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

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

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

\( A_{3}(5 ; 0 ; -6) \)

\( A_{4}(-10 ; 9 ; -7) \)
## Solution

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

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

$\overrightarrow{A_{1} A_{4}}=\{-10-2 ; 9-(-1) ;-7-(-2)\}=\{-12 ; 10 ;-5\}$

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

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

We will compute the mixed product:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
-1 & 3 & 3 \\
3 & 1 & -4 \\
-12 & 10 & -5
\end{array}\right|= \\
& =-1 \cdot\left|\begin{array}{cc}
1 & -4 \\
10 & -5
\end{array}\right|-3 \cdot\left|\begin{array}{cc}
3 & -4 \\
-12 & -5
\end{array}\right|+3 \cdot\left|\begin{array}{cc}
3 & 1 \\
-12 & 10
\end{array}\right|= \\
& =-1 \cdot 35-3 \cdot(-63)+3 \cdot 42=-35+189+126=280
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|280|=\frac{140}{3}=46 \frac{2}{3}
$$

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:

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

We will compute the vector product:

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
-1 & 3 & 3 \\
3 & 1 & -4
\end{array}\right|=i \cdot\left|\begin{array}{cc}
3 & 3 \\
1 & -4
\end{array}\right|-j\left|\begin{array}{cc}
-1 & 3 \\
3 & -4
\end{array}\right|+k \cdot\left|\begin{array}{cc}
-1 & 3 \\
3 & 1
\end{array}\right|= \\
& =i \cdot(-15)-j \cdot(-5)+k \cdot(-10)=\{-15 ; 5 ;-10\}
\end{aligned}
$$

We obtain:

$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{(-15)^{2}+5^{2}+(-10)^{2}}=\frac{1}{2} \cdot \sqrt{350}=\frac{5 \sqrt{14}}{2}
$$

$h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot \frac{140}{3}}{\frac{5 \sqrt{14}}{2}}=\frac{2 \cdot 28}{\sqrt{14}}=4 \sqrt{14}$

Volume of the tetrahedron: $46 \frac{2}{3}$

Height: $4 \sqrt{14}$

## Problem Kuznetsov Analytic Geometry $7-8$