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  • Please Solve R.D. Sharma class 12 Chapter 22 Algbra of Vectors Exercise Very Short Answer Question 10 Maths textbook Solution.

Please Solve R.D. Sharma class 12 Chapter 22 Algbra of Vectors  Exercise Very Short Answer Question 10 Maths textbook Solution.

Answers (1)

Answer: \frac{\vec{a}+\vec{b}+\vec{c}}{3}

Hint: You must know the rules of vector functions

Given: If \vec{a}+\vec{b}+\vec{c}are position vectors of vertices of a triangle, then write the position vector of centroid.

Solution: Let ABC be a triangle and D,E and F are the mid-points of sides BC, CA and AB respectively.

Also, Let \vec{a}+\vec{b}+\vec{c}are position vectors of A, B, C respectively.                          

Then position vectors of D, E and F are                                             

                \left(\frac{\vec{b}+\vec{c}}{2}\right) \cdot\left(\frac{\vec{c}+\vec{a}}{2}\right),\left(\frac{\vec{a}+\vec{b}}{2}\right) respectively.                                                            

The position vector of a point divides AD in the ratio of 2 is                                        

        

\frac{1 \cdot \vec{a}+2\left(\frac{\vec{b}+\vec{c}}{2}\right)}{3}=\frac{\vec{a}+\vec{b}+\vec{c}}{3}

 

Similarly, position vectors of the points divides BE, CF in the ratio of 2:1 are equal to

 

            \frac{\vec{a}+\vec{b}+\vec{c}}{3}

Thus, the points dividing AD in ratio 2:1 also divides BE, CF in the ratio.

Hence, medians of triangle are concurrent and the position of centroid is \frac{\vec{a}+\vec{b}+\vec{c}}{3}

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