The position vector of the point of intersection of the medians of a triangle, whose vertices are…

The position vector of the point of intersection of the medians of a triangle, whose vertices are $\mathrm{A}(1,2,3), \mathrm{B}(1,0,3)$ and $\mathrm{C}(4,1,-3)$ is
  1. $6 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
  2. $2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$
  3. $\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$
  4. $\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$

Solution

Let $\mathrm{M}$ be the mid point of $\mathrm{BC}$ and $\mathrm{N}$ be the mid point of $\mathrm{AC}$. $\therefore \mathrm{M}=\left(\frac{5}{2}, \frac{1}{2}, 0\right)$ We know that centroid G divides AM internally in the ratio $2: 1$ $\begin{aligned} & \therefore G=\frac{(1)(1)+(2)\left(\frac{5}{2}\right)}{2+1}, \frac{(1)(2)+(2)\left(\frac{1}{2}\right)}{2+1}, \frac{(1)(3)+0}{2+1} \\ & G=(2,1,1) \end{aligned}$

Asked in: MHT CET 2021 (22 Sep Shift 1)

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