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

\( A_{1}(3 ; 10 ;-1) \)
\( A_{2}(-2 ; 3 ;-5) \)

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

\( A_{4}(1 ;-1 ; 2) \)
## Solution

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

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{-2-3 ; 3-10 ;-5-(-1)\}=\{-5 ;-7 ;-4\} \\
& \vec{A}_{1} A_{3}=\{-6-3 ; 0-10 ;-3-(-1)\}=\{-9 ;-10 ;-2\} \\
& \overrightarrow{A_{1} A_{4}}=\{1-3 ;-1-10 ; 2-(-1)\}=\{-2 ;-11 ; 3\}
\end{aligned}
$$

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:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
-5 & -7 & -4 \\
-9 & -10 & -2 \\
-2 & -11 & 3
\end{array}\right|= \\
& =-5 \cdot\left|\begin{array}{cc}
-10 & -2 \\
-11 & 3
\end{array}\right|-(-7) \cdot\left|\begin{array}{cc}
-9 & -2 \\
-2 & 3
\end{array}\right|+(-4) \cdot\left|\begin{array}{cc}
-9 & -10 \\
-2 & -11
\end{array}\right|= \\
& =-5 \cdot(-52)+7 \cdot(-31)-4 \cdot 79=260-217-316=-273
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|-273|=\frac{91}{2}=45.5
$$

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:

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
-5 & -7 & -4 \\
-9 & -10 & -2
\end{array}\right|=i \cdot\left|\begin{array}{cc}
-7 & -4 \\
-10 & -2
\end{array}\right|-j\left|\begin{array}{cc}
-5 & -4 \\
-9 & -2
\end{array}\right|+k \cdot\left|\begin{array}{cc}
-5 & -7 \\
-9 & -10
\end{array}\right|= \\
& =i \cdot(-26)-j \cdot(-26)+k \cdot(-13)=\{-26 ; 26 ;-13\}
\end{aligned}
$$

We obtain:

$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{(-26)^{2}+26^{2}+(-13)^{2}}=\frac{1}{2} \cdot \sqrt{1521}=\frac{39}{2}=19.5
$$

Then:

$$
h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot 45.5}{19.5}=7
$$

Volume of the tetrahedron: 45.5

Height: 7

## Problem Kuznetsov Analytic Geometry 7-24