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

\( A_{1}(5 ; 2 ; 0) \)

\( A_{2}(2 ; 5 ; 0) \)

\( A_{3}(1 ; 2 ; 4) \)

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

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

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

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

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

We compute the scalar triple product:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
-3 & 3 & 0 \\
-4 & 0 & 4 \\
-6 & -1 & 1
\end{array}\right|= \\
& =-3 \cdot\left|\begin{array}{cc}
0 & 4 \\
-1 & 1
\end{array}\right|-3 \cdot\left|\begin{array}{cc}
-4 & 4 \\
-6 & 1
\end{array}\right|+0 \cdot\left|\begin{array}{cc}
-4 & 0 \\
-6 & -1
\end{array}\right|= \\
& =-3 \cdot 4-3 \cdot 20+0 \cdot 4=-12-60+0=-72
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|-72|=12
$$

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:

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
-3 & 3 & 0 \\
-4 & 0 & 4
\end{array}\right|=i \cdot\left|\begin{array}{cc}
3 & 0 \\
0 & 4
\end{array}\right|-j\left|\begin{array}{cc}
-3 & 0 \\
-4 & 4
\end{array}\right|+k \cdot\left|\begin{array}{cc}
-3 & 3 \\
-4 & 0
\end{array}\right|= \\
& =i \cdot 12-j \cdot(-12)+k \cdot 12=\{12 ; 12 ; 12\}
\end{aligned}
$$

We obtain:

$S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{12^{2}+12^{2}+12^{2}}=\frac{1}{2} \cdot \sqrt{432}=\frac{12 \sqrt{3}}{2}=6 \sqrt{3}$

Then:

$$
h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot 12}{6 \sqrt{3}}=\frac{3 \cdot 2}{\sqrt{3}}=2 \sqrt{3}
$$

Volume of the tetrahedron: 12

Height: $2 \sqrt{3}$

## Problem Kuznetsov Analytic Geometry $7-7$