## 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 \( A_{1} A_{2} A_{3} \).

\( A_{1}(-1 ; 2 ; 4) \)

\( A_{2}(-1 ;-2 ;-4) \)

\( A_{3}(3 ; 0 ;-1) \)

\( A_{4}(7 ;-3 ; 1) \)
## Solution

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

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{-1-(-1) ;-2-2 ;-4-4\}=\{0 ;-4 ;-8\} \\
& \vec{A}_{1} A_{3}=\{3-(-1) ; 0-2 ;-1-4\}=\{4 ;-2 ;-5\} \\
& \overrightarrow{A_{1} A_{4}}=\{7-(-1) ;-3-2 ; 1-4\}=\{8 ;-5 ;-3\}
\end{aligned}
$$

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:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
0 & -4 & -8 \\
4 & -2 & -5 \\
8 & -5 & -3
\end{array}\right|= \\
& =0 \cdot\left|\begin{array}{ll}
-2 & -5 \\
-5 & -3
\end{array}\right|-(-4) \cdot\left|\begin{array}{cc}
4 & -5 \\
8 & -3
\end{array}\right|+(-8) \cdot\left|\begin{array}{cc}
4 & -2 \\
8 & -5
\end{array}\right|= \\
& =0 \cdot(-19)--4 \cdot 28+-8 \cdot(-4)=0-(-112)+32=144
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|144|=24
$$

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:

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
0 & -4 & -8 \\
4 & -2 & -5
\end{array}\right|=i \cdot\left|\begin{array}{cc}
-4 & -8 \\
-2 & -5
\end{array}\right|-j\left|\begin{array}{cc}
0 & -8 \\
4 & -5
\end{array}\right|+k \cdot\left|\begin{array}{cc}
0 & -4 \\
4 & -2
\end{array}\right|= \\
& =i \cdot 4-j \cdot 32+k \cdot 16=\{4 ;-32 ; 16\}
\end{aligned}
$$

We obtain:

$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{4^{2}+(-32)^{2}+16^{2}}=\frac{1}{2} \cdot \sqrt{1296}=\frac{36}{2}=18
$$

Then:

$h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot 24}{18}=4$

Volume of the tetrahedron: 24

Height: 4

## Problem Kuznetsov Analytic Geometry $7-25$