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 its 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 have 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 question. The coordinate of the fourth vertex of the tetrahedron is:

(- 6, 2, 16)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(1, - 2, 13)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(- 2, 4, - 2)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(1, - 1, 1)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $(- 6, 2, 16)$
Given three vertices of tetrahedron are $(0, 0, 0), (6, 5, 1), (4, 1, 3)$
Now equation of plane passing through $(0, 0, 0)$ is given by,
$a x + b y + c z = 0$
Also this line passing through other two points
$⟹ 6 a + 5 b + c = 0, 4 a + b + 3 c = 0$
Solving these, we get $a = - c, b = c$
Hence required plane will be $- x + y + z = 0$
From the above equationn the coordinate of the fourth vertex of the tetrahedron is $(- 6, 2, 16)$

Suggest Corrections

Similar questions

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

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

Show that the coordinates of the centroid of the 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 $(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}, \frac{z_{1} + z_{2} + z_{3}}{3})$ .

ABCD is a parallelogram with vertices A (X₁, Y₁) , B(X₂, Y₂) and C (X₃, Y₃). Then the coordinates of the fourth vertex D in terms of the coordinates of A, B and C are

The coordinates of the mid-point of the line segment joining two points P(X₁, y₁, z₁) and Q(x₂, y₂, z₂) are:

Join BYJU'S Learning Program