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

\( A_{1}(-3 ; -5 ; 6) \)

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

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

\( A_{4}(-5 ; 2 ; -8) \)
## Solution

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

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{2-(-3) ; 1-(-5) ;-4-6\}=\{5 ; 6 ;-10\} \\
& \overrightarrow{A_{1} A_{3}}=\{0-(-3) ;-3-(-5) ;-1-6\}=\{3 ; 2 ;-7\} \\
& \overrightarrow{A_{1} A_{4}}=\{-5-(-3) ; 2-(-5) ;-8-6\}=\{-2 ; 7 ;-14\}
\end{aligned}
$$

According to the geometric meaning of the mixed product, we have:

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

We compute the mixed product:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1}} \overrightarrow{A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
5 & 6 & -10 \\
3 & 2 & -7 \\
-2 & 7 & -14
\end{array}\right|= \\
& =5 \cdot\left|\begin{array}{cc}
2 & -7 \\
7 & -14
\end{array}\right|-6 \cdot\left|\begin{array}{cc}
3 & -7 \\
-2 & -14
\end{array}\right|-10 \cdot\left|\begin{array}{cc}
3 & 2 \\
-2 & 7
\end{array}\right|= \\
& =5 \cdot(21)-6 \cdot(-56)-10 \cdot 25=105+336-250=191
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|191|=\frac{191}{6}
$$

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:

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
5 & 6 & -10 \\
3 & 2 & -7
\end{array}\right|=i \cdot\left|\begin{array}{cc}
6 & -10 \\
2 & -7
\end{array}\right|-j\left|\begin{array}{cc}
5 & -10 \\
3 & -7
\end{array}\right|+k \cdot\left|\begin{array}{cc}
5 & 6 \\
3 & 2
\end{array}\right|= \\
& =-22 \cdot i+5 \cdot j-8 \cdot k=\{-22 ; 5 ;-8\}
\end{aligned}
$$

$S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{(-22)^{2}+5^{2}+(-8)^{2}}=\frac{\sqrt{573}}{2}$

Then

$h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot \frac{191}{6}}{\frac{\sqrt{573}}{2}}=\frac{191}{\sqrt{3 \cdot 191}}=\sqrt{\frac{191}{3}}$

Volume of the tetrahedron: $\frac{191}{6}$

Height: $\sqrt{\frac{191}{3}}$

## Problem Kuznetsov Analytic Geometry $7-29$