Question

Calculate the volume of the tetrahedron whose vertices are the points A = (3, 2, 1),
B = (1, 2, 4), C = (4, 0, 3) and D = (1, 1, 7).

Answer 1

Answer on Question #62033 – Math – Geometry

Question

Calculate the volume of the tetrahedron whose vertices are the points $A = (3, 2, 1)$, $B = (1, 2, 4)$, $C = (4, 0, 3)$ and $D = (1, 1, 7)$.

Solution

The absolute value $\left| (\vec{a} \times \vec{b}) \cdot \vec{c} \right|$ of the scalar triple product of vectors $\vec{a}, \vec{b}, \vec{c}$ is the volume of the parallelepiped spanned by $\vec{a}, \vec{b}$ and $\vec{c}$. A tetrahedron is a pyramid, so the formula for the volume is

$$V = \frac{1}{3} A h,$$

where $A$ is base area (area triangle), $h$ is a height of pyramid. The basis of the parallelepiped is a parallelogram, whose area is twice the area of the triangle formed by the same vectors. The height of the pyramid and parallelepiped built on the same vectors are the same.

Hence

$$\left| (\vec{a} \times \vec{b}) \cdot \vec{c} \right| = 6 V_{\text{tetrahedron}}$$

and

$$V_{\text{tetrahedron}} = \frac{1}{6} \left| (\overrightarrow{AB} \times \overrightarrow{AC}) \cdot \overrightarrow{AD} \right|.$$

Vectors $\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}$ are given as follows:

$$\begin{array}{l} \overrightarrow{AB} = (x_B - x_A, y_B - y_A, z_B - z_A) = (1 - 3,2 - 2,4 - 1) = (-2,0,3); \\ \overrightarrow{AC} = (x_C - x_A, y_C - y_A, z_C - z_A) = (4 - 3,0 - 2,3 - 1) = (1, -2,2); \\ \overrightarrow{AD} = (x_D - x_A, y_D - y_A, z_D - z_A) = (1 - 3,1 - 2,7 - 1) = (-2, -1,6). \end{array}$$

The scalar triple product $(\overrightarrow{AB} \times \overrightarrow{AC}) \cdot \overrightarrow{AD}$ is

$$(\overrightarrow{AB} \times \overrightarrow{AC}) \cdot \overrightarrow{AD} = \left| \begin{array}{ccc} -2 & 0 & 3 \\ 1 & -2 & 2 \\ -2 & -1 & 6 \end{array} \right| = (-2) \cdot (-2) \cdot 6 + 0 \cdot 2 \cdot (-2) + 1 \cdot (-1) \cdot 3 - (-2) \cdot (-2) \cdot 3 - (-1) \cdot 2 \cdot (-2) - 1 \cdot 0 \cdot 6 = 24 + 0 - 3 - 12 - 4 - 0 = 24 - 19 = 5,$$

hence

$$\left| (\overrightarrow{AB} \times \overrightarrow{AC}) \cdot \overrightarrow{AD} \right| = 5.$$

Therefore, the volume of the tetrahedron equals

$$V_{\text{tetrahedron}} = \frac{1}{6} \cdot 5 = \frac{5}{6}.$$

Answer: $\frac{5}{6}$.

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