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

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

\( A_{2}(-1 ; 7 ; 1) \)

\( A_{3}(4 ;-8 ;-4) \)

\( A_{4}(1 ;-4 ; 6) \)
## Solution

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

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{-1-(-2) ; 7-0 ; 1-(-4)\}=\{1 ; 7 ; 5\} \\
& A_{1} A_{3}=\{4-(-2) ;-8-0 ;-4-(-4)\}=\{6 ;-8 ; 0\} \\
& \overrightarrow{A_{1} A_{4}}=\{1-(-2) ;-4-0 ; 6-(-4)\}=\{3 ;-4 ; 10\}
\end{aligned}
$$

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

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

We compute the scalar triple product:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
1 & 7 & 5 \\
6 & -8 & 0 \\
3 & -4 & 10
\end{array}\right|= \\
& =1 \cdot\left|\begin{array}{cc}
-8 & 0 \\
-4 & 10
\end{array}\right|-7 \cdot\left|\begin{array}{cc}
6 & 0 \\
3 & 10
\end{array}\right|+5 \cdot\left|\begin{array}{cc}
6 & -8 \\
3 & -4
\end{array}\right|= \\
& =1 \cdot(-80)-7 \cdot 60+5 \cdot 0=-80-420+0=-500
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|-500|=\frac{250}{3}=83 \frac{1}{3}
$$

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:

$$
\begin{aligned}
& A_{1} A_{2} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
1 & 7 & 5 \\
6 & -8 & 0
\end{array}\right|=i \cdot\left|\begin{array}{cc}
7 & 5 \\
-8 & 0
\end{array}\right|-j\left|\begin{array}{cc}
1 & 5 \\
6 & 0
\end{array}\right|+k \cdot\left|\begin{array}{cc}
1 & 7 \\
6 & -8
\end{array}\right|= \\
& =i \cdot 40-j \cdot(-30)+k \cdot(-50)=\{40 ; 30 ;-50\}
\end{aligned}
$$

We obtain:

$S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{40^{2}+30^{2}+(-50)^{2}}=\frac{1}{2} \cdot \sqrt{5000}=\frac{50 \sqrt{2}}{2}=25 \sqrt{2}$

Then:

$h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot \frac{250}{3}}{25 \sqrt{2}}=\frac{10}{\sqrt{2}}=5 \sqrt{2}$

Volume of the tetrahedron: $83 \frac{1}{3}$

Height: $5 \sqrt{2}$

## Problem Kuznetsov Analytic Geometry 7-9