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

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

\( A_{2}(-2 ; 0 ; 3) \)

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

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

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

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

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

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

We compute the mixed product:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1}} \overrightarrow{A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
-3 & 1 & 2 \\
1 & 2 & -2 \\
1 & -1 & -5
\end{array}\right|= \\
& =-3 \cdot\left|\begin{array}{cc}
2 & -2 \\
-1 & -5
\end{array}\right|-1 \cdot\left|\begin{array}{cc}
1 & -2 \\
1 & -5
\end{array}\right|+2 \cdot\left|\begin{array}{cc}
1 & 2 \\
1 & -1
\end{array}\right|= \\
& =-3 \cdot(-12)-1 \cdot(-3)+2 \cdot(-3)=36+3-6=33
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|33|=\frac{11}{2}=5.5
$$

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:

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
-3 & 1 & 2 \\
1 & 2 & -2
\end{array}\right|=i \cdot\left|\begin{array}{cc}
1 & 2 \\
2 & -2
\end{array}\right|-j\left|\begin{array}{cc}
-3 & 2 \\
1 & -2
\end{array}\right|+k \cdot\left|\begin{array}{cc}
-3 & 1 \\
1 & 2
\end{array}\right|= \\
& =i \cdot(-6)-j \cdot 4+k \cdot(-7)=\{-6 ;-4 ;-7\}
\end{aligned}
$$

We obtain:

$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{(-6)^{2}+(-4)^{2}+(-7)^{2}}=\frac{\sqrt{101}}{2}
$$

Then:

$h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot \frac{11}{2}}{\frac{\sqrt{101}}{2}}=\frac{33}{\sqrt{101}}$

Volume of the tetrahedron: 5.5

Height: $\frac{33}{\sqrt{101}}$

## Problem Kuznetsov Analytic Geometry 7-20