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

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

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

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

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

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

$$
\begin{aligned}
& A_{1} \overrightarrow{A_{2}}=\{0-(-2) ; 3-(-1) ; 2-(-1)\}=\{2 ; 4 ; 3\} \\
& A_{1} \overrightarrow{A_{3}}=\{3-(-2) ; 1-(-1) ;-4-(-1)\}=\{5 ; 2 ;-3\} \\
& \vec{A}_{1} A_{4}=\{-4-(-2) ; 7-(-1) ; 3-(-1)\}=\{-2 ; 8 ; 4\}
\end{aligned}
$$

According to the geometric meaning of the scalar triple product, we have:

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

We compute the scalar triple product:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
2 & 4 & 3 \\
5 & 2 & -3 \\
-2 & 8 & 4
\end{array}\right|= \\
& =2 \cdot\left|\begin{array}{cc}
2 & -3 \\
8 & 4
\end{array}\right|-4 \cdot\left|\begin{array}{cc}
5 & -3 \\
-2 & 4
\end{array}\right|+3 \cdot\left|\begin{array}{cc}
5 & 2 \\
-2 & 8
\end{array}\right|= \\
& =2 \cdot 32-4 \cdot 14+3 \cdot 44=64-56+132=140
\end{aligned}
$$

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:

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
2 & 4 & 3 \\
5 & 2 & -3
\end{array}\right|=i \cdot\left|\begin{array}{cc}
4 & 3 \\
2 & -3
\end{array}\right|-j\left|\begin{array}{cc}
2 & 3 \\
5 & -3
\end{array}\right|+k \cdot\left|\begin{array}{cc}
2 & 4 \\
5 & 2
\end{array}\right|= \\
& =i \cdot(-18)-j \cdot(-21)+k \cdot(-16)=\{-18 ; 21 ;-16\}
\end{aligned}
$$

We obtain:

$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{(-18)^{2}+21^{2}+(-16)^{2}}=\frac{\sqrt{1021}}{2}
$$

Then:

$$
h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot \frac{70}{3}}{\frac{\sqrt{1021}}{2}}=\frac{140}{\sqrt{1021}}
$$

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

140

Height: $\overline{\sqrt{1021}}$

## Problem Kuznetsov Analytic Geometry 7-28