A tetrahedron is a three dimensional figure bounded by non coplanar triangular planes. So a tetrahedron has four non-coplanar points as its vertices. Suppose a tetrehedron has points $A, B, C, D$ as it vertices which have coordinates $(x 1, y 1, z_{1}), (x_{2}, y_{2}, z_{2})$ , $(x_{3}, y_{3}, z_{3})$ and $(x_{4}, y_{4}, z_{4})$ respectively in a rectangular three dimensional space. Then the coordinates of its centroid are $[\frac{x_{1} + x_{2} + x_{3} + x_{4}}{4}, \frac{y_{1} + y_{2} + y_{3} + y_{4}}{4}, \frac{z_{1} + z_{2} + z_{3} + z_{4}}{4}]$ . Let $A$ tetrahedron has three of its vertices represented by the points $(0, 0, 0), (6, 5, 1)$ and $(4, 1, 3)$ and its centroid lies at the point $(1, 2, 5)$ . Now answer the following two questions. The equation of the traingular plane of tetrahedron that contains the given vertices

Find the coordinates of the centroid of a triangle whose vertices are (0, 6), (8, 12) and (8, 0). [2 MARKS]

Q. A tetrahedron is a three dimensional figure bounded by four non-coplanar triangular planes. So, a tetrahedron has four non-coplanar points as its vertices.
Let a tetrahedron has four points A, B, C, D whose co-ordinates are

(x_{i}, y_{i}, z_{i})

, i=1, 2, 3, 4 respectively in a rectangular three dimensional space. The co-ordinates of its centroid are given by

\frac{\sum_{i = 1}^{4} x_{i}}{4}, \frac{\sum_{i = 1}^{4} y_{i}}{4}, \frac{\sum_{i = 1}^{4} z_{i}}{4}

. The circumcenter of the tetrahedron is the center of a sphere passing through its vertices. So this point is equidistant from each of vertices of tetrahedron.
Let tetrahedron has three of its vertices represented by the points A (6, - 5, -1), B(- 4, 1, 3) & C(2, -4, 18) & its centroid lies at the point (1, -2, 5).
On the basis of above information answer the following questions.The co-ordinate of centre of the sphere circumscribe the tetrahedron is

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Four vertices of a tetrahedron are (0,0,0),(4,0,0),(0,−8,0) and (0,0,12),Its centroid has the coordinates

The correct option is C (1,−2,3) the centroid of the coordinates is (x1+x2+x3+x44,y1+y2+y3+y44,z1+z2+z3+z44)Thus by substituting the vertices we get =(0+4+0+04,0+0−8+04,0+0+0+124)=(44,−84,124)∴ the centroid of the coordinates is (1,−2,3)

Four vertices of a tetrahedron are $(0, 0, 0), (4, 0, 0), (0, - 8, 0)$ and $(0, 0, 12)$ ,Its centroid has the coordinates