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

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

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

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

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

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

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

According to the geometric meaning of the scalar triple 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 scalar triple product:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
-1 & 3 & -3 \\
1 & 3 & -1 \\
-6 & 3 & 3
\end{array}\right|= \\
& =-1 \cdot\left|\begin{array}{cc}
3 & -1 \\
3 & 3
\end{array}\right|-3\left|\begin{array}{cc}
1 & -1 \\
-6 & 3
\end{array}\right|-3 \cdot\left|\begin{array}{cc}
1 & 3 \\
-6 & 3
\end{array}\right|= \\
& =-1 \cdot 12-3 \cdot(-3)-3 \cdot 21=-12+9-63=-66
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|-66|=\frac{66}{6}=11
$$

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:

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
-1 & 3 & -3 \\
1 & 3 & -1
\end{array}\right|=i \cdot\left|\begin{array}{cc}
3 & -3 \\
3 & -1
\end{array}\right|-j\left|\begin{array}{cc}
-1 & -3 \\
1 & -1
\end{array}\right|+k \cdot\left|\begin{array}{cc}
-1 & 3 \\
1 & 3
\end{array}\right|= \\
& =6 \cdot i-4 \cdot j-6 \cdot k=\{6 ;-4 ;-6\}
\end{aligned}
$$

We obtain:

$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{6^{2}+(-4)^{2}+(-6)^{2}}=\frac{\sqrt{88}}{2}=\sqrt{22}
$$

Then:

$h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot 11}{\sqrt{22}}=3 \sqrt{\frac{11}{2}}$

Volume of the tetrahedron: $\mathbf{1 1}$

Height: $\sqrt{\frac{11}{2}}$

## Problem Kuznetsov Analytic Geometry 7-12