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  • Explain solution RD Sharma class 12 Chapter 22 Algebra of Vectors Exercise 22.6 question 9

Explain solution RD Sharma class 12 Chapter 22 Algebra of Vectors Exercise 22.6 question 9

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Answer:

Vector  O \vec{C}  will be  \frac{2}{3} \hat{\imath}+\frac{2}{3} \hat{j}+\frac{4}{3} \hat{k}

Hint

Use position vector formula.

Given:

Vertices of the triangle are  (1,-1,2),(2,1,3) \text { and }(-1,2,-1)

Solution:

Centroid of the triangle with vertices  (1,-1,2),(2,1,3)  and  (-1,2,-1)  is given by:

(x, y, z)=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}, \frac{z_{1}+z_{2}+z_{3}}{3}\right)

In vector algebra, x is consider as a co – efficient of  \hat{i}\text { and } \mathrm{y} as a co – efficient of  \hat{j}\text { and } \mathrm{z} as a co – efficient of  \hat{k}

So, the position vector of the centroid

\begin{aligned} &(\hat{i}, \hat{\jmath}, \hat{k})=\frac{1+2-1}{3}, \frac{-1+1+2}{3}, \frac{2+3-1}{3} \\ &(\hat{i}, \hat{\jmath}, \hat{k})=\frac{2}{3}, \frac{2}{3}, \frac{4}{3} \end{aligned}

So, the location of the centroid is   \left(\frac{2}{3}, \frac{2}{3}, \frac{4}{3}\right)

And the vector is,

O \vec{C}=\frac{2}{3} \hat{\imath}+\frac{2}{3} \hat{j}+\frac{4}{3} \hat{k}

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