The centroid of a tetrahedron with vertices P | vedclass

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Question

(C) The centroid of a tetrahedron with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,$(x_3, y_3, z_3)$,and $(x_4, y_4, z_4)$ is given by $\left(\frac{x

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) The centroid of a tetrahedron with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,$(x_3, y_3, z_3)$,and $(x_4, y_4, z_4)$ is given by $\left(\frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4}, \frac{z_1+z_2+z_3+z_4}{4}\right)$.Given vertices are $P(5, -7, 0)$,$Q(a, 5, 3)$,$R(4, -6, b)$,and $S(6, c, 2)$.The centroid is $\left(\frac{5+a+4+6}{4}, \frac{-7+5-6+c}{4}, \frac{0+3+b+2}{4}\right) = \left(\frac{15+a}{4}, \frac{-8+c}{4}, \frac{b+5}{4}\right)$.Equating this to the given centroid $(4, -3, 2)$:$\frac{15+a}{4} = 4$ $\Rightarrow 15+a = 16$ $\Rightarrow a = 1$.$\frac{-8+c}{4} = -3$ $\Rightarrow -8+c = -12$ $\Rightarrow c = -4$.$\frac{b+5}{4} = 2$ $\Rightarrow b+5 = 8$ $\Rightarrow b = 3$.Therefore,$2a + 3b + c = 2(1) + 3(3) + (-4) = 2 + 9 - 4 = 7$.

The centroid of a tetrahedron with vertices $P(5, -7, 0)$,$Q(a, 5, 3)$,$R(4, -6, b)$,and $S(6, c, 2)$ is $(4, -3, 2)$. Then the value of $2a + 3b + c$ is equal to:

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