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Need solution for RD Sharma maths class 12 chapter 22 Algebra of Vectors exercise Fill in the blanks question 19

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Answer:-0

Hint:-To solve this equation we use mid-point formula

Given:-If D,E,F are mid points of the side BC,CA \; and \; AB respectively of \Delta ABC  then  \overrightarrow{A D}+\overrightarrow{B E}+\overrightarrow{C \vec{F}}=---                                                                                                                                

Solution:-

In  \triangle A B C, D, E, F are mid pts.                                                                                                      

Let \vec{a}, \vec{b}, \vec{c}  be the position vector of  \vec{A}, \vec{B}, \vec{C}
Using mid point formula.
As D  is a mid point of BC

D=\frac{\vec{b}+\vec{c}}{2}

 Similarly E and F are the mid points of AC and AB respectively                   

\begin{aligned} &\therefore E=\frac{\vec{a}+\vec{c}}{2} \\\\ &F=\frac{\vec{a}+\vec{b}}{2} \end{aligned}

Now,

\overrightarrow{A D}+\overrightarrow{B E}+\overrightarrow{C F}=\overrightarrow{O D}-\overrightarrow{O A}+\overrightarrow{O E}-\overrightarrow{O B}+\overrightarrow{O F}-\overrightarrow{O C}

                                 \begin{aligned} &=\overrightarrow{O D}+\vec{O} E+\overrightarrow{O F}-[\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}] \\\\ &=\frac{\vec{b}+\vec{c}}{2}+\frac{\vec{a}+\vec{c}}{2}+\frac{\vec{a}+\vec{b}}{2}-(\vec{a}+\vec{b}+\vec{c}) \end{aligned}

                                 \begin{aligned} &=\frac{2(\vec{a}+\vec{b}+\vec{c})}{2}-(\vec{a}+\vec{b}+\vec{c}) \\\\ &=(\vec{a}+\vec{b}+\vec{c})-(\vec{a}+\vec{b}+\vec{c}) \\\\ &=0 \end{aligned}

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