If the origin is the centroid of the triangle | vedclass

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(A) The coordinates of the centroid of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ are given by $\left(\frac{x_

Vedclass · Accepted Answer

$a = -2, b = -\frac{16}{3}, c = 2$

Vedclass · Answer

(A) The coordinates of the centroid of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ are given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.Given the vertices $P(2a, 2, 6)$,$Q(-4, 3b, -10)$,and $R(8, 14, 2c)$,the centroid is:$\left(\frac{2a - 4 + 8}{3}, \frac{2 + 3b + 14}{3}, \frac{6 - 10 + 2c}{3}\right) = \left(\frac{2a + 4}{3}, \frac{3b + 16}{3}, \frac{2c - 4}{3}\right)$.Since the origin $(0, 0, 0)$ is the centroid,we equate the coordinates:$\frac{2a + 4}{3} = 0$ $\Rightarrow 2a = -4$ $\Rightarrow a = -2$.$\frac{3b + 16}{3} = 0$ $\Rightarrow 3b = -16$ $\Rightarrow b = -\frac{16}{3}$.$\frac{2c - 4}{3} = 0$ $\Rightarrow 2c = 4$ $\Rightarrow c = 2$.Thus,the values are $a = -2, b = -\frac{16}{3}, c = 2$.

If the origin is the centroid of the triangle $PQR$ with vertices $P(2a, 2, 6)$,$Q(-4, 3b, -10)$,and $R(8, 14, 2c)$,then find the values of $a, b$,and $c$.

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