The position vector of the point of intersect | vedclass

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(A) The centroid $G$ of a triangle with vertices $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$ is given by the formula:$G = \left( \fra

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$2 \hat{i} + \hat{j} + \hat{k}$

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(A) The centroid $G$ of a triangle with vertices $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$ is given by the formula:$G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right)$Given vertices are $A(1, 2, 3)$,$B(1, 0, 3)$,and $C(4, 1, -3)$.Substituting the values:$G = \left( \frac{1 + 1 + 4}{3}, \frac{2 + 0 + 1}{3}, \frac{3 + 3 - 3}{3} \right)$$G = \left( \frac{6}{3}, \frac{3}{3}, \frac{3}{3} \right)$$G = (2, 1, 1)$The position vector of the centroid is $2 \hat{i} + \hat{j} + \hat{k}$.

The position vector of the point of intersection of the medians (centroid) of a triangle,whose vertices are $A(1, 2, 3)$,$B(1, 0, 3)$,and $C(4, 1, -3)$ is

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