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

\( A_{1}(1 ; 3 ; 6) \)

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

\( A_{3}(-1 ; 0 ; 1) \)

\( A_{4}(-4 ; 6 ;-3) \)
## Solution

From vertex $A_{1 \text { we draw vectors: }}$
$\overrightarrow{A_{1} A_{2}}=\{2-1 ; 2-3 ; 1-6\}=\{1 ;-1 ;-5\}$

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

$\overrightarrow{A_{1} A_{4}}=\{-4-1 ; 6-3 ;-3-6\}=\{-5 ; 3 ;-9\}$

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

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

Let's compute the scalar triple product:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
1 & -1 & -5 \\
-2 & -3 & -5 \\
-5 & 3 & -9
\end{array}\right|= \\
& =1 \cdot\left|\begin{array}{cc}
-3 & -5 \\
3 & -9
\end{array}\right|-(-1) \cdot\left|\begin{array}{cc}
-2 & -5 \\
-5 & -9
\end{array}\right|+(-5) \cdot\left|\begin{array}{cc}
-2 & -3 \\
-5 & 3
\end{array}\right|= \\
& =1 \cdot 42+1 \cdot(-7)-5 \cdot(-21)=42-7+105=140
\end{aligned}
$$

We get:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|140|=\frac{70}{3}=23 \frac{1}{3}
$$

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:

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

Let's compute the vector product:

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
1 & -1 & -5 \\
-2 & -3 & -5
\end{array}\right|=i \cdot\left|\begin{array}{cc}
-1 & -5 \\
-3 & -5
\end{array}\right|-j\left|\begin{array}{cc}
1 & -5 \\
-2 & -5
\end{array}\right|+k \cdot\left|\begin{array}{cc}
1 & -1 \\
-2 & -3
\end{array}\right|= \\
& =i \cdot(-10)-j \cdot(-15)+k \cdot(-5)=\{-10 ; 15 ;-5\}
\end{aligned}
$$

We get:

$$
S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{(-10)^{2}+15^{2}+(-5)^{2}}=\frac{1}{2} \cdot \sqrt{350}=\frac{5 \sqrt{14}}{2}
$$

Then:
$h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot \frac{70}{3}}{\frac{5 \sqrt{14}}{2}}=\frac{2 \cdot 14}{\sqrt{14}}=2 \sqrt{14}$

Volume of the tetrahedron: $23 \frac{1}{3}$

Height: $2 \sqrt{14}$

## Problem Kuznetsov Analytic Geometry 7-1