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

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

\( A_{2}(4 ;-1 ; 0) \)

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

\( A_{4}(3 ; 4 ; 5) \)
## Solution

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

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

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

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

We compute the scalar triple product:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
5 & -3 & 3 \\
3 & -1 & 1 \\
4 & 2 & 8
\end{array}\right|= \\
& =5 \cdot\left|\begin{array}{cc}
-1 & 1 \\
2 & 8
\end{array}\right|-(-3) \cdot\left|\begin{array}{cc}
3 & 1 \\
4 & 8
\end{array}\right|+3 \cdot\left|\begin{array}{cc}
3 & -1 \\
4 & 2
\end{array}\right|= \\
& =5 \cdot(-10)+3 \cdot 20+3 \cdot 10=-50+60+30=40
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|40|=\frac{20}{3}=6 \frac{2}{3}
$$

Since

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

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

We compute the cross product:

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
5 & -3 & 3 \\
3 & -1 & 1
\end{array}\right|=i \cdot\left|\begin{array}{cc}
-3 & 3 \\
-1 & 1
\end{array}\right|-j\left|\begin{array}{cc}
5 & 3 \\
3 & 1
\end{array}\right|+k \cdot\left|\begin{array}{cc}
5 & -3 \\
3 & -1
\end{array}\right|= \\
& =i \cdot 0-j \cdot(-4)+k \cdot 4=\{0 ; 4 ; 4\}
\end{aligned}
$$

We obtain:

$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{0^{2}+4^{2}+4^{2}}=\frac{1}{2} \cdot \sqrt{32}=\frac{4 \sqrt{2}}{2}=2 \sqrt{2}
$$

Then:

$$
h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot \frac{20}{3}}{2 \sqrt{2}}=\frac{10}{\sqrt{2}}=5 \sqrt{2}
$$

Volume of the tetrahedron: $6 \frac{2}{3}$

Height: $5 \sqrt{2}$

## Problem Kuznetsov Analytic Geometry $7-18$