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

\( A_{1}(-1 ; -5 ; 2) \)

\( A_{2}(-6 ; 0 ; -3) \)

\( A_{3}(3 ; 6 ; -3) \)

\( A_{4}(-10 ; 6 ; 7) \)
## Solution

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

$\overrightarrow{A_{1} A_{2}}=\{-6-(-1) ; 0-(-5) ;-3-2\}=\{-5 ; 5 ;-5\}$

$\overrightarrow{A_{1} A_{3}}=\{3-(-1) ; 6-(-5) ;-3-2\}=\{4 ; 11 ;-5\}$

$\overrightarrow{A_{1} A_{4}}=\{-10-(-1) ; 6-(-5) ; 7-2\}=\{-9 ; 11 ; 5\}$

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

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

We compute the mixed product:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
-5 & 5 & -5 \\
4 & 11 & -5 \\
-9 & 11 & 5
\end{array}\right|= \\
& =-5 \cdot\left|\begin{array}{cc}
11 & -5 \\
11 & 5
\end{array}\right|-5 \cdot\left|\begin{array}{cc}
4 & -5 \\
-9 & 5
\end{array}\right|+(-5) \cdot\left|\begin{array}{cc}
4 & 11 \\
-9 & 11
\end{array}\right|= \\
& =-5 \cdot 110-5 \cdot(-25)-5 \cdot 143=-550+125-715=-1140
\end{aligned}
$$

We obtain:

$V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|-1140|=190$

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:

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

We compute the vector product:

$$
\begin{aligned}
& A_{1} A_{2} \times A_{1} A_{3}=\left|\begin{array}{ccc}
i & j & k \\
-5 & 5 & -5 \\
4 & 11 & -5
\end{array}\right|=i \cdot\left|\begin{array}{cc}
5 & -5 \\
11 & -5
\end{array}\right|-j\left|\begin{array}{cc}
-5 & -5 \\
4 & -5
\end{array}\right|+k \cdot\left|\begin{array}{cc}
-5 & 5 \\
4 & 11
\end{array}\right|= \\
& =i \cdot 30-j \cdot 45+k \cdot(-75)=\{30 ;-45 ;-75\}
\end{aligned}
$$

We obtain:

$S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{30^{2}+(-45)^{2}+(-75)^{2}}=\frac{1}{2} \cdot \sqrt{8550}=\frac{15 \sqrt{38}}{2}$

Then:

$$
h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot 190}{\frac{15 \sqrt{38}}{2}}=\frac{2 \cdot 38}{\sqrt{38}}=2 \sqrt{38}
$$

Volume of the tetrahedron: 190

Height: $2 \sqrt{38}$

## Problem Kuznetsov Analytic Geometry 7-5