## Problem Statement

Calculate the volume of the tetrahedron with vertices at points \( A_{1}, A_{2}, A_{3}, A_{4} \) and its height dropped from vertex \( A_{4} \) to the face \( A_{1} A_{2} A_{3} \).

\( A_{1}(1 ; 1 ;-1) \)
\( A_{2}(2 ; 3 ; 1) \)

\( A_{3}(3 ; 2 ; 1) \)

\( A_{4}(5 ; 9 ;-8) \)
## Solution

From vertex $A_{1}$, we draw vectors:

$\overrightarrow{A_{1} A_{2}}=\{2-1 ; 3-1 ; 1-(-1)\}=\{1 ; 2 ; 2\}$

$\overrightarrow{A_{1} A_{3}}=\{3-1 ; 2-1 ; 1-(-1)\}=\{2 ; 1 ; 2\}$

$\vec{A}_{1} A_{4}=\{5-1 ; 9-1 ;-8-(-1)\}=\{4 ; 8 ;-7\}$

According to the geometric meaning of the scalar triple product, we have:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot\left|\left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)\right|
$$

We compute the scalar triple product:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
1 & 2 & 2 \\
2 & 1 & 2 \\
4 & 8 & -7
\end{array}\right|= \\
& =1 \cdot\left|\begin{array}{cc}
1 & 2 \\
8 & -7
\end{array}\right|-2 \cdot\left|\begin{array}{cc}
2 & 2 \\
4 & -7
\end{array}\right|+2 \cdot\left|\begin{array}{cc}
2 & 1 \\
4 & 8
\end{array}\right|= \\
& =1 \cdot(-23)-2 \cdot(-22)+2 \cdot 12=-23+44+24=45
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|45|=\frac{15}{2}=7.5
$$

Since

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{3} \cdot S_{A_{1} A_{2} A_{3}} \cdot h \Rightarrow h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}
$$

According to the geometric meaning of the vector product:

$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot\left|\overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}\right|
$$

We compute the vector product:
$\overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}i & j & k \\ 1 & 2 & 2 \\ 2 & 1 & 2\end{array}\right|=i \cdot\left|\begin{array}{ll}2 & 2 \\ 1 & 2\end{array}\right|-j\left|\begin{array}{ll}1 & 2 \\ 2 & 2\end{array}\right|+k \cdot\left|\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right|=$

$=i \cdot 2-j \cdot(-2)+k \cdot(-3)=\{2 ; 2 ;-3\}$

We obtain:

$S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{2^{2}+2^{2}+(-3)^{2}}=\frac{\sqrt{17}}{2}$

Then:

$h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot \frac{15}{2}}{\frac{\sqrt{17}}{2}}=\frac{45}{\sqrt{17}}$

Volume of the tetrahedron: 7.5

Height: $\frac{45}{\sqrt{17}}$

## Problem Kuznetsov Analytic Geometry 7-15