 2\left( {AD}^2 + {CD}^2 \right)\]

 

 

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अध्याय 24: Scalar Or Dot Product - Exercise 24.2 [पृष्ठ ४६]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 24 Scalar Or Dot Product
Exercise 24.2 | Q 8 | पृष्ठ ४६

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