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

\( A_{1}(1 ; 5 ;-7) \)

\[
\begin{aligned}
& A_{2}(-3 ; 6 ; 3) \\
& A_{3}(-2 ; 7 ; 3) \\
& A_{4}(-4 ; 8 ;-12)
\end{aligned}
\]
## Solution

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

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{-3-1 ; 6-5 ; 3-(-7)\}=\{-4 ; 1 ; 10\} \\
& A_{1} A_{3}=\{-2-1 ; 7-5 ; 3-(-7)\}=\{-3 ; 2 ; 10\} \\
& \overrightarrow{A_{1} A_{4}}=\{-4-1 ; 8-5 ;-12-(-7)\}=\{-5 ; 3 ;-5\}
\end{aligned}
$$

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:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
-4 & 1 & 10 \\
-3 & 2 & 10 \\
-5 & 3 & -5
\end{array}\right|= \\
& =-4 \cdot\left|\begin{array}{cc}
2 & 10 \\
3 & -5
\end{array}\right|-\left|\begin{array}{cc}
-3 & 10 \\
-5 & -5
\end{array}\right|+10 \cdot\left|\begin{array}{cc}
-3 & 2 \\
-5 & 3
\end{array}\right|= \\
& =-4 \cdot(-40)-65+10 \cdot 1=160-65+10=105
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|105|=\frac{35}{2}=17.5
$$

Since

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

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:

$$
\begin{aligned}
& \vec{A}_{1} A_{2} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
-4 & 1 & 10 \\
-3 & 2 & 10
\end{array}\right|=i \cdot\left|\begin{array}{cc}
1 & 10 \\
2 & 10
\end{array}\right|-j\left|\begin{array}{cc}
-4 & 10 \\
-3 & 10
\end{array}\right|+k \cdot\left|\begin{array}{cc}
-4 & 1 \\
-3 & 2
\end{array}\right|= \\
& =-10 \cdot i+10 \cdot j-5 \cdot k=\{-10 ; 10 ;-5\}
\end{aligned}
$$

We obtain:

$S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{(-10)^{2}+10^{2}+(-5)^{2}}=\frac{15}{2}$

Then:

$h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot \frac{35}{2}}{\frac{15}{2}}=\frac{3 \cdot 35}{15}=7$

Volume of the tetrahedron: 17.5

Height: 7

## Problem Kuznetsov Analytic Geometry 7-16