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

\( A_{1}(1 ; 0 ; 2) \)

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

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

\( A_{4}(2 ; 1 ; 0) \)
## Solution

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

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{1-1 ; 2-0 ;-1-2\}=\{0 ; 2 ;-3\} \\
& \overrightarrow{A_{1} A_{3}}=\{2-1 ;-2-0 ; 1-2\}=\{1 ;-2 ;-1\} \\
& \overrightarrow{A_{1} A_{4}}=\{2-1 ; 1-0 ; 0-2\}=\{1 ; 1 ;-2\}
\end{aligned}
$$

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:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
0 & 2 & -3 \\
1 & -2 & -1 \\
1 & 1 & -2
\end{array}\right|= \\
& =0 \cdot\left|\begin{array}{cc}
-2 & -1 \\
1 & -2
\end{array}\right|-2 \cdot\left|\begin{array}{cc}
1 & -1 \\
1 & -2
\end{array}\right|+(-3) \cdot\left|\begin{array}{cc}
1 & -2 \\
1 & 1
\end{array}\right|= \\
& =0 \cdot 5-2 \cdot(-1)-3 \cdot 3=0+2-9=-7
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|-7|=\frac{7}{6}=1 \frac{1}{6}
$$

Since

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

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

We compute the vector product:

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

We obtain:

$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{(-8)^{2}+(-3)^{2}+(-2)^{2}}=\frac{\sqrt{77}}{2}
$$

Then:

$$
h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot \frac{7}{6}}{\frac{\sqrt{77}}{2}}=\frac{7}{\sqrt{77}}=\sqrt{\frac{7}{11}}
$$

Volume of the tetrahedron: $1 \frac{1}{6}$

Height: $\sqrt{\frac{7}{11}}$

## Problem Kuznetsov Analytic Geometry 7-22