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

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

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

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

\( A_{4}(-3 ; 0 ; 1) \)
## Solution

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

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

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

$\overrightarrow{A_{1} A_{4}}=\{-3-1 ; 0-2 ; 1-0\}=\{-4 ;-2 ; 1\}$

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:

$$
\begin{aligned}
& \left(\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{1} A_{3}}, \overrightarrow{A_{1} A_{4}}\right)=\left|\begin{array}{ccc}
0 & -3 & 2 \\
-1 & -1 & -1 \\
-4 & -2 & 1
\end{array}\right|= \\
& =0 \cdot\left|\begin{array}{cc}
-1 & -1 \\
-2 & 1
\end{array}\right|-(-3) \cdot\left|\begin{array}{cc}
-1 & -1 \\
-4 & 1
\end{array}\right|+2 \cdot\left|\begin{array}{cc}
-1 & -1 \\
-4 & -2
\end{array}\right|= \\
& =0 \cdot(-3)--3 \cdot(-5)+2 \cdot(-2)=0-15+(-4)=-19
\end{aligned}
$$

We obtain:

$$
V_{A_{1} A_{2} A_{3} A_{4}}=\frac{1}{6} \cdot|-19|=\frac{19}{6}
$$

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:

$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}} \times \overrightarrow{A_{1} A_{3}}=\left|\begin{array}{ccc}
i & j & k \\
0 & -3 & 2 \\
-1 & -1 & -1
\end{array}\right|=i \cdot\left|\begin{array}{cc}
-3 & 2 \\
-1 & -1
\end{array}\right|-j\left|\begin{array}{cc}
0 & 2 \\
-1 & -1
\end{array}\right|+k \cdot\left|\begin{array}{cc}
0 & -3 \\
-1 & -1
\end{array}\right|= \\
& =i \cdot 5-j \cdot 2+k \cdot(-3)=\{5 ;-2 ;-3\}
\end{aligned}
$$

We obtain:

$S_{A_{1} A_{2} A_{3}}=\frac{1}{2} \cdot \sqrt{5^{2}+(-2)^{2}+(-3)^{2}}=\frac{\sqrt{38}}{2}$

Then:

$h=\frac{3 V_{A_{1} A_{2} A_{3} A_{4}}}{S_{A_{1} A_{2} A_{3}}}=\frac{3 \cdot \frac{19}{6}}{\frac{\sqrt{38}}{2}}=\frac{19}{\sqrt{38}}=\sqrt{\frac{19}{2}}$

Volume of the tetrahedron: $\frac{19}{6}$

Height: $\sqrt{\frac{19}{2}}$

## Problem Kuznetsov Analytic Geometry 7-21