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

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

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

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

\( A_{4}(6 ; -3 ; 8) \)
## Solution

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

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

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

We compute the scalar triple product:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
1 & 2 & 0 \\
0 & 2 & 2 \\
5 & -2 & 6
\end{array}\right|= \\
& =1 \cdot\left|\begin{array}{cc}
2 & 2 \\
-2 & 6
\end{array}\right|-2 \cdot\left|\begin{array}{cc}
0 & 2 \\
5 & 6
\end{array}\right|+0 \cdot\left|\begin{array}{cc}
0 & 2 \\
5 & -2
\end{array}\right|= \\
& =1 \cdot 16-2 \cdot(-10)+0=16+20+0=36
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|36|=6
$$

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:

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

We obtain:

$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{4^{2}+(-2)^{2}+2^{2}}=\frac{\sqrt{24}}{2}=\sqrt{6}
$$

Then:

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

Volume of the tetrahedron: 6

Height: $3 \sqrt{6}$

## Problem Kuznetsov Analytic Geometry 7-31