Solve the problem.
-Find the vector from the origin to the center of mass of a thin triangular plate (uniform density) whose vertices are A(7, 9, 10), B(8, 3, 8), and C(9, 2, 9).
A) $- 1 \mathbf { i } + \frac { 8 } { 3 } j + \frac { 1 } { 3 } \mathbf { k }$
B) $- \frac { 1 } { 2 } \mathbf { i } + \frac { 4 } { 3 } \mathrm { j } + \frac { 1 } { 6 } \mathbf { k }$
C) $8 i + \frac { 14 } { 3 } j + 9 k$
D) $- \frac { 3 } { 2 } \mathbf { i } + 4 \mathbf { j } + \frac { 1 } { 2 } \mathbf { k }$

Question

Solve the problem.
-Find the vector from the origin to the center of mass of a thin triangular plate (uniform density) whose vertices are A(7, 9, 10), B(8, 3, 8), and C(9, 2, 9). 
A) $- 1 \mathbf { i } + \frac { 8 } { 3 } j + \frac { 1 } { 3 } \mathbf { k }$
B) $- \frac { 1 } { 2 } \mathbf { i } + \frac { 4 } { 3 } \mathrm { j } + \frac { 1 } { 6 } \mathbf { k }$
C) $8 i + \frac { 14 } { 3 } j + 9 k$
D) $- \frac { 3 } { 2 } \mathbf { i } + 4 \mathbf { j } + \frac { 1 } { 2 } \mathbf { k }$

Quizplus · Accepted Answer

The center of mass of a triangle with vertices at $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, and $C(x_3, y_3, z_3)$ is given by the average of the coordinates of the vertices. Thus, for vertices A(7, 9, 10), B(8, 3, 8), and C(9, 2, 9), the center of mass is at $\left(\frac{7+8+9}{3}, \frac{9+3+2}{3}, \frac{10+8+9}{3}\right) = \left(\frac{24}{3}, \frac{14}{3}, \frac{27}{3}\right) = (8, \frac{14}{3}, 9)$.

Solve the Problem −1i+83j+13k- 1 \mathbf { i } + \frac { 8 } { 3 } j + \frac { 1 } { 3 } \mathbf { k }−1i+38​j+31​k

Solve the Problem $- 1 \mathbf { i } + \frac { 8 } { 3 } j + \frac { 1 } { 3 } \mathbf { k }$