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

\( A_{1}(2 ; 3 ; 1) \)
\( A_{2}(4 ; 1 ;-2) \)

\( A_{3}(6 ; 3 ; 7) \)

\( A_{4}(7 ; 5 ;-3) \)
## Solution

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

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{4-2 ; 1-3 ;-2-1\}=\{2 ;-2 ;-3\} \\
& \vec{A}_{1} A_{3}=\{6-2 ; 3-3 ; 7-1\}=\{4 ; 0 ; 6\} \\
& \overrightarrow{A_{1} A_{4}}=\{7-2 ; 5-3 ;-3-1\}=\{5 ; 2 ;-4\}
\end{aligned}
$$

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

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

We compute the mixed product:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
2 & -2 & -3 \\
4 & 0 & 6 \\
5 & 2 & -4
\end{array}\right|= \\
& =2 \cdot\left|\begin{array}{cc}
0 & 6 \\
2 & -4
\end{array}\right|-(-2) \cdot\left|\begin{array}{cc}
4 & 6 \\
5 & -4
\end{array}\right|+(-3) \cdot\left|\begin{array}{cc}
4 & 0 \\
5 & 2
\end{array}\right|= \\
& =2 \cdot(-12)+2 \cdot(-46)-3 \cdot 8=-24-92-24=-140
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|-140|=\frac{70}{3}=23 \frac{1}{3}
$$

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:

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

We compute the vector product:
$A_{1} A_{2} \times A_{1} A_{3}=\left|\begin{array}{ccc}i & j & k \\ 2 & -2 & -3 \\ 4 & 0 & 6\end{array}\right|=i \cdot\left|\begin{array}{cc}-2 & -3 \\ 0 & 6\end{array}\right|-j\left|\begin{array}{cc}2 & -3 \\ 4 & 6\end{array}\right|+k \cdot\left|\begin{array}{cc}2 & -2 \\ 4 & 0\end{array}\right|=$
$=i \cdot(-12)-j \cdot 24+k \cdot 8=\{-12 ;-24 ; 8\}$

We obtain:

$S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{(-12)^{2}+(-24)^{2}+8^{2}}=\frac{1}{2} \cdot \sqrt{784}=\frac{28}{2}=14$

Then

$h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot \frac{70}{3}}{14}=5$

Volume of the tetrahedron: $23 \frac{1}{3}$

Height: 5

## Problem Kuznetsov Analytic Geometry 7-14